We can make an equation. Straight line. Equation of a straight line. How to write an equation according to the condition of the problem

Chemistry is the science of substances, their properties and transformations. .
That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does "nothing happens" mean? If a thunderstorm suddenly caught us in the field, and we all got wet, as they say, “to the skin”, then is this not a transformation: after all, the clothes were dry, but became wet.

If, for example, you take an iron nail, process it with a file, and then assemble iron filings (Fe) , then this is also not a transformation: there was a nail - it became powder. But if after that to assemble the device and hold obtaining oxygen (O 2): heat up potassium permanganate(KMpo 4) and collect oxygen in a test tube, and then place these iron filings heated “to red” in it, then they will flare up with a bright flame and, after combustion, will turn into a brown powder. And this is also a transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the state of clothing (dry, wet) change, these are not transformations. The fact is that the nail itself, as it was a substance (iron), remained so, despite its different form, and our clothes soaked up the water from the rain, and then it evaporated into the atmosphere. The water itself has not changed. So what are transformations in terms of chemistry?

From the point of view of chemistry, transformations are such phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It does not matter what form it took after being filed, but after being collected from it iron filings placed in an atmosphere of oxygen - it turned into iron oxide(Fe 2 O 3 ) . So, has something really changed? Yes, it has. There was a nail substance, but under the influence of oxygen a new substance was formed - element oxide gland. molecular equation this transformation can be represented by the following chemical symbols:

4Fe + 3O 2 = 2Fe 2 O 3 (1)

For a person uninitiated in chemistry, questions immediately arise. What is the "molecular equation", what is Fe? Why are there numbers "4", "3", "2"? What are the small numbers "2" and "3" in the formula Fe 2 O 3? This means that the time has come to sort things out in order.

Signs chemical elements.

Despite the fact that they begin to study chemistry in the 8th grade, and some even earlier, many people know the great Russian chemist D. I. Mendeleev. And of course, his famous "Periodic Table of Chemical Elements". Otherwise, more simply, it is called the "Mendeleev's Table".

In this table, in the appropriate order, the elements are located. To date, about 120 of them are known. The names of many elements have been known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we used these words without hesitation, identifying them with objects: an iron bolt, aluminum wire, oxygen in the atmosphere, a golden ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their respective elements. The whole paradox is that the element cannot be touched, picked up. How so? They are in the periodic table, but you can’t take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, however, as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it is characterized by its own electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element #1. Its atom consists of 1 proton and 1 electron. Helium is element number 2. Its atom consists of 2 protons and 2 electrons. Lithium is element number 3. Its atom consists of 3 protons and 3 electrons. Darmstadtium - element number 110. Its atom consists of 110 protons and 110 electrons.

Each element is denoted by a certain symbol, Latin letters, and has a certain reading in translation from Latin. For example, hydrogen has the symbol "N", read as "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has a symbol "Hg" and is read as "hydrargyrum". And so on. All these designations can be found in any chemistry textbook for the 8th grade. For us now, the main thing is to understand that when compiling chemical equations, you must operate on the specified element symbols.

Simple and complex substances.

Denoting various substances with single symbols of chemical elements (Hg mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if iron and sulfur substances interact, then the equation will take the following form:

Fe + S = FeS (2)

Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). And you should pay attention
special attention to the fact that all metals are denoted by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals - either by simple symbols: C, S, P or may have different indices that indicate their molecular structure: H 2 , Cl 2 , O 2 , J 2 , P 4 , S 8 . In the future, this will be very great importance when writing equations. It is not at all difficult to guess that complex substances are substances formed from atoms. different kind, For example,

1). Oxides:
aluminium oxide Al 2 O 3,

sodium oxide Na 2 O
copper oxide CuO,
zinc oxide ZnO
titanium oxide Ti2O3,
carbon monoxide or carbon monoxide (+2) CO
sulfur oxide (+6) SO 3

2). Reasons:
iron hydroxide(+3) Fe (OH) 3,
copper hydroxide Cu(OH)2,
potassium hydroxide or potassium alkali KOH,
sodium hydroxide NaOH.

3). Acids:
hydrochloric acid HCl
sulfurous acid H2SO3,
Nitric acid HNO3

4). Salts:
sodium thiosulfate Na 2 S 2 O 3,
sodium sulfate or Glauber's salt Na 2 SO 4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl 2

5). organic matter:
sodium acetate CH 3 COOHa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6

Finally, after we have clarified the structure of various substances, we can begin to write chemical equations.

Chemical equation.

The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations are almost the very essence of this science. For example, you can give such a simple equation in which the left and right sides will be equal to "2":

40: (9 + 11) = (50 x 2): (80 - 30);

And in chemical equations, the same principle: the left and right sides of the equation must correspond to the same number of atoms, the elements participating in them. Or, if provided ionic equation, then in it number of particles must also meet this requirement. A chemical equation is a conditional record of a chemical reaction using chemical formulas and mathematical signs. A chemical equation inherently reflects a particular chemical reaction, that is, the process of interaction of substances, during which new substances arise. For example, it is necessary write a molecular equation reactions that take part barium chloride BaCl 2 and sulfuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid Hcl:

ВаСl 2 + H 2 SO 4 = BaSO 4 + 2НCl (3)

First of all, it is necessary to understand that the large number “2” in front of the HCl substance is called the coefficient, and the small numbers “2”, “4” under the formulas ВаСl 2, H 2 SO 4, BaSO 4 are called indices. Both coefficients and indices in chemical equations play the role of multipliers, not terms. In order to correctly write a chemical equation, it is necessary arrange the coefficients in the reaction equation. Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the BaSO 4 substance there is 1 barium atom (Ba) 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). Whence it follows that on the right side of the equation the number of hydrogen and chlorine atoms is half that on the left side. Therefore, before the HCl formula on the right side of the equation, it is necessary to put the coefficient "2". If we now add the number of atoms of the elements involved in this reaction, both on the left and on the right, we get the following balance:

In both parts of the equation, the number of atoms of the elements participating in the reaction are equal, therefore it is correct.

Chemical equation and chemical reactions

As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are such phenomena in the process of which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:

1). Connection reactions
2). decomposition reactions.

The overwhelming majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with a single substance if it is not subjected to external influences (dissolution, heating, light). Nothing characterizes a chemical phenomenon, or reaction, as much as the changes that occur when two or more substances interact. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color changes, sedimentation, release of gaseous products, noise.

For clarity, we present several equations that reflect the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl 2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)

Cl 2 + 2Nа = 2NaCl (4)

CuCl 2 + Zn \u003d ZnCl 2 + Cu (5)

AgNO 3 + KCl \u003d AgCl + 2KNO 3 (6)

3HCl + Al(OH) 3 \u003d AlCl 3 + 3H 2 O (7)

Among the reactions of the compound, the following should be especially noted : substitution (5), exchange (6), and how special case exchange reactions - reaction neutralization (7).

Substitution reactions include those in which atoms of a simple substance replace the atoms of one of the elements in a complex substance. In example (5), zinc atoms replace copper atoms from the CuCl 2 solution, while zinc passes into the soluble ZnCl 2 salt, and copper is released from the solution in the metallic state.

Exchange reactions are reactions in which two compounds exchange their constituent parts. In the case of reaction (6) soluble salts AgNO 3 and KCl, when both solutions are drained, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are attached to NO 3 anions, and silver cations Ag + - to Cl - anions.

A special, particular case of exchange reactions is the neutralization reaction. Neutralization reactions are reactions in which acids react with bases to form salt and water. In example (7), hydrochloric acid HCl reacts with base Al(OH) 3 to form AlCl 3 salt and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl anions - from the acid. As a result, it happens hydrochloric acid neutralization.

Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex one. As reactions, one can cite those in the process of which 1) decompose. potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate is formed (K 2 MnO 4), manganese oxide(MnO 2) and oxygen (O 2); 3). calcium carbonate or marble; in the process are formed carbonicgas(CO 2) and calcium oxide(Cao)

2KNO 3 \u003d 2KNO 2 + O 2 (8)
2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 \u003d CaO + CO 2 (10)

In reaction (8), one complex and one simple substance is formed from a complex substance. In reaction (9) there are two complex and one simple. In reaction (10) there are two complex substances, but simpler in composition

All classes of complex substances undergo decomposition:

1). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)

2). Hydroxides: iron hydroxide 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O (12)

3). Acids: sulfuric acid H 2 SO 4 \u003d SO 3 + H 2 O (13)

4). Salts: calcium carbonate CaCO 3 \u003d CaO + CO 2 (14)

5). organic matter: alcoholic fermentation of glucose

C 6 H 12 O 6 \u003d 2C 2 H 5 OH + 2CO 2 (15)

According to another classification, all chemical reactions can be divided into two types: reactions that take place with the release of heat, they are called exothermic, and reactions that go with the absorption of heat - endothermic. The criterion for such processes is thermal effect of the reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interactions with oxygen methane combustion:

CH 4 + 2O 2 \u003d CO 2 + 2H 2 O + Q (16)

and to endothermic reactions - decomposition reactions, already given above (11) - (15). The Q sign at the end of the equation indicates whether heat is released during the reaction (+Q) or absorbed (-Q):

CaCO 3 \u003d CaO + CO 2 - Q (17)

You can also consider all chemical reactions according to the type of change in the degree of oxidation of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:

Ca +2 C +4 O 3 -2 \u003d Ca +2 O -2 + C +4 O 2 -2 (18)

And in reaction (16), the elements change their oxidation states:

2Mg 0 + O 2 0 \u003d 2Mg +2 O -2

These types of reactions are redox . They will be considered separately. To formulate equations for reactions of this type, it is necessary to use half-reaction method and apply electronic balance equation.

After bringing various types of chemical reactions, you can proceed to the principle of compiling chemical equations, in other words, the selection of coefficients in their left and right parts.

Mechanisms for compiling chemical equations.

Whatever the type of one or the other chemical reaction, its record (chemical equation) must comply with the condition of equality of the number of atoms before the reaction and after the reaction.

There are equations (17) that do not require adjustment, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right parts of the equation. What principles should be followed in such cases? Is there any system in the selection of coefficients? There is, and not one. These systems include:

1). Selection of coefficients according to given formulas.

2). Compilation according to the valencies of the reactants.

3). Compilation according to the oxidation states of the reactants.

In the first case, it is assumed that we know the formulas of the reactants both before and after the reaction. For example, given the following equation:

N 2 + O 2 →N 2 O 3 (19)

It is generally accepted that until the equality between the atoms of the elements before and after the reaction is established, the equal sign (=) is not put in the equation, but is replaced by an arrow (→). Now let's get down to the actual balancing. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). It is not necessary to equalize it by the number of nitrogen atoms, but by oxygen it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were three atoms. Let's make the following diagram:

before reaction after reaction
O 2 O 3

Let's define the smallest multiple between the given numbers of atoms, it will be "6".

O 2 O 3
\ 6 /

Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved:

N 2 + 3O 2 →N 2 O 3

We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:

N 2 + 3O 2 → 2N 2 O 3

The number of oxygen atoms in both the left and right parts of the equation became equal, respectively, 6 atoms:

But the number of nitrogen atoms in both sides of the equation will not match:

On the left side there are two atoms, on the right side there are four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, putting the coefficient "2":

Thus, the equality for nitrogen is observed and, in general, the equation will take the form:

2N 2 + 3O 2 → 2N 2 O 3

Now in the equation, instead of an arrow, you can put an equal sign:

2N 2 + 3O 2 \u003d 2N 2 O 3 (20)

Let's take another example. The following reaction equation is given:

P + Cl 2 → PCl 5

On the left side of the equation there is 1 phosphorus atom (P) and two chlorine atoms (Cl 2), and on the right side there is one phosphorus atom (P) and five oxygen atoms (Cl 5). It is not necessary to equalize it by the number of phosphorus atoms, but for chlorine it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were five atoms. Let's make the following diagram:

before reaction after reaction
Cl 2 Cl 5

Let's define the smallest multiple between the given numbers of atoms, it will be "10".

Cl 2 Cl 5
\ 10 /

Divide this number on the left side of the equation for chlorine by "2". We get the number "5", put it in the equation to be solved:

Р + 5Cl 2 → РCl 5

We also divide the number "10" for the right side of the equation by "5". We get the number "2", just put it in the equation to be solved:

Р + 5Cl 2 → 2РCl 5

The number of chlorine atoms in both the left and right parts of the equation became equal, respectively, 10 atoms:

But the number of phosphorus atoms in both sides of the equation will not match:

Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation, putting the coefficient "2":

Thus, the equality for phosphorus is observed and, in general, the equation will take the form:

2Р + 5Cl 2 = 2РCl 5 (21)

When writing equations by valency must be given definition of valency and set values ​​for the most famous elements. Valency is one of the previously used concepts, currently in a number of school programs not used. But with its help it is easier to explain the principles of compiling equations of chemical reactions. By valency is meant number chemical bonds, which one or another atom can form with another, or other atoms . Valence has no sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:

Where do these values ​​come from? How to apply them in the preparation of chemical equations? The numerical values ​​of the valencies of the elements coincide with their group number Periodic system chemical elements D. I. Mendeleev (Table 1).

