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

First, let's define: what is a linear equation and which one is called the simplest?

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

The simplest equation means the construction:

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

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Lead similar 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 fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. 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.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing 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 parentheses, if any (as in our last example);
2. Then combine similar
3. Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.

Then, as a rule, you need to give similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “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. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.

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

Scheme for solving simple linear equations

First, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the brackets, if any.
2. We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
3. We present similar terms.
4. We divide everything by the coefficient of “x”.

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

Solving real examples of simple linear equations

Task No. 1

The first step requires us 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. Note: we're talking about only about individual terms. Let's write it down:

We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:

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

So we got the answer.

Task No. 2

We can see the parentheses in this problem, so let's expand them:

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

Here are some similar ones:

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

Task No. 3

The third linear equation is more interesting:

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

There are several parentheses, but they are not multiplied by anything, they are simply preceded by various signs. Let's break them down:

We perform the second step already known to us:

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

Let's do the math:

We carry out the last step - divide everything by the coefficient of “x”:

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

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, 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, there may be zero among them - there is nothing wrong with that.

Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.

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

Understanding this simple fact will allow you to avoid making stupid and offensive 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 complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.

Example No. 1

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

Now let's take a look at privacy:

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

Here are some similar ones:

Obviously, this equation has no solutions, so we’ll write this in the answer:

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

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

Here are some similar ones:

Obviously, this linear equation has no solution, so we’ll write it this way:

\[\varnothing\],

or there are no roots.

Nuances of the solution

Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.

But I would like to draw your attention to another fact: how to work with parentheses and how to open 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: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply 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 not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence elementary transformations, where the inability to clearly and competently perform simple steps leads to the fact that high school students come to me and again learn to solve such simple equations.

Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on 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 No. 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 some privacy:

Here are some similar ones:

Let's complete 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, they canceled each other out, which makes the equation linear and not quadratic.

Task No. 2

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

Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:

Now let’s carefully perform the multiplication in each term:

Let’s move the terms with “X” to the left, and those without - to the right:

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

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, 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 will have four terms.

About the algebraic sum

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

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

Solving equations with fractions

To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:

1. Open the brackets.
2. Separate variables.
3. Bring similar ones.
4. Divide by the ratio.

Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and 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 done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:

1. Get rid of fractions.
2. Open the brackets.
3. Separate variables.
4. Bring similar ones.
5. Divide by the ratio.

What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.

Example No. 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 parentheses doesn't mean you have to multiply each one by "four." Let's write down:

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

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

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

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

We got final decision, let's move on to the second equation.

Example No. 2

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

Here we perform all the same actions:

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

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

The problem is solved.

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

Key points

Key findings are:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Don't worry if you see quadratic functions, most likely, in the process of further transformations they will decrease.
• There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and 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 and solve the examples presented there. Stay tuned, many more interesting things await you!

With this mathematical program you can solve a system of two linear equations with two variable method substitution and addition method.

The program not only gives the answer to the problem, but also gives detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.

This program may be useful for high school students in secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.

This way you can spend your own training and/or training their younger brothers or sisters, while the level of education in the field of problems being solved increases.

Rules for entering equations

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.

When entering equations you can use parentheses. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of elements.
For example: 6x+1 = 5(x+y)+2

You can use not only integers in equations, but also fractional numbers in the form of decimals and ordinary fractions.

Rules for entering decimal fractions.
Integer and fractional parts in decimals can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering numerical fraction The numerator is separated from the denominator by a division sign: /
Whole part separated from the fraction by an ampersand: &

Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)

Solve system of equations

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A little theory.

Solving systems of linear equations. Substitution method

The sequence of actions when solving a system of linear equations using the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression into another equation of the system instead of this variable;

$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$

Let's express y in terms of x from the first equation: y = 7-3x. Substituting the expression 7-3x into the second equation instead of y, we obtain the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$

It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$

Substituting the number 1 instead of x into the equality y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$

Pair (1;4) - solution of the system

Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.

Solving systems of linear equations by addition

Let's consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by substitution, we move from this system to another, equivalent system, in which one of the equations contains only one variable.

The sequence of actions when solving a system of linear equations using the addition method:
1) multiply the equations of the system term by term, selecting factors so that the coefficients for one of the variables become opposite numbers;
2) add the left and right sides of the system equations term by term;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.

Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$

In the equations of this system, the coefficients of y are opposite numbers. By adding the left and right sides of the equations term by term, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$

From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38\) we get an equation with the variable y: \(11-3y=38\). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)

Thus, we found the solution to the system of equations by addition: \(x=11; y=-9\) or \((11;-9)\)

Taking advantage of the fact that in the equations of the system the coefficients for y are opposite numbers, we reduced its solution to the solution equivalent system(by summing both sides of each of the equations of the original symbol), in which one of the equations contains only one variable.

