Equations with 1 variable. Equation with one variable. On the algebraic sum

In this video, we will analyze a whole set of linear equations that are solved using the same algorithm - that's why they are called the simplest.

To begin with, let's define: what is a linear equation and which of them should be called the simplest?

A linear equation is one in which there is only one variable, and only in the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest ones using the algorithm:

1. Open brackets, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Bring like terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$ .

Of course, this algorithm does not always help. The point is that sometimes, after all these machinations, the coefficient of the variable $x$ turns out to be zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when you get something like $0\cdot x=8$, i.e. on the left is zero, and on the right is a non-zero number. In the video below, we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

And now let's see how it all works on the example of real problems.

Examples of solving equations

Today we deal with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to open the brackets, if any (as in our last example);
2. Then bring similar
3. Finally, isolate the variable, i.e. everything that is connected with the variable - the terms in which it is contained - is transferred to one side, and everything that remains without it is transferred to the other side.

Then, as a rule, you need to bring similar on each side of the resulting equality, and after that it remains only to divide by the coefficient at "x", and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Usually, mistakes are made either when opening brackets, or when counting "pluses" and "minuses".

In addition, it happens that a linear equation has no solutions at all, or so that the solution is the entire number line, i.e. any number. We will analyze these subtleties in today's lesson. But we will start, as you already understood, with the most simple tasks.

Scheme for solving simple linear equations

To begin with, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the parentheses, if any.
2. Seclude variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
3. We present similar terms.
4. We divide everything by the coefficient at "x".

Of course, this scheme does not always work, it has certain subtleties and tricks, and now we will get to know them.

Solving real examples of simple linear equations

Task #1

In the first step, we are required to open the brackets. But they are not in this example, so we skip this step. In the second step, we need to isolate the variables. Please note: we are talking only about individual terms. Let's write:

We give similar terms on the left and on the right, but this has already been done here. Therefore, we proceed to the fourth step: divide by a factor:

\[\frac(6x)(6)=-\frac(72)(6)\]

Here we got the answer.

Task #2

In this task, we can observe the brackets, so let's expand them:

Both on the left and on the right, we see approximately the same construction, but let's act according to the algorithm, i.e. sequester variables:

Here are some like:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task #3

The third linear equation is already more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they just have different signs in front of them. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's calculate:

We perform the last step - we divide everything by the coefficient at "x":

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, then I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, zero can get in among them - there is nothing wrong with that.

Zero is the same number as the rest, you should not somehow discriminate it or assume that if you get zero, then you did something wrong.

Another feature is related to the expansion of parentheses. Please note: when there is a “minus” in front of them, we remove it, but in brackets we change the signs to opposite. And then we can open it according to standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such actions is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complicated and a quadratic function will appear when performing various transformations. However, you should not be afraid of this, because if, according to the author's intention, we solve a linear equation, then in the process of transformation all monomials containing a quadratic function will necessarily be reduced.

Example #1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some like:

Obviously, this equation has no solutions, so in the answer we write as follows:

\[\variety \]

or no roots.

Example #2

We perform the same steps. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some like:

Obviously, this linear equation has no solution, so we write it like this:

\[\varnothing\],

or no roots.

Nuances of the solution

Both equations are completely solved. On the example of these two expressions, we once again made sure that even in the simplest linear equations, everything can be not so simple: there can be either one, or none, or infinitely many. In our case, we considered two equations, in both there are simply no roots.

But I would like to draw your attention to another fact: how to work with brackets and how to expand them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by "x". Please note: multiply each individual term. Inside there are two terms - respectively, two terms and is multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can the bracket be opened from the point of view that there is a minus sign after it. Yes, yes: only now, when the transformations are done, we remember that there is a minus sign in front of the brackets, which means that everything below just changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is no coincidence that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and learn to solve such simple equations again.

Of course, the day will come when you will hone these skills to automatism. You no longer have to perform so many transformations each time, you will write everything in one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task #1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do a retreat:

Here are some like:

Let's do the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, however, they mutually canceled out, which makes the equation exactly linear, not square.

Task #2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's do the first step carefully: multiply every element in the first bracket by every element in the second. In total, four new terms should be obtained after transformations:

And now carefully perform the multiplication in each term:

Let's move the terms with "x" to the left, and without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

We have received a definitive answer.