For other elements valency values may have other values, but never greater than the number of the group in which they are located. Moreover, for even numbers of groups (IV and VI), the valences of elements take only even values, and for odd ones, they can have both even and odd values ​​(Table.2).

Of course, there are exceptions to the valency values ​​for some elements, but in each specific case, these points are usually specified. Now consider general principle compiling chemical equations for given valences for certain elements. More often this method acceptable in the case of compiling equations of chemical reactions of the combination of simple substances, for example, when interacting with oxygen ( oxidation reactions). Suppose you want to display the oxidation reaction aluminum. But recall that metals are denoted by single atoms (Al), and non-metals that are in a gaseous state - with indices "2" - (O 2). First, we write the general scheme of the reaction:

Al + O 2 → AlO

At this stage, it is not yet known what the correct spelling should be for alumina. And it is precisely at this stage that knowledge of the valencies of the elements will come to our aid. For aluminum and oxygen, we put them above the proposed formula for this oxide:

III II
Al O

After that, "cross"-on-"cross" these symbols of the elements will put the corresponding indices below:

III II
Al 2 O 3

Composition of a chemical compound Al 2 O 3 determined. The further scheme of the reaction equation will take the form:

Al + O 2 →Al 2 O 3

It remains only to equalize the left and right parts of it. We proceed in the same way as in the case of formulating equation (19). We equalize the number of oxygen atoms, resorting to finding the smallest multiple:

before reaction after reaction

O 2 O 3
\ 6 /

Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved. We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:

Al + 3O 2 → 2Al 2 O 3

In order to achieve equality for aluminum, it is necessary to adjust its amount on the left side of the equation by setting the coefficient "4":

4Al + 3O 2 → 2Al 2 O 3

Thus, the equality for aluminum and oxygen is observed and, in general, the equation will take the final form:

4Al + 3O 2 \u003d 2Al 2 O 3 (22)

Using the valency method, it is possible to predict which substance is formed in the course of a chemical reaction, what its formula will look like. Suppose nitrogen and hydrogen with the corresponding valences III and I entered into the reaction of the compound. Let's write the general reaction scheme:

N 2 + H 2 → NH

For nitrogen and hydrogen, we put down the valencies over the proposed formula of this compound:

As before, "cross"-on-"cross" for these element symbols, we put the corresponding indices below:

III I
N H 3

The further scheme of the reaction equation will take the form:

N 2 + H 2 → NH 3

Equalizing in the already known way, through the smallest multiple for hydrogen, equal to "6", we obtain the desired coefficients, and the equation as a whole:

N 2 + 3H 2 \u003d 2NH 3 (23)

When compiling equations for oxidation states reacting substances, it must be recalled that the degree of oxidation of an element is the number of electrons received or given away in the process of a chemical reaction. The oxidation state in compounds basically, numerically coincides with the values ​​of the element's valences. But they differ in sign. For example, for hydrogen, the valence is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II, and the oxidation state is (-2). For nitrogen, the valencies are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. . The oxidation states of the elements most commonly used in equations are shown in Table 3.

In the case of compound reactions, the principle of compiling equations in terms of oxidation states is the same as in compiling in terms of valencies. For example, let's give the reaction equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write the proposed equation:

Cl 2 + O 2 → ClO

We put the oxidation states of the corresponding atoms over the proposed ClO compound:

As in the previous cases, we establish that the desired compound formula will take the form:

7 -2
Cl 2 O 7

The reaction equation will take the following form:

Cl 2 + O 2 → Cl 2 O 7

Equalizing for oxygen, finding the smallest multiple between two and seven, equal to "14", we finally establish the equality:

2Cl 2 + 7O 2 \u003d 2Cl 2 O 7 (24)

A slightly different method must be used with oxidation states when compiling exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?

How do you know what happens in a reaction?

Indeed, how do you know: what reaction products can arise in the course of a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?

Ba (NO 3) 2 + K 2 SO 4 →?

Maybe VAC 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction, compounds are formed: BaSO 4 and KNO 3. And how is this known? And how to write formulas of substances? Let's start with what is most often overlooked: the very concept of "exchange reaction". This means that in these reactions, the substances change with each other in constituent parts. Since the exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will change are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). In general, the exchange reaction can be given in the following notation:

Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)

Where Kt1 and Kt2 are the metal cations (1) and (2), and An1 and An2 are the anions (1) and (2) corresponding to them. In this case, it must be taken into account that in compounds before and after the reaction, cations are always established in the first place, and anions in the second. Therefore, if it reacts potassium chloride And silver nitrate, both in solution

KCl + AgNO 3 →

then in the process of it substances KNO 3 and AgCl are formed and the corresponding equation will take the form:

KCl + AgNO 3 \u003d KNO 3 + AgCl (26)

In neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):

HCl + KOH \u003d KCl + H 2 O (27)

The oxidation states of metal cations and the charges of anions of acid residues are indicated in the table of the solubility of substances (acids, salts and bases in water). Metal cations are shown horizontally, and anions of acid residues are shown vertically.

Based on this, when compiling the equation for the exchange reaction, it is first necessary to establish the oxidation states of the particles receiving in this chemical process in its left part. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let's draw up the initial scheme for this reaction:

CaCl + NaCO 3 →

Ca 2+ Cl - + Na + CO 3 2- →

Having performed the already known “cross”-to-“cross” action, we determine the real formulas of the starting substances:

CaCl 2 + Na 2 CO 3 →

Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction:

CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl

We put down the corresponding charges over their cations and anions:

Ca 2+ CO 3 2- + Na + Cl -

Substance formulas are written correctly, in accordance with the charges of cations and anions. Let's make a complete equation by equating the left and right parts of it in terms of sodium and chlorine:

CaCl 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaCl (28)

As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:

VaON + NPO 4 →

We put the corresponding charges over cations and anions:

Ba 2+ OH - + H + RO 4 3- →

Let's define the real formulas of the starting substances:

Va (OH) 2 + H 3 RO 4 →

Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction, taking into account that in the exchange reaction, one of the substances must necessarily be water:

Ba (OH) 2 + H 3 RO 4 → Ba 2+ RO 4 3- + H 2 O

Let's determine the correct record of the formula of the salt formed during the reaction:

Ba (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

Equate the left side of the equation for barium:

3VA (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

Since on the right side of the equation the residue of phosphoric acid is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:

3VA (OH) 2 + 2H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O

It remains to match the number of hydrogen and oxygen atoms on the right side of the water. Since the total number of hydrogen atoms on the left is 12, on the right it must also correspond to twelve, therefore, before the water formula, it is necessary put a coefficient"6" (since there are already 2 hydrogen atoms in the water molecule). For oxygen, equality is also observed: on the left 14 and on the right 14. So, the equation has the correct form of writing:

3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + 6Н 2 O (29)

Possibility of chemical reactions

The world is made up of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, assert that a chemical reaction will correspond to it? There is a misconception that if the right arrange odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and drop into it zinc, then we can observe the process of hydrogen evolution:

Zn + H 2 SO 4 \u003d ZnSO 4 + H 2 (30)

But if copper is lowered into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.

Cu + H 2 SO 4 ≠

If concentrated sulfuric acid is taken, it will react with copper:

Cu + 2H 2 SO 4 \u003d CuSO 4 + SO 2 + 2H 2 O (31)

In reaction (23) between nitrogen and hydrogen gases, thermodynamic balance, those. how many molecules ammonia NH 3 is formed per unit time, the same number of them will decompose back into nitrogen and hydrogen. Shift in chemical equilibrium can be achieved by increasing the pressure and decreasing the temperature

N 2 + 3H 2 \u003d 2NH 3

If you take potassium hydroxide solution and pour on it sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:

KOH + Na 2 SO 4 ≠

Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be attributed to a substitution reaction:

NaCl + Br 2 ≠

What are the reasons for such discrepancies? The fact is that it is not enough just to correctly define compound formulas, it is necessary to know the specifics of the interaction of metals with acids, to skillfully use the table of solubility of substances, to know the rules of substitution in the series of activity of metals and halogens. This article outlines only the most basic principles of how arrange the coefficients in the reaction equations, How write molecular equations, How determine the composition of a chemical compound.

Chemistry, as a science, is extremely diverse and multifaceted. This article reflects only a small part of the processes taking place in the real world. Types, thermochemical equations, electrolysis, processes organic synthesis and many many others. But more on that in future articles.

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SECTION VI.

EQUALITY TRANSFORMATIONS.

___________

SOLUTION AND COMPILATION OF EQUATIONS OF THE 1ST DEGREE

§ 5. Compilation of an equation with one unknown.

Any arithmetic problem consists in the fact that according to several known quantities and according to the given ratios between these known quantities and others, unknown ones, unknown ones are found. Algebra provides a special way to solve arithmetic problems. This method is based on the fact that the verbally expressed conditions of arithmetic problems can be translated into algebraic language, i.e. are expressible by means of algebraic formulas.

The translation of verbally expressed conditions of the problem into algebraic language is generally called formulating.

To compose an equation with one unknown according to the conditions of the problem means to translate these conditions into algebraic language in such a way that the entire set of these conditions is expressed by one equation containing one unknown. For this, it is necessary that the number of separate independent conditions of the problem be equal to the number of unknowns implied in it.

Due to the extreme diversity of problems, the methods for compiling equations corresponding to these problems are extremely diverse. General rules for compiling equations no. But there is one general indication that guides our reasoning when translating the conditions of the problem into algebraic language and allows us from the very beginning of the reasoning to follow the right path to achieve the final goal. This general indication, or the general principle of compiling an equation, we will express it as follows:

To compose an equation with one unknown according to the conditions of the problem, you need:

1) choose between the unknowns, which are either directly indicated in the problem, or implied, some one taken as the first, and designate this unknown with some letter, for example, X ;

2) using this designation and the designations given in the problem, express all the quantities that are directly mentioned in the problem or that are implied, observing that when compiling such expressions, all the numbers given in the problem and all related to dains or to unknown values ​​of the condition;

3) after such an application of all the conditions, find between the composed or simply written expressions two such that, by virtue of one of the given conditions, must be equal to each other, and connect these expressions with an equal sign.

Let's apply this principle to the solution of two problems:

Task 1 i. The number of coins in one wallet is half that in the other. If you lay out six coins from the first, and add eight coins to the second, then the number of coins in the first will be seven times less than in the second. Find out how many coins are in each wallet?

In this problem, several known and several unknown quantities are indicated. Let's take the first unknown number of coins of the first purse as the first unknown number and denote it by X. Then we will deal with the designation of all quantities, which include the conditions of the problem.

The number of coins in the first wallet is X . The ratio of the numbers of coins in the second and first wallets 2 . So the number of coins of the second wallet 2X.

Take out from the first 6 coins Therefore, in the first wallet there are coins X -6 .

In the second add 8 coins. Therefore, in the second wallet you will get coins 2X +8 . The new ratio between the numbers of coins of the second and first purse is . It is also equal 7 . On this basis, we compose an equation , solving which, we get x= 10 , after which it is not difficult to determine the other unknowns that we mentioned here.

If we took the second purse as the first unknown number of coins and denoted it to distinguish it from the previous designation through at , then, as is easy to see, another equation would be obtained, namely ( at + 8 ):( at / 2 -6 )=7 , which also solves the problem and gives the answer at=20 .

One could take for the first unknown the number of coins that appeared in the first purse after the calculation from it 6 coins; then, denoting this unknown by z and going in the same way as we went in the first equation, we would get the equation , where z = 4 .

But one could also change the very way in which the equation is related, for example, by taking into account the changed ratio between the numbers of coins, and basing the formulation of the equation on what is known about the original ratio. In this case, the equation would be written as follows:

The number of coins of the first wallet after the calculation is z . Posted 6 coins. So the initial number of coins of the first wallet z + 6. Changed ratio between numbers of coins 7 . Therefore, the changed number of coins of the second wallet 7z. was added 8 coins. Therefore, the initial number of coins of the second purse 7z. - 8 . The initial ratio between the numbers of coins is It is equal to 2 . On this basis, we have an equation that is compatible with the previous one, although it differs from it in form.

If, going this second way, we took for the first unknown number of coins of the second purse after adding to it 8 coins, then, denoting this unknown for difference through And , we would get the equation ( And -8 ):( And / 7 + 6 )=2 , where And =28 .

These clarifications show that, guided by the same general rule for writing equations, we still get in each problem a variety of ways to achieve this goal. the best way the one that expresses the conditions of the problem more simply and leads faster both to the compilation and to the solution of the equation is considered. In this case, the first and third methods are equally convenient for solving the equation, but the first is still simpler and therefore better than the others.