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Linear equations are a fairly harmless and understandable topic. school mathematics. But, oddly enough, the number of errors out of the blue when solving linear equations is only slightly less than in other topics - quadratic equations, logarithms, trigonometry and others. The causes of most errors are banal identical transformations of equations. First of all, this is confusion in signs when transferring terms from one part of the equation to another, as well as errors when working with fractions and fractional odds. Yes Yes! Fractions also appear in linear equations! All around. Below we will definitely analyze such evil equations.)

Well, let’s not pull the cat by the tail and let’s start figuring it out, shall we? Then we read and delve into it.)

What is a linear equation? Examples.

Typically the linear equation looks like this:

ax + b = 0,

Where a and b are any numbers. Any kind: integers, fractions, negative, irrational - there can be any!

For example:

7x + 1 = 0 (here a = 7, b = 1)

x – 3 = 0 (here a = 1, b = -3)

x/2 – 1.1 = 0 (here a = 1/2, b = -1.1)

In general, you understand, I hope.) Everything is simple, like in a fairy tale. For the time being... And if you look closely at general record ax+b=0 take a closer look and think a little? After all, a and b are any numbers! And if we have, say, a = 0 and b = 0 (any numbers can be taken!), then what will we get?

0 = 0

But that's not all the fun! What if, say, a = 0, b = -10? Then it turns out to be some kind of nonsense:

0 = 10.

Which is very, very annoying and undermines the trust in mathematics that we have gained through sweat and blood... Especially during tests and exams. But out of these incomprehensible and strange equalities, you also need to find X! Which doesn’t exist at all! And here, even well-prepared students can sometimes fall into what is called a stupor... But don’t worry! IN this lesson We will also consider all such surprises. And we will also definitely find an X from such equalities.) Moreover, this same X can be found very, very simply. Yes Yes! Surprising but true.)

Okay, that's understandable. But how can you tell by the appearance of the task that it is a linear equation and not some other equation? Unfortunately, it is not always possible to recognize the type of equation just by appearance. The point is that not only equations of the form ax + b = 0 are called linear, but also any other equations that, in one way or another, can be reduced to this form by identical transformations. How do you know if it adds up or not? Until you can hardly solve the example - almost not at all. This is upsetting. But for some types of equations, you can immediately tell with confidence whether it is linear or not with one quick glance.

To do this, let us turn once again to general structure any linear equation:

ax + b = 0

Please note: in the linear equation Always only variable x is present in the first degree and some numbers! That's all! Nothing else. At the same time, there are no X’s in the square, in the cube, under the root, under the logarithm and other exotic things. And (most importantly!) there are no fractions with X in the denominators! But fractions with numbers in the denominators or division per number- easily!

For example:

This is a linear equation. The equation contains only X's to the first power and numbers. And there are no more X's high degrees- squared, cubed, and so on. Yes, there are fractions here, but at the same time the denominators of the fractions contain only numbers. Namely, two and three. In other words, there is no division by x.

And here is the equation

It can no longer be called linear, although here, too, there are only numbers and X’s to the first power. Because, among other things, there are also fractions with X's in the denominators. And after simplifications and transformations, such an equation can become anything: linear, quadratic - anything.

How to solve linear equations? Examples.

So how do you solve linear equations? Read on and be surprised.) The entire solution of linear equations is based on just two main things. Let's list them.

1) A set of elementary actions and rules of mathematics.

These are using parentheses, opening parentheses, working with fractions, working with negative numbers, multiplication tables, and so on. This knowledge and skills are necessary not only for solving linear equations, but for all mathematics in general. And if you have problems with this, remember junior classes. Otherwise you will have a hard time...

2)

There are only two of them. Yes Yes! Moreover, these very basic identity transformations underlie the solution of not only linear, but generally any mathematical equations! In a word, the solution to any other equation - quadratic, logarithmic, trigonometric, irrational, etc. – as a rule, it begins with these very basic transformations. But the solution of linear equations, in fact, ends with them (transformations). Ready answer.) So don’t be lazy and take a look at the link.) Moreover, linear equations are also analyzed in detail there.

Well, I think it's time to start looking at examples.

To begin with, as a warm-up, let's look at some basic stuff. Without any fractions or other bells and whistles. For example, this equation:

x – 2 = 4 – 5x

This is a classic linear equation. All X's are at most in the first power and there is no division by X anywhere. The solution scheme in such equations is always the same and terribly simple: all terms with X's must be collected on the left, and all terms without X's (i.e. numbers) must be collected on the right. So let's start collecting.

To do this, we launch the first identity transformation. We need to move -5x to the left, and move -2 to the right. With a change of sign, of course.) So we transfer:

x + 5x = 4 + 2

Here you go. Half the battle is done: the X's have been collected into a pile, and so have the numbers. Now we present similar ones on the left, and we count them on the right. We get:

6x = 6

What do we need now? complete happiness? Yes, so that the pure X remains on the left! And the six gets in the way. How to get rid of it? Now we run the second identity transformation - divide both sides of the equation by 6. And - voila! The answer is ready.)

x = 1

Of course, the example is completely primitive. To general idea catch. Well, let's decide something more significant. For example, let's look at this equation:

Let's look at it in detail.) This is also a linear equation, although it would seem that there are fractions here. But in fractions there is division by two and there is division by three, but there is no division by an expression with an X! So let's decide. Using the same identical transformations, yes.)