Nuances of the solution

The most important remark about these two equations is this: as soon as we start multiplying brackets in which there is more than a term, then this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we get four terms.

On the algebraic sum

With the last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: we subtract seven from one. In algebra, we mean by this the following: to the number "one" we add another number, namely "minus seven." This algebraic sum differs from the usual arithmetic sum.

As soon as when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

In conclusion, let's look at a couple more examples that will be even more complex than the ones we just looked at, and in order to solve them, we will have to slightly expand our standard algorithm.

Solving equations with a fraction

To solve such tasks, one more step will have to be added to our algorithm. But first, I will remind our algorithm:

1. Open brackets.
2. Separate variables.
3. Bring similar.
4. Divide by a factor.

Alas, this wonderful algorithm, for all its efficiency, is not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on the left and on the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be performed both before the first action and after it, namely, to get rid of fractions. Thus, the algorithm will be as follows:

1. Get rid of fractions.
2. Open brackets.
3. Separate variables.
4. Bring similar.
5. Divide by a factor.

What does it mean to "get rid of fractions"? And why is it possible to do this both after and before the first standard step? In fact, in our case, all fractions are numeric in terms of the denominator, i.e. everywhere the denominator is just a number. Therefore, if we multiply both parts of the equation by this number, then we will get rid of fractions.

Example #1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two brackets doesn't mean you have to multiply each of them by "four". Let's write:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's open it:

We perform seclusion of a variable:

We carry out the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, we pass to the second equation.

Example #2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

Problem solved.

That, in fact, is all that I wanted to tell today.

Key points

The key findings are as follows:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Do not worry if somewhere you have quadratic functions, most likely, in the process of further transformations, they will be reduced.
• The roots in linear equations, even the simplest ones, are of three types: one single root, the entire number line is a root, there are no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site, solve the examples presented there. Stay tuned, there are many more interesting things waiting for you!

In this lesson you will learn how to solve linear equations and understand how to do two kinds of transformations to make solving linear equations EASIER!

How many apples did each friend get?

Each of us, without hesitation, will answer: "Each friend got an apple."

But now I propose to think about it all the same ... Yes, yes. It turns out that when answering such a simple question, you decide in your head linear equation!

or verbally - three friends were given apples each, based on the fact that Vasya has all the apples.

And now you have decided linear equation.

Now let's give this term a mathematical definition.

What are "linear equations"

Linear Equation - is an algebraic equation whose total degree of its constituent polynomials is. It looks like this:

Where and are any numbers and

For our case with Vasya and apples, we will write:

- “if Vasya gives all three friends the same number of apples, he will have no apples left”

"Hidden" linear equations, or the importance of identical transformations

Despite the fact that at first glance everything is extremely simple, when solving equations, you need to be careful, because linear equations are called not only equations of the form, but also any equations that are transformations and simplifications are reduced to this type.

For example:

We see that it is on the right, which, in theory, already indicates that the equation is not linear.

Moreover, if we open the brackets, we will get two more terms in which it will be, but don't jump to conclusions!

Before judging whether the equation is linear, it is necessary to make all the transformations and thus simplify the original example.

In this case, transformations can change appearance, but not the very essence of the equation.

In other words, these transformations must be identical or equivalent.

There are only two such transformations, but they play a very, VERY important role in solving problems. Let's consider both transformations on concrete examples.

Move left - right.

Let's say we need to solve the following equation:

Also in primary school we were told: "with X - to the left, without X - to the right."

What expression with x is on the right?

Right, not how not.

And this is important, because if this seemingly simple question is misunderstood, the wrong answer comes out.

And what is the expression with x on the left?

Right, .

Now that we have dealt with this, we transfer all terms with unknowns to left side, and all that is known - to the right.

And remembering that if there is no sign before the number, for example, then the number is positive, that is, it is preceded by the sign "".

Moved? What did you get?

All that remains to be done is to bring like terms. We present:

So, we have successfully parsed the first identical transformation, although I am sure that you already knew it and actively used it without me.

The main thing - do not forget about the signs for numbers and change them to the opposite when transferring through the equal sign!

Multiplication-division.

Let's start right away with an example

We look and think: what do we not like in this example?

The unknown is all in one part, the known in the other, but something is stopping us ...

And this is something - a four, because if it weren't there, everything would be perfect - x is equal to the number- just the way we want it!