In applying this rule of formulating equations, it must be remembered that in any correctly expressed equation, every given number and every expressed condition must be taken into account.

Task 2. From the city A a traveler comes out, passing in the day along 20 verst. Two days later, she leaves the city to meet him. IN another traveler who passes daily by 30 verst. Distance between A And IN equals 190 verst. The question is, when and where will the two travelers meet?

1st way. Let us take as the first unknown time of movement of the first traveler from the exit from A before the meeting, and for the last condition is that the distance between A And IN equals 190 verst. Then we will argue like this:

Let's assume that the first one went before the meeting X days. Every day he walked 20 verst. So he went through 20X verst.

The second one came out later 2 day. So he went to meet X -2 day. Every day he walked 30 verst. Therefore, he went through 30 (X -2 ) verst. Together both travelers passed [ 20X + 30 (X -2 )] verst. All distance between A And IN equals 190 verst. Based on this, we find the equation

20X + 30 (X -2 ) =190 ,

where x= 5 . From this we see that the first traveler went 5 days and passed 100 verst, the second was 3 days and passed 90 verst.

2nd way. Let us take as the first unknown distance traveled by the first traveler from the exit to the meeting, and for the last condition that the second traveler left later than the first on 2 day. Then the discussion goes like this:

We believe that the first passed before the meeting at verst. Every day he walked 20 verst. So he walked all at / 20 days.

The second passed all ( 190 -at ) verst. Every day he walked 30 verst. So he walked only days.

The difference between the times of motion of both is and is equal to 2 . Therefore, we find the equation , where at =100 .

3rd way. The first unknown is the time of movement of the second traveler from the exit from IN see you, the last condition is that the first traveler passes daily 20 verst.

Let's assume that the second one goes to the meeting z days. So the first one will pass z +2 ) of the day. Walking daily through 30 verst, the second will pass only 30z verst. Since both need to pass 190 miles, then the first will have to do ( 190 -30z ) verst. To do this, he must do a mile a day. Since this expression is 20 , then the equation is obtained, whence z = 3.

4th way. The first unknown is the distance traveled by the second traveler before meeting, the last condition is that the second travels daily 10 versts more than the first.

We believe that the second passed before the meeting And verst. So the first one still had to go ( 190 -And ) verst. Since before the release of the second it has already passed 40 miles, then after the exit of the second he still had to go ( 150 -And ) verst. The difference in distances traveled simultaneously by both is ( 2And-150 ) verst. The time of their common movement is And / 30 days. Therefore, the second day passes more than the first by ( 2And-150 ) : And / 30 verst. Since this expression is 10 , then you get the equation ( 2And-150 ) : And / 30 =10 , which gives And = 90 .

The previous explanations show that the variety of ways to compose equations in the same problem depends both on the order of successively denoted quantities, and on the order of successively taken into account conditions.

231. Two persons have together 38 rubles, and the first has 6 rubles more money than the second. How much money does each have?

231. Two persons have together 114 rubles, and the first has 18 rubles more money than the second. How much money does each have?

232. In one house, there are 15 fewer windows than in the other; in total, there are 51 windows in both houses. How many windows are in each?

232. In one house there are 6 less windows than in another; in total, both houses have 62 windows. How many windows are in each?

233. There are 81 rubles in two wallets. In the first, there is half as much money as in the second. How much money is in each?

233. There are 72 rubles in two wallets. In the first, there is five times less money than in the second. How much money is in each?

234. Father older than son three times, and the sum of the years of both of them is 48 years. Determine the age of both.

234. The father is twice as old as the son, and the sum of both years is 13 years. Determine the age of both.

235. The son is four times younger than the father, and the difference between their years is 27 years. How much to fly each?

235. The son is five times younger than the father, and the difference between their years is 32 years. How old is each?

236. There are 47 apples in three baskets, and the first and second baskets are equally divided, and the third has 2 more apples than each of the others. How many apples are in each basket?

236. There are 110 apples in three baskets, and the first and third are equally divided, and the second has 4 less apples than each of the others. How many apples are in each basket?

237. The three pieces of silver weigh together 48 pounds. The first is 12 pounds heavier than the second, and the third is 9 pounds heavier than the first. How much does each piece weigh?

237. Three pieces of silver weigh together 33 pounds. The first is 5 pounds lighter than the second, and the third is 2 pounds lighter than the first. How much does each piece weigh?

238. The son is 20 years younger than the father and 5 years older than the daughter. The sum of all three years is 60 years. How old is each

238. Mother is 21 years older than her son and 7 years younger than his father. The sum of all three years is 64 years. How old is each?

239. There are a total of 66 books on three shelves, with three times as many on the bottom and twice as many on the middle as on the top. How many books are on each shelf?

239. There are only 60 books on three shelves, and on the bottom there are six times more, and on the top five times more than on the middle one. How many books are on each shelf?

240. The forest, the garden and the meadow together cost 10,800 rubles. The meadow is 2 times more expensive than the garden, and the forest is 3 times more expensive than the meadow. What does each of them cost separately?

240. A forest, a garden and a meadow cost together 17,600 rubles. A forest is 3 times more expensive than a garden, and a meadow is 4 times more expensive than a forest. What does each of them cost separately?

241. Divide the number 21 into two parts so that the multiple of the ratio of the first part to the second is equal to the fraction 3/4.

241. Divide the number 48 into two parts so that the multiple ratio of the second part to the first is equal to the fraction 5/3.

242. Divide the number 88 into two parts so that the quotients of dividing the first part by 5 and the second by 6 are equal.

242. Divide the number 55 into two parts so that the quotients of dividing the first part by 7, a. the second by 4 were equal.

243. The sum of two numbers is 85 and their difference is 15. Find both numbers.

243. The sum of two numbers is 72, and their difference is 8. Find both numbers.

244. The difference of two numbers is 8, and their multiple ratio is equal to the fraction 3 / 2. Find these numbers.

244. The difference of two numbers is 12, and their multiple ratio is equal to the fraction 5/3. Find these numbers.

245. Divide the number 46 into two parts so that the difference of the quotients from dividing the first part by 3 and the second by 7 is equal to 2.

245. Divide the number 59 into two parts so that the difference of the quotients from dividing the first part by 3 and the second by 5 is equal to 1.

246. Divide the number 75 into two parts so that the larger part is three times the difference between the two parts.

246. Divide the number 56 into two parts so that the smaller part is three times the difference between the two parts.

247. The sum of two numbers is 64. When dividing a larger number by a smaller one, the quotient is 3 and the remainder is 4. Find these numbers.

247. The sum of two numbers is 45. When a larger number is divided by a smaller one, the quotient is 5 and the remainder is 3. Find these numbers.

248. The difference of two numbers is 35. When dividing a larger number by a smaller one, the quotient is 4 and the remainder is 2. Find these numbers.

248. Difference of two numbers 23. When dividing a larger number by a smaller one, the quotient is 2 and the remainder is 11. Find these numbers.

249. One of the two unknown numbers is greater than the other by 5. If you divide the smaller number by 4 and the larger number by 3, then the first quotient will be 4 less than the second. Find both numbers.

249. One of two unknown numbers is greater than the other by 15. If we divide more by 9, and less by 2, then the first quotient will be 3 less than the second. Find both numbers.

250. One of the two unknown numbers is less than the other by 6. If you divide the larger number in half, then the resulting quotient will be three units less than the other number. Find both numbers.

250. One of the two unknown numbers is less than the other by 18. If you divide the larger number by three, then the resulting quotient will be two units greater than the other number. Find both numbers.

251. One tank has twice as much water as the other; if you pour 16 buckets from the first into the second, then there will be equal amounts of water in both. How much water is in each?

251. There is three times more water in one reservoir than in another; if you pour 22 buckets from the first into the second, then both will contain equal amounts of water. How much water is in each?

252. In the market, two traders have only 220 eggs; if the second of them gave the first 14 eggs, then the number of eggs for each of them would be the same. How many eggs does each have?

252. In the market, two traders have only 186 eggs; if the second of them gave the first 10 eggs, then the number of eggs for each of them would be the same. How many eggs does each have?

253. Someone has 4 times more rubles in his right pocket than in his left; if he transfers 6 rubles from his right pocket to his left, then there will be only 3 times more money in the right than in the left. How much money is in each pocket?

253. Someone has 3 times more rubles in his right pocket than in his left; If, however, 5 rubles are transferred from the left pocket to the right, then there will be five times more money in the right than in the left. How much money is in each pocket?

254. When two workers were paid at the factory, the first of them received 12 rubles more than the second for work, and after that the second worker paid him 2 rubles. debt. It turned out that the first brought home three times more money than the second. How much did each earn?

254. When calculating two workers at the factory, the first of them received 20 rubles less than the second for work, but at the same time the second worker returned 2 rubles to him. debt. It turned out that the first took home half the money of the second. How much did each earn?

255. One boy has 30 kopecks, the other 11 kopecks. How many times should they give one kopeck each so that the first has twice as much money as the second?

255. One boy has 48 kopecks, another has 22 kopecks. How many times do they have to spend one kopeck each so that the first has three times more money than the second?

256. The father is 40 years old and the son is 12 years old. How many years ago was the father five times older than the son?

256. The father is 49 years old and the son is 11 years old. In how many years will the father be three times as old as the son?

257. One landowner has four times as many sheep as another. If both bought 9 sheep each, then the first would have three times more sheep than the second. How many sheep does each have?

257. One landowner has three times less sheep than another. If both sold 10 sheep each, then the first would have five times fewer sheep than the second. How many sheep does each have?

258. The father is 39 years older than his son, and in 7 years he will be 4 times older than his son. How old is the one and the other?

258. Father and son together are 88 years old, and 8 years ago the father was 7 times older than his son. How old is the one and the other?

259. One tank has 48 buckets and the other has 22 buckets of water. Twice as much water was drained from the first as from the second, and then three times more water remained in the first than in the second. How many buckets are poured out of each?

259. There are 42 buckets in one tank and 8 buckets of water in another. Three times more water was poured into the first than into the second, and then four times more water turned out to be in the first than in the second. How many buckets are poured into each?

260. Two persons, playing cards separately, had at the beginning of the game - the first 72 rubles, the second 21 rubles. The first lost three times as much as the second won. After the game, the first player had twice as much money as the second player. How much did the second win and lose the first?

260. Two persons, playing cards separately, had at the beginning of the game - the first 25 rubles, the second 12 rubles. The first won twice as much as the second lost. After the game, the first player turned out to have five times more money than the second. How much did the second lose and the first win?

261. The peddler sold for the first time a part of 2/7 of the number of apples he had, for the second time p of the same number; then he had only 8 apples left. How many apples did he have?

261. The pedlar sold the first time 1/9 of the number of apples he had, the second time 5/6 of the same number; then he had only 4 apples left. How many apples did he have?

262. First, a third of the total amount of water was poured out of the water tank, then 5/6 of the rest, and then only 6 buckets remained. How much water was in the tank?

262. First, 3/5 of the total amount was poured out of the water tank, then 3/4 of the rest, and then only 5 buckets remained. How much water was in the tank?

263. In one society there were 40 men, women and children. The number of women was 3/5 of the number of men, and the number of children was 2/3 of the number of men and women together. How many men, women and children were there?

263. In one society there were 72 men, women and children. The number of men was 2/3 of the number of women, and the number of children was 4/5 of the number of men and women together. How many men, women and children were there?

264. For 30 arshins of cloth of two varieties, only 128 rubles were paid; a yardstick of the first grade costs 4 1/2 rubles, and a yardstick of the second grade costs 4 rubles. How many arshins of both grades were bought?

264. Only 120 rubles were paid for 27 arshins of cloth of two grades; arshin of the first grade costs 5 rubles; arshin of the second 3 p. 75 k.. How many arshins of both certificates were bought?

265. The tea merchant sold 38 pounds of tea of ​​two varieties, at a price of 3 r. per pound of the first grade and 1 p. 60 kopecks per pound of the second grade, and at the same time earned 22 rubles more for the entire first grade than for the second. How many teas of both varieties were sold?

265. A tea merchant sold 110 pounds of tea of ​​two varieties, at a price of 4 1/2 r. per pound of the first grade and 2 p. 25 k. for a pound of the second grade, and at the same time gained 45 rubles less for the first grade than for the second. How many teas of both varieties were sold?

266. The contractor hired an employee on the condition of paying him 90 kopecks. for each working day and deduct 40 kopecks from it. for each non-working day. After 12 days, the worker received 6 r. 90 k.. How many days did he work?

266. The contractor hired an employee on the condition of paying him 80 kopecks. for each working day and deduct 50 kopecks from it. for each non-working day. After 50 days, the worker received 21 rubles. 80 in .. How many days did he skip?

267. A And IN they play billiards with the condition that the winner of the game receives 76 k from the loser; after 20 games it turned out that IN won only 4 r. 50 k.. How many games did he win?

267 A And IN they play billiards with the condition that the winner of the game receives 50 k from the loser; after 12 games it turned out that A won only 2 times. How many games did he lose?

268. Two couriers left at the same time from two cities located at a distance of 300 miles, and are traveling towards one another. The first travels 12 versts per hour, the second 13 versts. When will they meet?

268. Two couriers left at the same time from two cities located at a distance of 280 miles, and go towards one another. The first travels 11 versts per hour, the second 17 versts. When will they meet?

269. from two stations railway, located at a distance of 77 versts, two trains leave simultaneously and go in the same direction at speeds of 31 1/2 versts and 18 2/3 versts per hour, the first following the second. When will he catch up?

269. From two railway stations located at a distance of 38 versts, two trains leave simultaneously and go in the same direction at speeds of 25 1/4 versts and 20 1/2 versts per hour, the first following the second. When will he catch up?

270. A passenger train leaves the station at 12 noon, making 32 in. at one o'clock. After 45 minutes, a courier train leaves the same station, making 42 in. at one o'clock. At what time will the courier train overtake the passenger train?

270. A passenger train leaves the station at 9 o'clock in the morning, making 28 in. at one o'clock. An hour and a quarter later, a courier train leaves the same station, making 40 volts. at one o'clock. At what time will the courier train overtake the passenger train?

271. What capital should be given in growth at 6% in order to make a profit of 224 rubles in 1 year and 2 months?

271. What capital must be given up for growth at 8% in order to receive a profit of 182 rubles in 7 months?

272. How much interest must be given to capital growth of 4400 rubles in order to receive a profit of 280 rubles in 1 year 5 months. 50 k.?

272. By how much interest must a capital of 1,800 rubles be paid in interest in order to receive a profit of 93 rubles in 11 months. 60 k.?

273. The merchant, having sold the goods for 299 rubles, gained 15% of the profit. What does the product cost him?

273. A merchant, having sold goods for 161 rubles, received 7 1/2% of the profit. What does the product cost him?

274. When selling goods in the amount of 429 p. received at a loss of 2 1/2%. What is the price of a product?

274. When selling goods in the amount of 366 rubles. received at a loss 8 1 / 2 % What is the cost of the goods?

275. On the bill 10 months before the due date, 1120 rubles were paid, with commercial accounting at 8%. Find the currency of the bill.

275. On a bill for 1 year 3 months before the due date, 839 rubles were paid. 60 kop. with commercial accounting at 7%. Find the currency of the bill.

276. The pool fills with one pipe at 3 o'clock, the other at 5 o'clock. How long will it take to fill up if both pipes are opened at the same time?

276. The pool is filled with one pipe at 7 1/2 o'clock, the other at 5 o'clock. How long will it take to fill up if both pipes are opened at the same time?

277. The pool is filled with one pipe at 4 o'clock, and through the other it can all flow out at 6 o'clock. At what time will the pool be filled with the simultaneous action of both pipes?

277. The pool is filled with one pipe at 2 1 / 3 hours, and through the other it can all flow out at 2 hours 48 hours. How long will the pool be filled with the simultaneous action of both pipes?

278. Two workers together finish work at 3 hours 36 minutes; the first one can perform it at 6 o'clock. At what time will the second person do the same work?

278. Two workers together finish work at 12 o'clock; the first one can perform it at 20 o'clock. At what time will the second one do the same work?

279. There are three pipes in the pool; water enters through the first two, flows out through the third. Through the first pipe, the pool can be filled at 3 o'clock, through the second at 2 o'clock, and through the third, all the water can flow out of the pool at 6 o'clock. At what time will the pool be filled if all three pipes are opened?

279. There are three pipes in the basin; water enters through the first two, flows out through the third. Through the first pipe, the pool can be filled at 2 o'clock, through the second at 5 o'clock, and through the third, all the water can flow out of the pool at 10 o'clock. At what time will the pool be filled if all three pipes are opened?

280. Of the three pipes drawn into the pool, the first fills it at 5 o'clock, the second fills it at 15 o'clock, and through the third the entire pool flows out at 3 o'clock. How long will it take for a full pool to drain if all pipes are active at the same time?

280. Of the three pipes drawn into the pool, the first fills it at 6 o'clock, the second fills it at 18 o'clock, and through the third the entire pool flows out at 3 o'clock. How long will it take for the full pool to drain if all the pipes operate simultaneously?

281. The second train of the railway goes from A V IN at an average speed of 30 miles per hour, then returns from IN V A at a speed of 28 miles per hour. He makes the whole trip there and back at 2 1/2 hours. How many miles from A before IN?

281. The second train of the railway goes from A V IN at an average speed of 24 miles per hour, then returns from IN V A at a speed of 30 miles per hour. He makes the whole trip there and back at 11 1/4 hours. How many miles from A before IN?

282. From A V IN a train came out, passing at an hour of 20 versts. After 8 hours the train leaves IN V A, passing 30 in. at one o'clock. Distance AB equals 350 V.. At what distance from A trains meet?

282. From A V IN a train came out, passing at the hour of 24 versts. The train leaves in 5 hours. IN V A passing 28 c. at one o'clock. Distance AB equal to 380 in., At what distance from IN trains meet?

283. The sum of three numbers is 70. The second number, when divided by the first, gives a quotient of 2 and a remainder of 1, the third, when divided by the second, gives a quotient of 3 and a remainder of 3. Find these numbers.

283. The sum of three numbers is 60. The second number, when divided by the first, gives a quotient of 3 and a remainder of 2; the third, when divided by the second, gives a quotient of 2 and a remainder of 4. Find the numbers.

284. Find a number which, when divided by 5, leaves a remainder of 2, and when divided by 8, gives a remainder of 5, knowing that the first quotient is three times greater than the second.

284. Find a number which, when divided by 7, leaves a remainder of 2, and when divided by 9, gives a remainder of 4, knowing moreover. that the first quotient is two greater than the second.

285. Someone, wanting to distribute the money he had with him to the poor, calculated that if each was given 15 kopecks, then 10 kopecks would not be enough for him, and if each were given 13 kopecks, then 6 kopecks would remain extra. How many beggars were there and how much money?

285. Someone, wanting to distribute the money he had with him to the poor, calculated that if everyone was given 8 kopecks, then 4 kopecks would remain. superfluous, and if everyone is given 9 kopecks, then 2 kopecks will not be enough .. How many beggars were there and how much money?

286. The engineer places telegraph poles at some distance. If he put them at a distance of 25 fathoms from one another, then 150 more pillars would have to be made, and if he increased the distance between the pillars by 5 fathoms, then 70 pillars would be extra. How great is the distance and how many poles are made?

286. An engineer places telegraph poles at some distance. If he had placed them at a distance of 30 fathoms from one another, then he would have left 100 extra pillars, and if he reduced the distance of the pillars by 4 fathoms, then another 180 pillars would have to be made. How great is the distance and how many poles are made?

287. Someone, when hiring a servant, promised him for a year of service to pay money and 144 rubles. and give clothes. The servant paid off after 7 months and received clothes and 54 rubles in payment. What did the clothes cost?

287. When hiring a servant, someone promised him for 7 months of service to pay 75 rubles in money and give him clothes. The servant paid off after 5 months and received clothes and 45 rubles in payment. What is the cost of clothes?

288. Paid for 46 pounds of sugar for 195 rubles. more than 73 pounds of tea; 9 poods of sugar cost 30 rubles less than 37 pounds of tea. What is a pound of tea and a pood of sugar worth?

288. Paid for 21 pounds of tea for 238 rubles less than for 40 pounds of sugar; 15 pounds of tea cost 2 rubles. more expensive than 4 poods of sugar. What is a pound of tea and a pood of sugar worth?

289. The landowner hired two peasants for the same daily wage. One of them for 40 days, he gave 7 p. 50 kopecks in money and 3 1/2 quarters of oats, another 4 rubles in 24 days. 80 k. in cash and 2 quarters of oats. What is a quarter of an oat worth?

289. The landowner hired two peasants for the same daily wage. He gave 14 rubles to one of them in 56 days. money and 8 quarters of oats, another for 88 days 13 p. 50 k. in cash and 15 quarters of oats. What is the cost of a quarter of an oat?

290. Paid for 25 arshins of cloth and 21 arshins. velvet 247 rubles. It is known that 10 arsh. velvet cost 18 rubles more than 13 arshins of cloth. What is worth an arshin of both?

290. Paid for 15 arshins of velvet and 52 arshins. cloth 276 rubles. It is known that 2 arsh. velvet cost 17 rubles less than 11 arsh. cloth. What is worth an arshin of both?

291. The sum of the digits of a two-digit number is 12. If 18 is subtracted from the desired number, then you get a number indicated by the same digits, but written in reverse order. Find this number.

291. The difference between the digits of units and tens of some two-digit number is equal to 3. If 27 is added to the desired number, then we get a number indicated by the same digits, but written in reverse order. Find this number.

292. In some two-digit number, the number of tens is twice the number of units. If we rearrange the digits of this number, then we get a number less than the desired one by 36. Find this number.

292. In some two-digit number, the number of tens is three times less than the number of units. If we rearrange the digits of this number, we get a number greater than the desired one by 36. Find this number.

293. A playing chess with IN and wins three out of every four games against him, then plays with WITH and wins two out of every three games against the latter. Total A played 21 games and won 15 of them. How many games did he play with IN and with WITH?

293. A playing chess with IN and loses to him three out of every eight games, then plays with WITH and loses two out of every five games to the last. All in all A played 26 games and lost 10 of them. How many games did he play with IN and with WITH?

294. What time is it now if 1/5 of the number of hours since noon is 1/3 of the number of hours until midnight?

294. What time is it now, if 1/11 of the number of hours that have passed since noon is equal to 1/13 of the number of hours left until midnight?

295. Find the weight of the fish, knowing that the tail weighs 2 pounds, the head weighs as much as the tail and half of the body, and the body weighs as much as the head and tail.

295. Find the weight of the fish, knowing that its head weighs 7 pounds, the tail weighs as much as the head and half of the body weigh, and the body weighs as much as the tail and head.

296. A certain amount must be divided by two persons so that the parts of the first and second are related to each other like the numbers 5 and 3, and that part of the first is 50 rubles. over 5/9 of the total. How big is each part?

296. A certain amount must be divided between two persons so that the parts of the first and second are related to each other as the numbers 7 and 4, and that the part of the second is 21 rubles. less than 5/12 of the whole sum. How big is each part?

297. The product was sold at a loss for 420 rubles; if it were sold for 570 rubles, then the profit received would be 5 times more than the loss incurred. What is the price of a product?

297. Goods sold at a profit for 520 rubles; if it had been sold for 320 rubles, then there would have been a loss amounting to 3/7 of the proceeds. What is the price of a product?

298. The numbers of arshins of calico contained in three pieces are related as 2:3:5. If you cut off 4 arshins from the first piece, 6 arshins from the second. and from the third 10 arsh., then the remaining amount of the whole chintz will be 5/6 of the previous amount. How many arshins are in each piece?

298. The number of arshins of calico contained in three pieces is 3:5:8. If cut off from the first 10 arshins, from the second 20 arshins. and from the third 30 arsh., then the remaining amount of the entire chintz will be 5/8 of the previous amount. How many arshins are in each piece?

299. First, half of all the water in it and half a bucket were poured out of the reservoir, then half of the remainder and half a bucket, finally another half of the remainder and half a bucket; after that, 6 buckets remained in the tank. How much water was there in the beginning?

299. A third of the water that was in it and a third of a bucket were poured out of the reservoir, then a third of the remainder and a third of the bucket, finally another third of the remainder and a third of the bucket; after that, 7 buckets remained in the tank. How much water was there at the beginning?

300. Several persons divide some amount as follows; the first one gets 100 r. and a fifth of the remainder, the second 200 rubles and a fifth of the new balance, the third 300 rubles and a fifth of the remainder, etc. It turned out that the entire amount was divided into equal parts. How big is this amount, how many participants are in the division and how much did each get?

300. Several persons divide a certain amount as follows: the first receives 50 rubles and a sixth of the balance, the second 100 rubles and a sixth of the new balance, the third 150 rubles and a sixth of the remainder, etc. It turned out that the entire amount was divided into equal parts . How big is this amount, how many participants are in the division and how much did each get?

The following tasks differ from the previous ones in that the data is expressed implicitly, namely in letters. These tasks belong to the same types as the previous ones. When solving them, the most important of those methods that were used earlier are repeated, but, due to the implicit form of the data, the reasoning is more general and at the same time more abstract. In the new exercises, just as in the previous ones, one must first of all take care to express through the main unknown and through the given designations all the quantities that are directly mentioned in the problem or that are implied in it, and in this case one must consistently take into account attention to all the designations given in the problem, and all the conditions related to the data and to those sought, when in this way all the conditions will be used in the case, then the thought of how to compose the required equation will itself appear.