How can you get rid of it?

We cannot transfer to the right, because then we need to transfer the entire multiplier (we cannot take it and tear it away from it), and transferring the entire multiplier also does not make sense ...

It's time to remember about the division, in connection with which we will divide everything just into!

All - this means both the left and the right side. So and only so!

What do we get?

Here is the answer.

Let's now look at another example:

Guess what to do in this case? That's right, multiply the left and right sides by! What answer did you get? Right. .

Surely you already knew everything about identical transformations. Consider that we just refreshed this knowledge in your memory and it's time for something more - For example, to solve our big example:

As we said earlier, looking at it, you cannot say that this equation is linear, but we need to open the brackets and perform identical transformations. So let's get started!

To begin with, we recall the formulas for abbreviated multiplication, in particular, the square of the sum and the square of the difference. If you don’t remember what it is and how brackets are opened, I strongly recommend reading the topic, as these skills will be useful to you when solving almost all the examples found on the exam.
Revealed? Compare:

Now it's time to bring like terms. Do you remember how we are in the same primary school did they say “we don’t put flies with cutlets”? Here I am reminding you of this. We add everything separately - factors that have, factors that have, and other factors that do not have unknowns. As you bring like terms, move all unknowns to the left, and everything that is known to the right. What did you get?

As you can see, the x-square has disappeared, and we see a completely ordinary linear equation. It remains only to find!

And finally, I will say one more very important thing about identical transformations - identical transformations are applicable not only for linear equations, but also for square, fractional rational and others. You just need to remember that when transferring factors through the equal sign, we change the sign to the opposite, and when dividing or multiplying by some number, we multiply / divide both sides of the equation by the same number.

What else did you take away from this example? That looking at an equation it is not always possible to directly and accurately determine whether it is linear or not. You must first completely simplify the expression, and only then judge what it is.

Linear equations. 3 examples

Here are a couple more examples for you to practice on your own - determine if the equation is linear and if so, find its roots:

Answers:

1. Is.

2. Is not.

Let's open the brackets and give like terms:

Let's make an identical transformation - we divide the left and right parts into:

We see that the equation is not linear, so there is no need to look for its roots.

3. Is.

Let's make an identical transformation - multiply the left and right parts by to get rid of the denominator.

Think why is it so important to? If you know the answer to this question, we proceed to the further solution of the equation, if not, be sure to look into the topic so as not to make mistakes in more difficult examples. By the way, as you can see, a situation where it is impossible. Why?
So let's go ahead and rearrange the equation:

If you coped with everything without difficulty, let's talk about linear equations with two variables.

Linear Equations with Two Variables

Now let's move on to a slightly more complicated one - linear equations with two variables.

Linear equations with two variables look like:

Where, and are any numbers and.

As you can see, the only difference is that one more variable is added to the equation. And so everything is the same - there are no x squared, there is no division by a variable, etc. and so on.

What would be a real life example for you...

Let's take the same Vasya. Suppose he decides that he will give each of his 3 friends the same number of apples, and keep the apples for himself.

How many apples does Vasya need to buy if he gives each friend an apple? What about? What if by?

The dependence of the number of apples that each person will receive on the total number of apples that need to be purchased will be expressed by the equation:

• - the number of apples that a person will receive (, or, or);
• - the number of apples that Vasya will take for himself;
• - how many apples Vasya needs to buy, taking into account the number of apples per person.

Solving this problem, we get that if Vasya gives one friend an apple, then he needs to buy pieces, if he gives apples - and so on.

And generally speaking. We have two variables.

Why not plot this dependence on a graph?

We build and mark the value of ours, that is, points, with coordinates, and!

As you can see, and depend on each other linearly, hence the name of the equations - “ linear».

We abstract from apples and consider graphically different equations.

Look carefully at the two constructed graphs - a straight line and a parabola, given by arbitrary functions:

Find and mark the corresponding points on both figures.
What did you get?

You can see that on the graph of the first function alone corresponds one, that is, and linearly depend on each other, which cannot be said about the second function.

Of course, you can object that on the second graph, x also corresponds to - , but this is only one point, that is special case, since you can still find one that matches more than just one.

And the constructed graph does not resemble a line in any way, but is a parabola.

I repeat, one more time: the graph of a linear equation must be a STRAIGHT line.