301. Difference of two numbers s q . Find both numbers.

301. Difference of two numbers d , the multiple ratio of the larger to the smaller q . Find both numbers.

302. Divide a number A into three parts so that the first part is greater than the second by a number T and less than a third P once.

302. Divide a number A into three parts so that the first part is less than the second by a number T and more than a third P once.

303. One number in A times less than the other. If you add to the first number T , and to the second P , then the first sum will be in b times less than the second. Find these numbers.

303. One number in A times less than the other. If we subtract from the first T , and from the second P , then the first difference will be in b times more than the second. Find these numbers.

304. The number of a fraction is less than its denominator by a number A ; If, however, the fractions are subtracted from both members by b T / P . Find terms of a fraction.

304. The numerator of a fraction is greater than its denominator by a number A . If we add to both members of the fraction by b , then you get a fraction equal to the fraction T / P . Find terms of a fraction.

305. Divide a number A R times more than the second and q times less than a third.

305. Divide a number A into three parts so that the first was. V R times less than the second and q times more than a third.

306. The denominator of a fraction is the largest of its numerator in A once. If we add to the numerator the number b and subtract the number from the denominator With , then you get a fraction equal to the fraction k /l . Find terms of a fraction.

306. The denominator of a fraction is less than its numerator in A once. If we subtract the number from the numerator b and add a number to the denominator With , then learn a fraction equal to the fraction k /l . Find terms of a fraction.

307. Divide a number T into two parts so that the difference between the quotients from dividing the first part by A and second on b would love r.

307. Divide a number T into two parts so that the sum of the quotients from dividing the first part by A and second on b would equal s .

308. An employee receives for each working day A kopecks, and for each non-working one they deduct b kopecks. After the lapse of P days, the net income of the worker is equal to s rubles. How many working days and how many non-working days?

308. An employee receives for each working day A kopecks, and for each non-working one they deduct from it b kopecks. After the lapse of P days, the employee must pay 5 rubles himself. How many working days and how many non-working days?

309. Difference of two numbers d . Dividing the minuend by the subtrahend gives the quotient q and a remainder equal to half the difference. Find those numbers

309. Difference of two numbers d . Dividing the minuend by the subtrahend gives the remainder r and a quotient equal to half the difference. Find these numbers.

310. For a few arshins of cloth. paid A rubles; if we bought more cloth With b

310. Paid for a few arshins of cloth A rubles; if we bought cloth for less With arshin, then you would have to pay b rubles. How many arshins were bought?

311. What number, when multiplied by a , will increase by the number T ?

311. What number, being divided by A , decrease by the number T ?

312. When selling a house for m rubles received R percent loss. What did it cost the seller himself?

312. When selling a house for T ruble received R percent profit. What did it cost the seller himself?

313. Two couriers leave at the same time from two places A And IN and travel in the same direction A To IN and so on. ІІfirst passes in an hour A verst, second b verst. Distance AB equals d verst. When and how far from A Will the first courier overtake the second?

313. Two couriers leave at the same time from two places A And IN and go towards one another. The first one passes in an hour A verst, second b verst. Distance AB equals d verst. When. and how far from A will both couriers meet?

314. The front wheel of the carriage has a circumference of A feet, rear circumference b ft. How far must the carriage travel in order for the front wheel to make P high revs in reverse?

314. The front wheel of the carriage has a circle on A feet less than the rear. How far must the carriage travel for the front wheel to make T , and the back P revolutions?

315. Two pipes are led into the pool, both of which fill it, the first with a separate action in A hours, the second also with a separate action in b hours. At what time will the pool be filled with the simultaneous action of both pipes?

315. Two pipes are led into the pool, of which the first, with a separate action, fills it in A hours, and the second also, in a separate action, pours the entire pool into b hours. How long will it take to fill the pool with simultaneous operation of both pipes?

316. Crew wheel circumference A times the circumference of the front wheel. The crew passed T feet, and in doing so, the front wheel made To revolutions more than the rear. Determine the circumference of both wheels and the number of revolutions.

316. The circumference of the front wheel on A feet less than the rear circumference. The crew passed T feet, and at the same time the rear wheel did in To times less revolutions than the front. Determine the circumference of both wheels and the number of revolutions.

317. The population of one city increases annually by R % compared to the population of the previous year. Currently in the city T

317. The population of one city decreases annually by R % compared to the population of the previous year. Currently in the city T residents. How many people were there 3 years ago?

318. Two workers, working at the same time, finish their work in A hours. One first will do the same job in b , times faster than one second. At what time will each worker finish work?

318. Two workers, working at the same time, finish work in A hours. One first will do the same job in b , times slower than one second. What time does each worker finish work?

319. The boatman, rowing down the river, swims P sazhen in t hours; rowing against the current, he uses And more hours to swim the same distance. Determine the hourly flow rate.

319. A boatman, rowing against the current, swims P sazhen in t hours; paddling downstream, he uses And hours less to swim the same distance. Determine the hourly flow rate.

320. body A moving at a speed v meters per second. How fast should the other body move? IN, coming from the same place t seconds earlier if it was overtaken by the body A through And seconds after the start of the movement of this body?

320. Body A moving at a speed v meters per second. How fast should the other body move? IN coming from the same place And seconds later if it catches up with the body A through and seconds after the start of its movement?

321. Of the two varieties of goods, at a price of A rubles and b rubles per pound, compiled d T rubles per pound received s rubles loss. How many pounds of both sorts went into making the mixture?

321. Of two varieties of goods, at a price of A rubles and b rubles per pound, compiled d pounds of mix. When selling this mixture by T rubles per pound received s rubles profit. How many pounds of both sorts went into making the mixture?

322. B pool, accommodating T buckets, two pipes were laid. The first pours into the pool A buckets per hour. The second pours the entire pool into b hours. At what time will the pool be filled with simultaneous operation of both pipes?

322. To the pool containing T buckets, two pipes were laid. The first fills the entire pool in A hours. The second in an hour pours out of the pool b buckets. At what time will the pool be filled with simultaneous operation of both pipes?

323. Divide a number A into three parts so that the first relates to the second, as t:p , and the second to the third, as p: q.

323. Divide a number A into three parts so that the second relates to the first, as t:p , and the third to the second, as p: q.

324. From two places A And IN P sazhen, two boats are sailing towards each other, driven by rowers with the same strength. The first, floating downstream, travels the entire distance AB V t hours; the second, swimming against the current, uses the same distance more time for And hours. Determine the hourly flow rate.

324. From two places A And IN on the river, separated from one another by P sazhen, two boats are sailing towards each other, driven by rowers with the same strength. The first, swimming against the current, travels the entire distance AB V t hours; the second, going with the flow, uses less time for the same distance And hours. Determine the hourly flow rate.

325. Determine the capitals of three persons, knowing that the first and the second have together T rubles, the second with the third P rubles, and that the capital of the first R times less than the capital of the third.

325. Determine the capitals of three persons, knowing that the first and the third have together T rubles, the second with the third P rubles, and that the capital of the first R times the capital of the second.

326. Two bodies move towards each other from two places at a distance d meters. The first moves at a speed v meters per second. With what speed should the second body move if it has reached the h seconds later than the first and should go before the meeting of everything P seconds?

326. Two bodies move towards one another from two places at a distance d meters. The first moves at a speed v meters per second. With what speed must the second body move if it has reached h seconds before the first and should go until the meeting of all P seconds?

327. Promissory note discounted commercially R % behind P years before the deadline, gives more mathematical consideration, also made according to R % and for P years, on A rubles. Find the currency of the week.

327. A bill discounted commercially R % behind P years, stands on T rubles cheaper than with mathematical accounting, also made according to R % and for P years What is the amount of the bill?

328. Two couriers leave places A And B located at a distance d verst, and they go towards, passing at the first hour u version and second v versts; departure of the first A took place on h IN. Determine when and where the couriers will meet?

328. Two couriers leave places A And B located at a distance d verst, and they both go in the same direction, passing at an hour or one And verst and second v versts; departure first from A took place on h hours before the departure of the second B. Determine when and where the first courier will overtake the second?

329. Divide a number A into such three parts, that if you attach to the first T , the second is first reduced by m , and then multiply by P , and divide the third into P , then the results will be the same.

329. Divide a number A into such three parts that if the first is reduced by T , first increase the second by T , then multiply by P , and divide the third into P , then the results will be the same.

330. There are three pipes in the pool. A, B And WITH. Through A And WITH water flows through IN A And IN the pool fills up T hours, under action A And C V P hours, under action IN And WITH V R hours. At what time will the pool be filled with the simultaneous action of all three pipes?

330. Three pipes are led into the pool A, B And WITH. Through A water flows through IN And WITH follows. With the joint action of pipes A And IN the pool fills up T hours, under action A And WITH V P clock, pipes IN And WITH pour the whole pool into R hours. How long will it take for the entire pool to drain if all three pipes operate simultaneously?

To compose an equation means to express in mathematical form the relationship between the data (known) of the problem and the required (unknown) values ​​of it. Sometimes this connection is so clearly contained in the formulation of the problem that the formulation of the equation is simply a literal retelling of the problem, in the language of mathematical signs.

Example 1. Petrov received 160 rubles for his work. more than half of the amount that Ivanov received. Together they received 1120 rubles. How much did Petrov and Ivanov get for their work? Let x be Ivanov's earnings. Half of his earnings are 0.5x; Petrov's monthly salary is 0.5x + 160 together they earn 1120 rubles; the mathematical notation of the last phrase would be

(0.5x + 160) + x = 1120.

The equation has been made. Solving it according to the once established rules, we find Ivanov's earnings x = 640 rubles; Petrov's earnings are 0.5x + 160=480 (rubles).

More often, however, it happens that the connection between the data and the sought quantities is not indicated directly in the problem; it must be set based on the conditions of the task. In practical problems, this is almost always the case. The example just given is contrived; In real life, such tasks are almost never encountered.

Therefore, it is impossible to give completely exhaustive instructions for compiling an equation. However, at first it is useful to be guided by the following. Let us take for the value of the desired value (or several values) some randomly taken number (or several numbers) and set ourselves the task of checking whether we have guessed the correct solution of the problem or not. If we were able to carry out this test and find out either that our guess is correct or that it is incorrect (the second is most likely to happen, of course), then we can immediately compose the desired equation (or several equations). Namely, we will write down the very actions that we performed to check, but instead of a randomly taken number, we will introduce an alphabetic sign of an unknown value. We get the required equation.

Example 2. A piece of an alloy of copper and zinc with a volume of 1 dm3 weighs 8.14 kg. How much copper is in the alloy? (specific weight of copper 8.9 kg/dm3; zinc - 7.0 kg/dm3).

Let us take at random a number expressing the desired volume of copper, for example, 0.3 dm3. Let's check if we have successfully taken this number. Since 1 kg / dm3 of copper weighs 8.9 kg, then 0.3 dm3 weighs 8.9 * 0.3 = 2.67 (kg). The volume of zinc in the alloy is 1 - 0.3 = 0.7 (dm3). Its weight is 7.0 0.7 = 4.9 (kg). The total weight of zinc and copper is 2.67 + + 4.9 = 7.57 (kg). Meanwhile, the weight of our piece, according to the condition of the problem, is 8.14 kg. Our guess is invalid. But on the other hand, we will immediately get an equation whose solution will give the correct answer. Instead of a randomly taken number of 0.3 dm3, we denote the volume of copper (in dm3) through x. Instead of the product 8.9 0.3 = 2.67 we take the products 8.9 x. This is the weight of the copper in the alloy. Instead of 1 - 0.3 = 0.7 we take 1 - x; this is the amount of zinc. Instead of 7.0 0.7 = 4.9 we take 7.0 (1 - x); this is the weight of zinc. Instead of 2.67 + 4.9 we take 8.9 x + 7.0 (1 - x); this is the combined weight of zinc and copper. By condition, it is equal to 8.14 kg; so 8.9 x + 7.0 (1 - x) = 8.14.

Solving this equation gives x = 0.6. Checking a randomly chosen solution can be done in various ways; accordingly, for the same problem, different types of equations can be obtained; all of them, however, will give the same solution for the desired value, such equations are called equivalent to each other.

Of course, after gaining skills in compiling equations, there is no need to randomly check the number taken: you can take for the value of the desired value not a number, but some letter (x, y, etc.) and act as if this letter ( unknown) was the number we are going to test.

The solution of the problem usually comes down to finding the value of some quantity by logical reasoning and calculations. For example, find the speed, time, distance, mass of an object or the amount of something.

This problem can be solved using an equation. To do this, the desired value is denoted through a variable, then, by logical reasoning, they compose and solve an equation. Having solved the equation, they check whether the solution of the equation satisfies the conditions of the problem.

Lesson content

Writing Expressions Containing the Unknown

The solution of the problem is accompanied by the compilation of an equation for this problem. On initial stage studying problems, it is desirable to learn how to compose literal expressions describing one or another life situation. This stage is not difficult and can be studied in the process of solving the problem itself.