With the fact that the equation will not be linear if we go to any extent - this is understandable using the example of a parabola, although for yourself you can build a few more simple graphs, for example or.

But I assure you - none of them will be a STRAIGHT LINE.

Do not believe? Build and then compare with what I got:

And what happens if we divide something by, for example, some number?

Will there be a linear dependence and?

We will not argue, but we will build! For example, let's plot a function graph.

Somehow it doesn’t look like a straight line built ... accordingly, the equation is not linear.

Let's summarize:

1. Linear Equation - is an algebraic equation in which the total degree of its constituent polynomials is equal.
2. Linear Equation with one variable looks like:
   , where and are any numbers;
   Linear Equation with two variables:
   , where, and are any numbers.
3. It is not always immediately possible to determine whether an equation is linear or not. Sometimes, in order to understand this, it is necessary to perform identical transformations, move similar terms to the left / right, not forgetting to change the sign, or multiply / divide both sides of the equation by the same number.

LINEAR EQUATIONS. BRIEFLY ABOUT THE MAIN

1. Linear equation

This is an algebraic equation in which the total degree of its constituent polynomials is equal.

2. Linear equation with one variable looks like:

Where and are any numbers;

3. Linear equation with two variables looks like:

Where, and are any numbers.

4. Identity transformations

To determine whether the equation is linear or not, it is necessary to make identical transformations:

• move left/right like terms, not forgetting to change the sign;
• multiply/divide both sides of the equation by the same number.

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Linear equations. Solution, examples.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

Linear equations.

Linear equations are not the best difficult topic school mathematics. But there are some tricks there that can puzzle even a trained student. Shall we figure it out?)

A linear equation is usually defined as an equation of the form:

ax + b = 0 Where a and b- any numbers.

2x + 7 = 0. Here a=2, b=7

0.1x - 2.3 = 0 Here a=0.1, b=-2.3

12x + 1/2 = 0 Here a=12, b=1/2

Nothing complicated, right? Especially if you do not notice the words: "where a and b are any numbers"... And if you notice, but carelessly think about it?) After all, if a=0, b=0(any numbers are possible?), then we get a funny expression:

But that's not all! If, say, a=0, A b=5, it turns out something quite absurd:

What strains and undermines confidence in mathematics, yes ...) Especially in exams. But of these strange expressions, you also need to find X! Which doesn't exist at all. And, surprisingly, this X is very easy to find. We will learn how to do it. In this lesson.

How to recognize a linear equation in appearance? It depends on what appearance.) The trick is that linear equations are called not only equations of the form ax + b = 0 , but also any equations that are reduced to this form by transformations and simplifications. And who knows if it is reduced or not?)

A linear equation can be clearly recognized in some cases. Say, if we have an equation in which there are only unknowns in the first degree, yes numbers. And the equation doesn't fractions divided by unknown , it is important! And division by number, or a numeric fraction - that's it! For example:

This is a linear equation. There are fractions here, but there are no x's in the square, in the cube, etc., and there are no x's in the denominators, i.e. No division by x. And here is the equation

cannot be called linear. Here x's are all in the first degree, but there is division by expression with x. After simplifications and transformations, you can get a linear equation, and a quadratic one, and anything you like.

It turns out that it is impossible to find out a linear equation in some intricate example until you almost solve it. It's upsetting. But in assignments, as a rule, they don’t ask about the form of the equation, right? In tasks, equations are ordered decide. This makes me happy.)

Solution of linear equations. Examples.

The entire solution of linear equations consists of identical transformations of equations. By the way, these transformations (as many as two!) underlie the solutions all equations of mathematics. In other words, the decision any The equation begins with these same transformations. In the case of linear equations, it (the solution) on these transformations ends with a full-fledged answer. It makes sense to follow the link, right?) Moreover, there are also examples of solving linear equations.

Let's start with the simplest example. Without any pitfalls. Let's say we need to solve the following equation.

x - 3 = 2 - 4x

This is a linear equation. Xs are all to the first power, there is no division by X. But, actually, we don't care what the equation is. We need to solve it. The scheme here is simple. Collect everything with x's on the left side of the equation, everything without x's (numbers) on the right.