Consider several situations that can be written using a mathematical expression.

Task 1. Father's age x years. Mom is two years younger. The son is 3 times younger than the father. Record the age of each using expressions.

Solution:

Task 2. Father's age x years, mother is 2 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. Record the age of each using expressions.

Solution:

Task 3. Father's age x years, mother is 3 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. How old is each if the combined age of father, mother, son and daughter is 92?

Solution:

In this problem, in addition to writing expressions, it is necessary to calculate the age of each family member.

First, we write down the age of each family member using expressions. Per variable x let's take the age of the father, and then using this variable we will compose the remaining expressions:

Now let's determine the age of each family member. To do this, we need to write and solve an equation. We have all the components of the equation ready. It remains only to collect them together.

The total age of 92 years was obtained by adding the ages of dad, mom, son and daughter:

For each age, we made a mathematical expression. These expressions will be the components of our equation. Let's assemble our equation according to this scheme and the table that was given above. That is, the words dad, mom, son, daughter will be replaced by the expression corresponding to them in the table:

Expression for mother's age x − 3 for clarity, it was taken in brackets.

Now let's solve the resulting equation. To begin with, you can open the brackets where possible:

To free the equation from fractions, multiply both sides by 3

We solve the resulting equation using the known identical transformations:

We found the value of the variable x. This variable was responsible for the age of the father. So the age of the father is 36 years.

Knowing the age of the father, you can calculate the ages of the rest of the family. To do this, you need to substitute the value of the variable x in those expressions that are responsible for the age of a particular family member.

In the problem, it was said that the mother is 3 years younger than the father. We denoted her age through the expression x−3. Variable value x is now known, and in order to calculate the age of the mother, it is necessary in the expression x − 3 instead of x substitute the found value 36

x - 3 \u003d 36 - 3 \u003d 33 years old mom.

Similarly, the age of the remaining family members is determined:

Examination:

Task 4. A kilogram of apples is worth x rubles. Write down an expression that calculates how many kilograms of apples you can buy for 300 rubles.

Solution

If a kilogram of apples costs x rubles, then for 300 rubles you can buy a kilogram of apples.

Example. A kilogram of apples costs 50 rubles. Then for 300 rubles you can buy, that is, 6 kilograms of apples.

Task 5. On x rubles, 5 kg of apples were bought. Write down an expression that calculates how many rubles one kilogram of apples costs.

Solution

If for 5 kg of apples was paid x rubles, then one kilogram will cost rubles

Example. For 300 rubles, 5 kg of apples were bought. Then one kilogram of apples will cost, that is, 60 rubles.

Task 6. Tom, John and Leo went to the cafeteria during recess and bought a sandwich and a mug of coffee. The sandwich is worth x rubles, and a mug of coffee - 15 rubles. Determine the cost of a sandwich if it is known that 120 rubles were paid for everything?

Solution

Of course, this problem is as simple as three pennies and can be solved without resorting to an equation. To do this, subtract the cost of three cups of coffee (15 × 3) from 120 rubles, and divide the result by 3

But our goal is to write an equation for the problem and solve this equation. So the cost of a sandwich x rubles. Bought only three. So, having tripled the cost, we get an expression describing how many rubles were paid for three sandwiches

3x - cost of three sandwiches

And the cost of three cups of coffee can be written as 15 × 3. 15 is the cost of one mug of coffee, and 3 is a multiplier (Tom, John and Leo) that triples this cost.

According to the condition of the problem, 120 rubles were paid for everything. We already have an approximate scheme of what needs to be done:

We already have expressions describing the cost of three sandwiches and three cups of coffee. These are expressions 3 x and 15×3. Using the scheme, we will write an equation and solve it:

So, the cost of one sandwich is 25 rubles.

The problem is solved correctly only if the equation for it is compiled correctly. Unlike ordinary equations, by which we learn to find roots, equations for solving problems have their own specific application. Each component of such an equation can be described in verbal form. When compiling an equation, it is imperative to understand why we include one or another component in its composition and why it is needed.

It is also necessary to remember that the equation is an equality, after solving which the left side will have to be equal to the right side. The resulting equation should not contradict this idea.

Imagine that the equation is a balance with two bowls and a screen showing the state of the balance.

IN this moment the screen shows an equal sign. It is clear why the left bowl is equal to the right bowl - there is nothing on the bowls. We write the state of the scales and the absence of something on the bowls using the following equality:

0 = 0

Let's put a watermelon on the left scale:

The left bowl outweighed the right bowl and the screen sounded the alarm, showing the not equal sign (≠). This sign indicates that the left bowl is not equal to the right bowl.

Now let's try to solve the problem. Let it be required to find out how much the watermelon weighs, which lies on the left bowl. But how do you know? After all, our scales are designed only to check whether the left bowl is equal to the right one.

Equations come to the rescue. Recall that by definition the equation is equality A that contains the variable whose value you want to find. In this case, the scales play the role of this very equation, and the mass of the watermelon is a variable whose value must be found. Our goal is to get this equation right. Understand, align the scales so that you can calculate the mass of the watermelon.

To level the scales, you can put some heavy object on the right bowl. For example, let's put a weight of 7 kg there.

Now, on the contrary, the right bowl outweighed the left. The screen still shows that the bowls are not equal.

Let's try to put a weight of 4 kg on the left bowl

Now the scales have leveled out. The figure shows that the left bowl is at the level of the right bowl. And the screen shows an equal sign. This sign indicates that the left bowl is equal to the right bowl.

Thus, we have obtained an equation - an equality containing an unknown. The left pan is the left side of the equation, consisting of the 4 components and the variable x(mass of the watermelon), and the right bowl is the right side of the equation, consisting of component 7.

Well, it is not difficult to guess that the root of the equation 4 + x\u003d 7 is 3. So the mass of the watermelon is 3 kg.

The same is true for other tasks. To find some unknown value, various elements are added to the left or right side of the equation: terms, factors, expressions. In school problems, these elements are already given. It remains only to correctly structure them and build an equation. In this example, we were engaged in selection, trying weights of different masses in order to calculate the mass of a watermelon.

Naturally, the data that is given in the problem must first be brought to a form in which they can be included in the equation. Therefore, as they say "Whether you like it or not, you have to think".

Consider the following problem. The age of the father is equal to the age of the son and daughter together. The son is twice as old as the daughter and twenty years younger than the father. How old is each?

The daughter's age can be expressed as x. If the son is twice as old as the daughter, then his age will be indicated as 2 x. The condition of the problem says that together the age of the daughter and son is equal to the age of the father. So the father's age will be denoted by the sum x + 2x

You can add like terms in an expression. Then the father's age will be denoted as 3 x

Now let's make an equation. We need to get an equality in which we can find the unknown x. Let's use weights. On the left bowl we put the age of the father (3 x), and on the right bowl the age of the son (2 x)

It is clear why the left bowl outweighed the right one and why the screen shows the sign (≠) . After all, it is logical that the age of the father is greater than the age of the son.

But we need to balance the scales so that we can calculate the unknown x. To do this, you need to add some number to the right bowl. What number is indicated in the problem. The condition stated that the son was 20 years younger than the father. So 20 years is the same number that needs to be put on the scales.

The scales will even out if we add these 20 years to the right side of the scale. In other words, let's raise the son to the age of the father

Now the scales have leveled out. It turned out the equation , which is easily solved:

x we marked the age of the daughter. Now we have found the value of this variable. Daughter 20 years old.

And finally, we calculate the age of the father. In the problem, it was said that it is equal to the sum of the ages of the son and daughter, that is, (20 + 40) years.

Let's return to the middle of the task and pay attention to one point. When we put the age of the father and the age of the son on the scale, the left bowl outweighed the right

But we solved this problem by adding another 20 years to the right bowl. As a result, the scales leveled out and we got the equality

But it was possible not to add these 20 years to the right bowl, but subtract them from the left. We would get equality in this case

This time the equation is . The root of the equation is still 20

That is, the equations And are equivalent. And we remember that equivalent equations have the same roots. If you look closely at these two equations, you can see that the second equation is obtained by transferring the number 20 from the right side to the left side with the opposite sign. And this action, as indicated in the previous lesson, does not change the roots of the equation.

You also need to pay attention to the fact that at the beginning of solving the problem, the ages of each family member could be denoted through other expressions.

Let's say the son's age is denoted by x and since he is two older than the daughter, then the age of the daughter is indicated by (understand to make her twice as young as the son). And the age of the father, since it is the sum of the ages of the son and daughter, is denoted through the expression . And finally, to build a logically correct equation, you need to add the number 20 to the age of the son, because the father is twenty years older. The result is a completely different equation. . Let's solve this equation

As you can see, the answers to the problem have not changed. My son is still 40 years old. The daughters are still years old, and the father is 40 + 20 years old.

In other words, the problem can be solved various methods. Therefore, one should not despair that it is not possible to solve this or that problem. But you need to keep in mind that there are the most simple ways to solve the problem. There are various routes to the city center, but there is always the most convenient, fastest and safest route.

Examples of problem solving

Task 1. There are 30 notebooks in two packs. If 2 notebooks were transferred from the first bundle to the second, then there would be twice as many notebooks in the first bundle as in the second. How many notebooks were in each pack?

Solution

Denote by x the number of notebooks that were in the first pack. If there were 30 notebooks in total, and the variable x this is the number of notebooks from the first pack, then the number of notebooks in the second pack will be denoted by the expression 30 − x. That is, from the total number of notebooks, we subtract the number of notebooks from the first pack and thereby obtain the number of notebooks from the second pack.

and add these two notebooks to the second pack

Let's try to make an equation from the existing expressions. We put both packs of notebooks on the scales

The left bowl is heavier than the right. This is because the condition of the problem says that after two notebooks were taken from the first bundle and placed in the second, the number of notebooks in the first bundle became twice as large as in the second.

To equalize the scales and get the equation, double the right side. To do this, multiply it by 2

It turns out an equation. Let's solve this equation:

We denoted the first pack by the variable x. Now we have found its meaning. Variable x equal to 22. So there were 22 notebooks in the first pack.

And we denoted the second pack through the expression 30 − x and since the value of the variable x Now we know, we can calculate the number of notebooks in the second pack. It is equal to 30 − 22, that is, 8 pieces.

Task 2. Two people were peeling potatoes. One peeled two potatoes a minute, and the other three potatoes. Together they cleared 400 pieces. How long did each work if the second worked 25 minutes more than the first?

Solution

Denote by x time of the first person. Since the second person worked 25 minutes more than the first, his time will be denoted by the expression

The first worker peeled 2 potatoes per minute, and since he worked x minutes, then in total he cleared 2 x potatoes.

The second person peeled three potatoes per minute, and since he worked for minutes, he peeled potatoes in total.

Together they peeled 400 potatoes

From the available components, we will compose and solve the equation. On the left side of the equation there will be potatoes peeled by each person, and on the right side of their sum:

At the beginning of the solution of this problem through the variable x we marked the time of work of the first person. Now we have found the value of this variable. The first person worked 65 minutes.

And the second person worked for minutes, and since the value of the variable x now it is known, then you can calculate the time of the second person - it is equal to 65 + 25, that is, 90 minutes.

Problem from Andrey Petrovich Kiselev's Algebra Textbook. A mixture of 32 kg was made from the varieties of tea. A kilogram of the first grade costs 8 rubles, and of the second grade 6 rubles. 50 kop. How many kilograms are taken of both varieties, if a kilogram of the mixture costs (without profit or loss) 7 rubles. 10 kopecks?

Solution

Denote by x a lot of tea of ​​the first grade. Then the mass of tea of ​​the second grade will be denoted through the expression 32 − x

A kilogram of tea of ​​the first grade costs 8 rubles. If these eight rubles are multiplied by the number of kilograms of tea of ​​the first grade, then it will be possible to find out how much the rubles cost x kg of tea of ​​the first grade.

A kilogram of second-class tea costs 6 rubles. 50 kop. If these 6 rubles. 50 kop. multiply by 32 − x, then you can find out how many rubles cost 32 − x kg of tea of ​​the second grade.

The condition says that a kilogram of the mixture costs 7 rubles. 10 kop. In total, 32 kg of the mixture was prepared. Multiply 7 rubles. 10 kop. at 32 we can find out how much 32 kg of the mixture costs.

The expressions from which we will compose the equation now take the following form:

Let's try to make an equation from the existing expressions. Let's put the cost of mixtures of teas of the first and second grades on the left pan of the scales, and put the cost of 32 kg of the mixture on the right pan, that is, the total cost of the mixture, which includes both varieties of tea:

At the beginning of the solution of this problem through the variable x we designated the mass of tea of ​​the first grade. Now we have found the value of this variable. Variable x equals 12.8. This means that 12.8 kg of tea of ​​the first grade was taken to prepare the mixture.