To do this, you need to transfer - 4x to the left side, with a change of sign, of course, but - 3 - to the right. By the way, this is first identical transformation of equations. Surprised? So, they didn’t follow the link, but in vain ...) We get:

x + 4x = 2 + 3

We give similar, we consider:

What do we need to be completely happy? Yes, so that there is a clean X on the left! Five gets in the way. Get rid of the five with second identical transformation of equations. Namely, we divide both parts of the equation by 5. We get a ready-made answer:

An elementary example, of course. This is for a warm-up.) It is not very clear why I recalled identical transformations here? OK. We take the bull by the horns.) Let's decide something more impressive.

For example, here is this equation:

Where do we start? With X - to the left, without X - to the right? Could be so. Little steps along the long road. And you can immediately, in a universal and powerful way. Unless, of course, in your arsenal there are identical transformations of equations.

I ask you a key question: What do you dislike the most about this equation?

95 people out of 100 will answer: fractions ! The answer is correct. So let's get rid of them. So we start right away with second identical transformation. What do you need to multiply the fraction on the left by so that the denominator is completely reduced? That's right, 3. And on the right? By 4. But math allows us to multiply both sides by the same number. How do we get out? Let's multiply both sides by 12! Those. to a common denominator. Then the three will be reduced, and the four. Do not forget that you need to multiply each part entirely. Here's what the first step looks like:

Expanding the brackets:

Note! Numerator (x+2) I took in brackets! This is because when multiplying fractions, the numerator is multiplied by the whole, entirely! And now you can reduce fractions and reduce:

Opening the remaining parentheses:

Not an example, but pure pleasure!) Now we recall the spell from lower grades: with x - to the left, without x - to the right! And apply this transformation:

Here are some like:

And we divide both parts by 25, i.e. apply the second transformation again:

That's all. Answer: X=0,16

Take note: to bring the original confusing equation to a pleasant form, we used two (only two!) identical transformations- translation left-right with a change of sign and multiplication-division of the equation by the same number. This is the universal way! We will work in this way any equations! Absolutely any. That is why I keep repeating these identical transformations all the time.)

As you can see, the principle of solving linear equations is simple. We take the equation and simplify it with the help of identical transformations until we get the answer. The main problems here are in the calculations, and not in the principle of the solution.

But ... There are such surprises in the process of solving the most elementary linear equations that they can drive into a strong stupor ...) Fortunately, there can be only two such surprises. Let's call them special cases.

Special cases in solving linear equations.

Surprise first.

Suppose you come across an elementary equation, something like:

2x+3=5x+5 - 3x - 2

Slightly bored, we transfer with X to the left, without X - to the right ... With a change of sign, everything is chin-chinar ... We get:

2x-5x+3x=5-2-3

We believe, and ... oh my! We get:

In itself, this equality is not objectionable. Zero is really zero. But X is gone! And we must write in the answer, what x is equal to. Otherwise, the solution doesn't count, yes...) A dead end?

Calm! In such doubtful cases, the most general rules save. How to solve equations? What does it mean to solve an equation? This means, find all values ​​of x that, when substituted into the original equation, will give us the correct equality.

But we have the correct equality already happened! 0=0, where really?! It remains to figure out at what x's this is obtained. What values ​​of x can be substituted into original equation if these x's still shrink to zero? Come on?)

Yes!!! Xs can be substituted any! What do you want. At least 5, at least 0.05, at least -220. They will still shrink. If you don't believe me, you can check it.) Substitute any x values ​​in original equation and calculate. All the time the pure truth will be obtained: 0=0, 2=2, -7.1=-7.1 and so on.

Here is your answer: x is any number.

The answer can be written in different mathematical symbols, the essence does not change. This is a completely correct and complete answer.

Surprise second.

Let's take the same elementary linear equation and change only one number in it. This is what we will decide:

2x+1=5x+5 - 3x - 2

After the same identical transformations, we get something intriguing:

Like this. Solved a linear equation, got a strange equality. Mathematically speaking, we have wrong equality. And in simple terms, this is not true. Rave. But nevertheless, this nonsense is quite a good reason for the correct solution of the equation.)

Again, we think from general rules. What x, when substituted into the original equation, will give us correct equality? Yes, none! There are no such xes. Whatever you substitute, everything will be reduced, nonsense will remain.)

Here is your answer: there are no solutions.

This is also a perfectly valid answer. In mathematics, such answers often occur.

Like this. Now, I hope, the loss of Xs in the process of solving any (not only linear) equation will not bother you at all. The matter is familiar.)

Now that we have dealt with all the pitfalls in linear equations, it makes sense to solve them.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

First you need to understand what it is.

There is a simple definition linear equation, which is given in an ordinary school: "an equation in which a variable occurs only in the first degree." But it is not entirely true: the equation is not linear, it is not even reduced to such, it is reduced to quadratic.

A more precise definition is: linear equation is an equation that equivalent transformations can be reduced to the form where title="a,b in bbR, ~a0">. На деле мы будем приводить это уравнение к виду путём переноса в правую часть и деления обеих частей уравнения на . Осталось разъяснить, какие уравнения и как мы можем привести к такому виду, и, самое главное, что дальше делать с ними, чтобы решить его.!}

In fact, in order to understand whether an equation is linear or not, it must first be simplified, that is, brought to a form where its classification will be unambiguous. Remember, you can do anything with the equation that does not change its roots - this is equivalent transformation. Of the simplest equivalent transformations, we can distinguish:

1. parenthesis expansion
2. bringing similar
3. multiplication and/or division of both sides of the equation by a non-zero number
4. addition and/or subtraction from both parts of the same number or expression*

You can do these transformations painlessly, without thinking about whether you "spoil" the equation or not.
*A particular interpretation of the last transformation is the "transfer" of terms from one part to another with a change of sign.

Example 1:
(open brackets)
(add to both parts and subtract / transfer with a change of sign of the number to the left, and variables to the right)
(Give similar ones)
(divide by 3 both sides of the equation)

So we got an equation that has the same roots as the original one. We remind the reader that "solve equation" means to find all its roots and prove that there are no others, and "root of the equation"- this is a number that, when substituted for the unknown, will turn the equation into a true equality. Well, in the last equation, finding a number that turns the equation into the correct equality is very simple - this is the number. No other number will make this equation an identity. Answer:

Example 2:
(multiply both sides of the equation by , making sure we don't multiply by : title="x3/2"> и title="x3">. То есть если такие корни получатся, то мы их обязаны будем выкинуть.)!}
(open brackets)
(move terms)
(Give similar ones)
(divide both parts by )

This is how all linear equations are solved. For younger readers, most likely, this explanation seemed complicated, so we offer the version "linear equations for grade 5"

When solving linear equations, we strive to find a root, that is, a value for a variable that will turn the equation into a correct equality.

To find the root of the equation you need equivalent transformations bring the equation given to us to the form

\(x=[number]\)

This number will be the root.

That is, we transform the equation, making it easier with each step, until we reduce it to a completely primitive equation “x = number”, where the root is obvious. The most commonly used in solving linear equations are the following transformations:

For example: add \(5\) to both sides of the equation \(6x-5=1\)

\(6x-5=1\) \(|+5\)
\(6x-5+5=1+5\)
\(6x=6\)

Please note that we could get the same result faster - simply by writing the five on the other side of the equation and changing its sign in the process. Actually, this is exactly how the school “transfer through equals with a change of sign to the opposite” is done.

2. Multiplying or dividing both sides of an equation by the same number or expression.

For example: Divide the equation \(-2x=8\) by minus two

\(-2x=8\) \(|:(-2)\)
\(x=-4\)

Usually this step is done at the very end, when the equation has already been reduced to \(ax=b\), and we divide by \(a\) to remove it from the left.

3. Using the properties and laws of mathematics: opening brackets, reducing like terms, reducing fractions, etc.

Add \(2x\) left and right

Subtract \(24\) from both sides of the equation

Again, we present like terms

Now we divide the equation by \ (-3 \), thereby removing before the x on the left side.

Answer : \(7\)

Answer found. However, let's check it out. If the seven is really a root, then substituting it instead of x in the original equation should result in the correct equality - the same numbers on the left and right. We try.

Examination:
\(6(4-7)+7=3-2\cdot7\)
\(6\cdot(-3)+7=3-14\)
\(-18+7=-11\)
\(-11=-11\)

Agreed. This means that the seven is indeed the root of the original linear equation.

Do not be lazy to check the answers you found by substitution, especially if you are solving an equation on a test or exam.

The question remains - how to determine what to do with the equation at the next step? How exactly to convert it? Share something? Or subtract? And what exactly to subtract? What to share?