And through expression 32 − x we denoted the mass of tea of ​​the second grade, and since the value of the change x now known, we can calculate the mass of tea of ​​the second grade. It is equal to 32 − 12.8, that is, 19.2. This means that 19.2 kg of second grade tea was taken to prepare the mixture.

Task 3. A cyclist traveled a distance at a speed of 8 km/h. He had to return by another road, which was 3 km longer than the first, and although returning, he was traveling at a speed of 9 km / h, he used time for more than minutes. How long were the roads?

Solution

Some tasks may cover topics that the person may not have studied. This task belongs to such a range of tasks. It deals with the concepts of distance, speed and time. Accordingly, in order to solve such a problem, you need to have an idea about the things that are said in the problem. In our case, we need to know what is the distance, speed and time.

The task is to find the distances of two roads. We must write an equation that will allow us to calculate these distances.

Consider the relationship between distance, speed and time. Each of these quantities can be described using a literal equation:

We will use the right side of one of these equations to draw up our equation. To find out which one, you need to return to the text of the task and pay attention to the following point:

Attention should be paid to the moment where the cyclist on the way back took more than a minute of time. This hint tells us that we can use the equation, namely its right side. This will allow us to write an equation that contains the variable S .

So let's denote the length of the first road as S. The cyclist traveled this path at a speed of 8 km/h. The time for which he covered this path will be denoted by the expression, since time is the ratio of the distance traveled to the speed

The way back for the cyclist was 3 km longer. Therefore, its distance will be denoted by the expression S+ 3 . A cyclist traveled this road at a speed of 9 km/h. So the time for which he overcame this path will be denoted by the expression .

Now let's make an equation from the existing expressions

The right bowl is heavier than the left. This is because the problem says that the cyclist spent more time on the way back.

To equalize the scales, add these same minutes to the left side. But first, let's convert minutes to hours, since in the problem the speed is measured in kilometers per hour, and not in meters per minute.

To convert minutes to hours, you need to divide them by 60

Minutes make hours. Add these hours to the left side of the equation:

It turns out the equation . Let's solve this equation. To get rid of fractions, both parts of the part can be multiplied by 72. Further, using the known identical transformations, we find the value of the variable S

Through a variable S we marked the distance of the first road. Now we have found the value of this variable. Variable S is 15. So the distance of the first road is 15 km.

And we denoted the distance of the second road through the expression S+ 3 , and since the value of the variable S Now we know, we can calculate the distance of the second road. This distance is equal to the sum of 15 + 3, that is, 18 km.

Task 4. Two cars are driving down the highway at the same speed. If the first one increases the speed by 10 km / h, and the second one reduces the speed by 10 km / h, then the first one will cover the same distance in 2 hours as the second one in 3 hours. At what speed do cars go?

Solution

Denote by v the speed of each car. Further in the problem, hints are given: increase the speed of the first car by 10 km/h, and decrease the speed of the second car by 10 km/h. Let's use this hint

It is further stated that at such speeds (increased and decreased by 10 km / h), the first car will cover the same distance in 2 hours as the second in 3 hours. Phrase "as many" can be understood as "the distance traveled by the first car will be equals distance traveled by the second car.

The distance, as we remember, is determined by the formula. We are interested in the right side of this literal equation - it will allow us to write an equation containing a variable v .

So, at speed v + 10 km/h the first car will pass 2(v+10) km, and the second will pass 3(v − 10) km. Under this condition, the cars will cover the same distances, so to get the equation, it is enough to connect these two expressions with an equal sign. Then we get the equation. Let's solve it:

In the condition of the problem, it was said that the cars go at the same speed. We denoted this speed by the variable v. Now we have found the value of this variable. Variable v equals 50. So the speed of both cars was 50 km/h.

Task 5. In 9 hours downstream the ship travels the same distance as in 11 hours upstream. Find the speed of the boat if the speed of the river is 2 km/h.

Solution

Denote by v own speed of the ship. The speed of the river flow is 2 km/h. In the course of the river, the speed of the boat will be v + 2 km/h, and against the current - (v − 2) km/h.

The condition of the problem states that in 9 hours the ship travels the same distance along the river as in 11 hours against the current. Phrase "same way" can be understood as the distance traveled by the boat along the river in 9 hours, equals distance traveled by the ship against the current of the river in 11 hours. That is, the distances will be the same.

The distance is determined by the formula . Let's use the right side of this literal equation to write our own equation.

So, in 9 hours, the ship will pass along the river 9(v + 2) km, and in 11 hours upstream - 11(v − 2) km. Since both expressions describe the same distance, we equate the first expression to the second. As a result, we get the equation . Let's solve it:

So the speed of the ship is 20 km/h.

When solving problems, a useful habit is to determine in advance on which solution is sought for it.

Assume that the task required finding the time it would take for a pedestrian to cover a given path. We denoted the time through the variable t, then we made an equation containing this variable and found its value.

From practice, we know that the time of movement of an object can take both integer values ​​and fractional values, for example, 2 hours, 1.5 hours, 0.5 hours. Then we can say that the solution to this problem is sought on the set rational numbers Q, since each of the values ​​2 h, 1.5 h, 0.5 h can be represented as a fraction.

Therefore, after an unknown quantity has been denoted by a variable, it is useful to indicate to which set this quantity belongs. In our example, the time t belongs to the set of rational numbers Q

t ∈ Q

You can also introduce a constraint on the variable t, indicating that it can only accept positive values. Indeed, if the object has spent a certain time on the way, then this time cannot be negative. Therefore, next to the expression t∈ Q specify that its value must be greater than zero:

t∈ R, t > 0

If we solve the equation, we get a negative value for the variable t, then it will be possible to conclude that the problem was solved incorrectly, since this solution will not satisfy the condition t∈ Q , t> 0 .

Another example. If we were solving a problem in which it was required to find the number of people to perform a particular job, then we would denote this number through a variable x. In such a problem, the solution would be sought on the set natural numbers

x ∈ N

Indeed, the number of people is an integer, such as 2 people, 3 people, 5 people. But not 1.5 (one whole person and half a person) or 2.3 (two whole people and another three tenths of a person).

Here one could indicate that the number of people must be greater than zero, but the numbers included in the set of natural numbers N are themselves positive and greater than zero. This set does not negative numbers and the number 0. Therefore, the expression x > 0 can be omitted.

Task 6. To repair the school, a team arrived in which there were 2.5 times more painters than carpenters. Soon the foreman included four more painters in the team, and transferred two carpenters to another object. As a result, there were 4 times more painters in the brigade than carpenters. How many painters and how many carpenters were in the brigade initially

Solution

Denote by x carpenters who arrived for repairs initially.

The number of carpenters is an integer greater than zero. Therefore, we point out that x belongs to the set of natural numbers

x∈N

There were 2.5 times more painters than carpenters. Therefore, the number of painters will be denoted as 2.5x.

And the number of painters will increase by 4

Now the number of carpenters and painters will be denoted by the following expressions:

Let's try to make an equation from the existing expressions:

The right bowl is larger, because after adding four more painters to the team, and moving two carpenters to another object, the number of painters in the team turned out to be 4 times more than carpenters. To equalize the scales, you need to increase the left bowl by 4 times:

Got an equation. Let's solve it:

Through a variable x the initial number of carpenters was designated. Now we have found the value of this variable. Variable x equals 8. So 8 carpenters were in the brigade initially.

And the number of painters was indicated through the expression 2.5 x and since the value of the variable x now it is known, then you can calculate the number of painters - it is equal to 2.5 × 8, that is, 20.

We return to the beginning of the task and make sure that the condition is met x∈ N. Variable x equals 8, and the elements of the set of natural numbers N these are all numbers starting with 1, 2, 3 and so on ad infinitum. The same set includes the number 8, which we found.

8 ∈N

The same can be said about the number of painters. The number 20 belongs to the set of natural numbers:

20 ∈N

To understand the essence of the problem and correct compilation equation, it is not necessary to use the scale model with bowls. You can use other models: segments, tables, diagrams. You can come up with your own model that would describe the essence of the problem well.

Task 9. 30% of milk was poured from the can. As a result, 14 liters remained in it. How many liters of milk were in the can originally?

Solution

The desired value is the initial number of liters in the can. Draw the number of liters as a line and label this line as X

It is said that 30% of the milk was poured out of the can. We select in the figure approximately 30%

A percentage, by definition, is one hundredth of something. If 30% of the milk was poured out, then the remaining 70% remained in the can. These 70% account for 14 liters indicated in the problem. Select the remaining 70% in the figure

Now you can make an equation. Let's remember how to find the percentage of a number. To do this, the total amount of something is divided by 100 and the result is multiplied by the desired percentage. Note that 14 liters, which is 70%, can be obtained in the same way: the initial number of liters X divide by 100 and multiply the result by 70. Equate all this to the number 14

Or get a simpler equation: write 70% as 0.70, then multiply by X and equate this expression to 14

This means that initially there were 20 liters of milk in the can.

Task 9. They took two alloys of gold and silver. In one, the ratio of these metals is 1: 9, and in the other, 2: 3. How much of each alloy should be taken to get 15 kg of a new alloy in which gold and silver would be related as 1: 4?

Solution

Let's first try to find out how much gold and silver will be contained in 15 kg of the new alloy. The task says that the content of these metals should be in a ratio of 1: 4, that is, gold should be one part of the alloy, and silver should be four parts. Then the total number of parts in the alloy will be 1 + 4 = 5, and the mass of one part will be 15: 5 = 3 kg.

Let's determine how much gold will be contained in 15 kg of alloy. To do this, multiply 3 kg by the number of parts of gold:

3 kg × 1 = 3 kg

Let's determine how much silver will be contained in 15 kg of alloy:

3 kg × 4 = 12 kg

This means that an alloy weighing 15 kg will contain 3 kg of gold and 12 kg of silver. Now back to the original alloys. You need to use each of them. Denote by x the mass of the first alloy, and the mass of the second alloy can be denoted by 15 − x

Let's express as a percentage all the relationships that are given in the problem and fill in the following table with them:

In the first alloy, gold and silver are in a ratio of 1: 9. Then the total parts will be 1 + 9 = 10. Of these, there will be gold , and silver .

Let's transfer this data to the table. 10% will be entered in the first line in the column "percentage of gold in the alloy", 90% will also be entered in the first line of the column "percentage of silver in the alloy", and in the last column "weight of alloy" enter a variable x, since this is how we denoted the mass of the first alloy:

We do the same with the second alloy. Gold and silver in it are in a ratio of 2: 3. Then there will be 2 + 3 = 5 parts in total. Of these, gold will be , and silver .

Let's transfer this data to the table. 40% will be entered in the second line in the column "percentage of gold in the alloy", 60% will also be entered in the second line of the column "percentage of silver in the alloy", and in the last column "weight of alloy" enter the expression 15 − x, because this is how we denoted the mass of the second alloy:

Let's fill in the last line. The resulting alloy weighing 15 kg will contain 3 kg of gold, which is alloy, and silver will be alloy. In the last column we write down the mass of the resulting alloy 15

You can now write equations using this table. We remember. If we separately add up the gold of both alloys and equate this amount to the mass of gold of the resulting alloy, we can find out what the value is x.

The first gold alloy had 0.10 x, and in the second gold alloy it was 0.40(15 − x) . Then in the resulting alloy, the mass of gold will be the sum of the masses of gold of the first and second alloys, and this mass is 20% of the new alloy. And 20% of the new alloy is 3 kg of gold, calculated by us earlier. As a result, we obtain the equation 0,10x+ 0.40(15 − x) = 3 . Let's solve this equation:

Initially through x we have designated the mass of the first alloy. Now we have found the value of this variable. Variable x is equal to 10. And we denoted the mass of the second alloy through 15 − x, and since the value of the variable x now it is known, then we can calculate the mass of the second alloy, it is equal to 15 − 10 = 5 kg.

This means that to obtain a new alloy weighing 15 kg in which gold and silver would be treated as 1: 4, you need to take 10 kg of the first and 5 kg of the second alloy.

The equation could be made using the second column of the resulting table. Then we would get the equation 0,90x+ 0.60(15 − x) = 12. The root of this equation is also 10

Task 10. There is ore from two layers with a copper content of 6% and 11%. How much low-grade ore should be taken to get it when mixed with rich 20 tons with a copper content of 8%?

Solution

Denote by x mass of poor ore. Since you need to get 20 tons of ore, then 20 rich ore will be taken − x. Since the copper content in poor ore is 6%, then in x tons of ore will contain 0.06 x tons of copper. In rich ore, the copper content is 11%, and in 20 - x tons of rich ore will contain 0.11(20 − x) tons of copper.

In the resulting 20 tons of ore, the copper content should be 8%. This means that 20 tons of copper ore will contain 20 × 0.08 = 1.6 tons.

Add expressions 0.06 x and 0.11(20 − x) and equate this sum to 1.6. We get the equation 0,06x + 0,11(20 − x) = 1,6

Let's solve this equation:

This means that to obtain 20 tons of ore with a copper content of 8%, you need to take 12 tons of poor ore. The rich will take 20 − 12 = 8 tons.

Task 11. Increasing average speed from 250 to 300 m/min, the athlete began to run the distance 1 minute faster. What is the length of the distance?

Solution

The length of the distance (or the distance of the distance) can be described by the following letter equation:

Let's use the right side of this equation to write our own equation. Initially, the athlete ran the distance at a speed of 250 meters per minute. At this speed, the length of the distance will be described by the expression 250 t

Then the athlete increased her speed to 300 meters per minute. At this speed, the length of the distance will be described by the expression 300t

Note that the length of the distance is a constant value. From the fact that the athlete increases the speed or reduces it, the length of the distance will remain unchanged.

This allows us to equate the expression 250 t to expression 300 t, since both expressions describe the length of the same distance

250t = 300t

But the task says that at a speed of 300 meters per minute, the athlete began to run the distance 1 minute faster. In other words, at a speed of 300 meters per minute, the travel time will decrease by one. Therefore, in equation 250 t= 300t on the right side, the time must be reduced by one:

At a speed of 250 meters per minute, the athlete runs the distance in 6 minutes. Knowing the speed and time, you can determine the length of the distance:

S= 250 × 6 = 1500 m

And at a speed of 300 meters per minute, the athlete runs the distance for t− 1 , that is, in 5 minutes. As mentioned earlier, the length of the distance does not change:

S= 300 × 5 = 1500 m

Task 12. A rider overtakes a pedestrian who is 15 km ahead of him. In how many hours will the rider catch up with the pedestrian if every hour the first rider travels 10 km, and the second travels only 4 km?

Solution

This task is . It can be solved by determining the approach speed and dividing the initial distance between the rider and the pedestrian by this speed.

The closing speed is determined by subtracting the lower speed from the larger one:

10 km/h − 4 km/h = 6 km/h (speed of approach)

Every hour the distance of 15 kilometers will be reduced by 6 kilometers. To find out when it will decrease completely (when the rider catches up with the pedestrian), you need to divide 15 by 6

15:6 = 2.5 h

2,5 h it's two whole hours and half an hour. And half an hour is 30 minutes. So the rider will overtake the pedestrian in 2 hours and 30 minutes.

Let's solve this problem using the equation.

After that, after him, a rider set out on the road at a speed of 10 km / h. And the walking speed is only 4 km/h. This means that the rider will overtake the pedestrian after some time. We need to find this time.

When the rider catches up with the pedestrian, it will mean that they passed together same distance. The distance traveled by the rider and the pedestrian is described by the following equation:

Let's use the right side of this equation to write our own equation.

The distance traveled by the rider will be described by the expression 10 t. Since the pedestrian set off before the rider and managed to overcome 15 km, the distance traveled by him will be described by the expression 4 t + 15 .

By the time the rider catches up with the pedestrian, both of them will have covered the same distance. This allows us to equate the distances traveled by the rider and the walker:

The result is a simple equation. Let's solve it:

Tasks for independent solution

Problem 1. A passenger train arrives from one city to another 45 minutes faster than a freight train. Calculate the distance between cities if the speed of the passenger train is 48 km/h and the speed of the freight train is 36 km/h.

Solution

The train speeds in this problem are measured in kilometers per hour. Therefore, we will convert the 45 minutes indicated in the task into hours. 45 minutes is 0.75 hours

Let us denote the time during which a freight train arrives in the city through the variable t. Since the passenger train arrives in this city 0.75 hours faster, the time of its movement will be denoted by the expression t - 0,75

Passenger train overcame 48( t - 0.75) km, and commodity 36 t km. Since we are talking about the same distance, we equate the first expression to the second. As a result, we obtain the equation 48(t - 0.75) = 36t . Let's solve it:

Now let's calculate the distance between cities. To do this, the speed of a freight train (36 km / h) is multiplied by the time of its movement t. Variable value t now known - it is equal to three hours

36 × 3 = 108 km

To calculate the distance, you can also use the speed of the passenger train. But in this case the value of the variable

Variable value t equals 1.2. So the cars met after 1.2 hours.

Answer: the cars met after 1.2 hours.

Task 3. There are a total of 685 workers in three workshops of the plant. In the second shop there are three times more workers than in the first, and in the third - 15 workers less than in the second shop. How many workers are in each shop?

Solution

Let x workers were in the first shop. In the second workshop there were three times more than in the first, so the number of workers in the second workshop can be denoted by the expression 3 x. The third shop had 15 fewer workers than the second. Therefore, the number of workers in the third workshop can be denoted by the expression 3 x - 15 .

The problem says that there were 685 workers in total. Therefore, we can add the expressions x, 3x, 3x - 15 and equate this sum to the number 685. As a result, we obtain the equation x + 3x + ( 3x - 15) = 685

Through a variable x the number of workers in the first workshop was indicated. Now we have found the value of this variable, it is equal to 100. So there were 100 workers in the first shop.

In the second workshop there were 3 x workers, i.e. 3 × 100 = 300. And in the third workshop there were 3 x - 15, i.e. 3 × 100 − 15 = 285

Answer: in the first workshop there were 100 workers, in the second - 300, in the third - 285.

Task 4. Two repair shops within a week should repair 18 motors according to the plan. The first workshop completed the plan by 120%, and the second by 125%, so 22 motors were repaired within a week. What weekly engine repair plan did each workshop have?

Solution

Let x the motors were to be repaired by the first workshop. Then the second workshop had to renovate 18 − x motors.

Since the first workshop completed its plan by 120%, this means that it has repaired 1.2 x motors. And the second workshop fulfilled its plan by 125%, which means that it repaired 1.25 (18 − x) motors.

The task says that 22 motors were repaired. Therefore, we can add the expressions 1,2x and 1.25(18 − x) , then equate this sum to the number 22. As a result, we obtain the equation 1,2x + 1,25(18− x) = 22

Through a variable x the number of motors that the first workshop was supposed to repair was indicated. Now we have found the value of this variable, it is equal to 10. So the first workshop had to repair 10 motors.

And through the expression 18 − x the number of motors that the second workshop was supposed to repair was indicated. So the second workshop had to repair 18 − 10 = 8 motors.

Answer: the first workshop was to repair 10 motors, and the second 8 motors.

Problem 5. The price of the goods has increased by 30% and is now 91 rubles. How much was the product before the price increase?

Solution

Let x rubles worth of goods before the price increase. If the price has increased by 30% it means that it has increased by 0.30 x rubles. After the price increase, the goods began to cost 91 rubles. Add x with 0.30 x and equate this sum to 91. As a result, we obtain the equation Decreasing the number by 10% resulted in 45. Find the original value of the number. x -

Answer: to get a 12% salt solution, you need to add 0.25 kg of a 20% solution to 1 kg of a 10% solution.

Problem 12. Two solutions of salt in water are given, the concentrations of which are equal to 20% and 30%. How many kilograms of each solution must be mixed in one vessel to obtain 25 kg of a 25.2% solution?

Solution

Let x kg of the first solution must be taken. Since it is required to prepare 25 kg of solution, the mass of the second solution can be denoted by the expression 25 − x.

The first solution will contain 0.20x kg of salt, and the second will contain 0.30(25 − x) kg of salt. In the resulting solution, the salt content will be 25 × 0.252 = 6.3 kg. Add the expressions 0.20x and 0.30(25 − x), then equate this sum to 6.3. As a result, we obtain the equation

So the first solution needs to be taken 12 kg, and the second 25 - 12 = 13 kg.

Answer: the first solution you need to take 12 kg, and the second 13 kg.

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Properties of a straight line in Euclidean geometry.

There are infinitely many lines that can be drawn through any point.

Through any two non-coinciding points, there is only one straight line.

Two non-coincident lines in the plane either intersect at a single point, or are

parallel (follows from the previous one).

There are three options in 3D space. relative position two straight lines:

• lines intersect;

• straight lines are parallel;

• straight lines intersect.

Straight line- algebraic curve of the first order: in the Cartesian coordinate system, a straight line

is given on the plane by an equation of the first degree (linear equation).

General equation of a straight line.

Definition. Any line in the plane can be given by a first order equation

Ah + Wu + C = 0,

and constant A, B not equal to zero at the same time. This first order equation is called general

straight line equation. Depending on the values ​​of the constants A, B And WITH The following special cases are possible:

. C = 0, A ≠ 0, B ≠ 0- the line passes through the origin

. A = 0, B ≠0, C ≠0 ( By + C = 0)- straight line parallel to the axis Oh

. B = 0, A ≠ 0, C ≠ 0 ( Ax + C = 0)- straight line parallel to the axis OU

. B = C = 0, A ≠ 0- the line coincides with the axis OU

. A = C = 0, B ≠ 0- the line coincides with the axis Oh

The equation of a straight line can be represented in various forms depending on any given

initial conditions.

Equation of a straight line by a point and a normal vector.

Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)

perpendicular to the line given by the equation

Ah + Wu + C = 0.

Example. Find the equation of a straight line passing through a point A(1, 2) perpendicular to the vector (3, -1).

Solution. Let's compose at A \u003d 3 and B \u003d -1 the equation of the straight line: 3x - y + C \u003d 0. To find the coefficient C

we substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C \u003d 0, therefore

C = -1. Total: the desired equation: 3x - y - 1 \u003d 0.

Equation of a straight line passing through two points.

Let two points be given in space M 1 (x 1 , y 1 , z 1) And M2 (x 2, y 2 , z 2), Then straight line equation,

passing through these points:

If any of the denominators zero, the corresponding numerator should be set equal to zero. On

plane, the equation of a straight line written above is simplified:

If x 1 ≠ x 2 And x = x 1, If x 1 = x 2 .

Fraction = k called slope factor straight.

Example. Find the equation of a straight line passing through the points A(1, 2) and B(3, 4).

Solution. Applying the above formula, we get:

Equation of a straight line by a point and a slope.

If the general equation of a straight line Ah + Wu + C = 0 bring to the form:

and designate , then the resulting equation is called

equation of a straight line with slope k.

The equation of a straight line on a point and a directing vector.

By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task

a straight line through a point and a direction vector of a straight line.

Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition

Aα 1 + Bα 2 = 0 called direction vector of the straight line.

Ah + Wu + C = 0.

Example. Find the equation of a straight line with direction vector (1, -1) and passing through point A(1, 2).

Solution. We will look for the equation of the desired straight line in the form: Ax + By + C = 0. According to the definition,

coefficients must satisfy the conditions:

1 * A + (-1) * B = 0, i.e. A = B.

Then the equation of a straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.

at x=1, y=2 we get C/ A = -3, i.e. desired equation:

x + y - 3 = 0

Equation of a straight line in segments.

If in the general equation of the straight line Ah + Wu + C = 0 C≠0, then, dividing by -C, we get:

or , where

geometric sense coefficients in that the coefficient a is the coordinate of the intersection point

straight with axle Oh, A b- the coordinate of the point of intersection of the line with the axis OU.

Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this straight line in segments.

C \u003d 1, , a \u003d -1, b \u003d 1.

Normal equation of a straight line.

If both sides of the equation Ah + Wu + C = 0 divide by number , which is called

normalizing factor, then we get

xcosφ + ysinφ - p = 0 -normal equation of a straight line.

The sign ± of the normalizing factor must be chosen so that μ * C< 0.

R- the length of the perpendicular dropped from the origin to the line,

A φ - the angle formed by this perpendicular with the positive direction of the axis Oh.

Example. Given the general equation of a straight line 12x - 5y - 65 = 0. Required to write various types of equations

this straight line.

The equation of this straight line in segments:

The equation of this line with slope: (divide by 5)

Equation of a straight line:

cos φ = 12/13; sin φ= -5/13; p=5.

It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,

parallel to the axes or passing through the origin.

Angle between lines on a plane.

Definition. If two lines are given y \u003d k 1 x + b 1, y \u003d k 2 x + b 2, That sharp corner between these lines

will be defined as

Two lines are parallel if k 1 = k 2. Two lines are perpendicular

If k 1 \u003d -1 / k 2 .

Theorem.

Direct Ah + Wu + C = 0 And A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients are proportional

A 1 \u003d λA, B 1 \u003d λB. If also С 1 \u003d λС, then the lines coincide. Coordinates of the point of intersection of two lines

are found as a solution to the system of equations of these lines.

The equation of a line passing through a given point is perpendicular to a given line.

Definition. A line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b

represented by the equation:

The distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then the distance to the line Ah + Wu + C = 0 defined as:

Proof. Let the point M 1 (x 1, y 1)- the base of the perpendicular dropped from the point M for a given

direct. Then the distance between the points M And M 1:

(1)

Coordinates x 1 And 1 can be found as a solution to the system of equations:

The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicularly

given line. If we transform the first equation of the system to the form:

A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.