The answer is simple:

Your goal is to bring the equation to the form \(x=[number]\), that is, on the left x without coefficients and numbers, and on the right - only a number without variables. So see what's stopping you and do the opposite of what the interfering component does.

To understand this better, let's take a step-by-step solution to the linear equation \(x+3=13-4x\).

Let's think: how does this equation differ from \(x=[number]\)? What's stopping us? What's wrong?

Well, firstly, the triple interferes, since there should be only a lone X on the left, without numbers. And what does the trio do? Added to xx. So, to remove it - subtract the same trio. But if we subtract a triple from the left, then we must subtract it from the right so that the equality is not violated.

\(x+3=13-4x\) \(|-3\)
\(x+3-3=13-4x-3\)
\(x=10-4x\)

Fine. Now what's stopping you? \(4x\) on the right, because it should only contain numbers. \(4x\) subtracted- remove adding.

\(x=10-4x\) \(|+4x\)
\(x+4x=10-4x+4x\)

Now we give like terms on the left and right.

It's almost ready. It remains to remove the five on the left. What is she doing"? multiplies on x. So we remove it division.

\(5x=10\) \(|:5\)
\(\frac(5x)(5)\) \(=\)\(\frac(10)(5)\)
\(x=2\)

The solution is complete, the root of the equation is two. You can check by substitution.

notice, that most often there is only one root in linear equations. However, two special cases may occur.

Special case 1 - there are no roots in a linear equation.

Example . Solve the equation \(3x-1=2(x+3)+x\)

Solution :

Answer : no roots.

In fact, the fact that we will come to such a result was seen earlier, even when we got \(3x-1=3x+6\). Think about it: how can \(3x\) be equal, from which \(1\) was subtracted, and \(3x\) to which \(6\) was added? Obviously, no way, because they did the same thing different actions! It is clear that the results will differ.

Special case 2 - a linear equation has an infinite number of roots.

Example . Solve the linear equation \(8(x+2)-4=12x-4(x-3)\)

Solution :

Answer : any number.

By the way, this was noticeable even earlier, at the stage: \(8x+12=8x+12\). Indeed, left and right are the same expressions. Whatever x you substitute, there will be the same number both there and there.

More complex linear equations.

The original equation does not always immediately look like a linear one, sometimes it is “disguised” as other, more complex equations. However, in the process of transformation, the masking subsides.

Example . Find the root of the equation \(2x^(2)-(x-4)^(2)=(3+x)^(2)-15\)

Solution :

\(2x^(2)-(x-4)^(2)=(3+x)^(2)-15\)		It would seem that there is an x ​​squared here - this is not a linear equation! But don't rush. Let's Apply
\(2x^(2)-(x^(2)-8x+16)=9+6x+x^(2)-15\)		Why is the result of expansion \((x-4)^(2)\) in parentheses, but the result of \((3+x)^(2)\) is not? Because there is a minus before the first square, which will change all the signs. And in order not to forget about it, we take the result in brackets, which we now open.
\(2x^(2)-x^(2)+8x-16=9+6x+x^(2)-15\)		We give like terms
\(x^(2)+8x-16=x^(2)+6x-6\)
\(x^(2)-x^(2)+8x-6x=-6+16\)		Again, here are similar ones.
		Like this. It turns out that the original equation is quite linear, and x squared is nothing more than a screen to confuse us. :) We complete the solution by dividing the equation by \(2\), and we get the answer.

Answer : \(x=5\)

Example . Solve the linear equation \(\frac(x+2)(2)\) \(-\) \(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6 )\)

Solution :

\(\frac(x+2)(2)\) \(-\) \(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6)\)		The equation does not look like a linear one, some fractions ... However, let's get rid of the denominators by multiplying both parts of the equation by the common denominator of all - six
\(6\cdot\)\((\frac(x+2)(2)\) \(-\) \(\frac(1)(3))\) \(=\) \(\frac( 9+7x)(6)\)\(\cdot 6\)		Open bracket on the left
\(6\cdot\)\(\frac(x+2)(2)\) \(-\) \(6\cdot\)\(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6)\) \(\cdot 6\)		Now we reduce the denominators
\(3(x+2)-2=9+7x\)		Now it looks like a regular linear one! Let's solve it.
		By transferring through equals, we collect x's on the right, and numbers on the left
		Well, dividing by \ (-4 \) the right and left parts, we get the answer

Answer : \(x=-1.25\)