What should we do first? With X's - to the left, without X's - to the right? In principle, this is possible. Fly to Sochi via Vladivostok.) Or you can take the shortest route, immediately using a universal and powerful method. If you know the identity transformations, of course.)

To begin with I ask key question: What stands out and dislikes most about this equation? 99 out of 100 people will say: fractions! And they will be right.) So let’s get rid of them first. Safe for the equation itself.) Therefore, let's start right away with second identity transformation- from multiplication. What should we multiply the left side by so that the denominator is successfully reduced? That's right, a two. What about the right side? For three! But... Mathematics is a capricious lady. She, you see, requires multiplying both sides only for the same number! Multiplying each part by its own number doesn’t work... What are we going to do? Something... Look for a compromise. So that we can satisfy our desires (to get rid of fractions) and not offend mathematics.) Let’s multiply both parts by six!) That is, by common denominator all fractions included in the equation. Then in one fell swoop both the two and the three will be reduced!)

So let's multiply. The entire left side and the entire right side! Therefore, we use parentheses. This is what the procedure itself looks like:

Now we open these same brackets:

Now, representing 6 as 6/1, let's multiply six by each of the fractions on the left and right. This is the usual multiplication of fractions, but so be it, I’ll describe it in detail:

And here - attention! I put the numerator (x-3) in brackets! This is all because when multiplying fractions, the numerator is multiplied entirely, entirely! And the x-3 expression must be worked as one integral structure. But if you write the numerator like this:

6x – 3,

But we have everything right and we need to finalize it. What to do next? Open the parentheses in the numerator on the left? In no case! You and I multiplied both sides by 6 to get rid of fractions, and not to worry about opening parentheses. On at this stage we need reduce our fractions. With a feeling of deep satisfaction, we reduce all the denominators and get an equation without any fractions, in a ruler:

3(x-3) + 6x = 30 – 4x

And now the remaining brackets can be opened:

3x – 9 + 6x = 30 – 4x

The equation keeps getting better and better! Now let’s remember again about the first identical transformation. With a straight face we repeat the spell from junior classes: with X - to the left, without X - to the right. And apply this transformation:

3x + 6x + 4x = 30 + 9

We present similar ones on the left and count on the right:

13x = 39

It remains to divide both parts by 13. That is, apply the second transformation again. We divide and get the answer:

x = 3

The job is done. As you can see, in given equation we had to apply the first transformation once (transfer of terms) and the second twice: at the beginning of the solution we used multiplication (by 6) in order to get rid of fractions, and at the end of the solution we used division (by 13) to get rid of the coefficient in front of the X. And the solution to any (yes, any!) linear equation consists of a combination of these same transformations in one sequence or another. Where exactly to start depends on the specific equation. In some places it is more profitable to start with transfer, and in others (as in this example) with multiplication (or division).

We work from simple to complex. Let's now consider outright cruelty. With a bunch of fractions and parentheses. And I’ll tell you how not to overstrain yourself.)

For example, here's the equation:

We look at the equation for a minute, are horrified, but still pull ourselves together! The main problem is where to start? You can add fractions on the right side. You can subtract fractions in parentheses. You can multiply both parts by something. Or divide... So what is still possible? Answer: everything is possible! Mathematics does not prohibit any of the listed actions. And no matter what sequence of actions and transformations you choose, the answer will always be the same - the correct one. Unless, of course, at some step you violate the identity of your transformations and, thereby, make mistakes...

And, in order not to make mistakes, in such sophisticated examples as this one, it is always most useful to evaluate it appearance and figure out in your mind: what can be done in the example so that maximum simplify it in one step?

So let's figure it out. On the left are sixes in the denominators. Personally, I don't like them, and they are very easy to remove. Let me multiply both sides of the equation by 6! Then the sixes on the left will be successfully reduced, the fractions in brackets will not go anywhere yet. Well, that's okay. We'll deal with them a little later.) But on the right, we have the denominators 2 and 3 cancelling. It is with this action (multiplying by 6) that we achieve maximum simplifications in one step!

After multiplication, our whole evil equation becomes like this:

If you don’t understand exactly how this equation came about, then you haven’t understood the analysis of the previous example well. And I tried, by the way...

So, let's reveal:

Now the most logical step would be to isolate the fractions on the left, and send 5x to the right side. At the same time, we will present similar ones on the right side. We get:

Much better already. Now the left side has prepared itself for multiplication. What should we multiply the left side by so that both the five and the four are reduced at once? On 20! But we also have disadvantages on both sides of the equation. Therefore, it will be most convenient to multiply both sides of the equation not by 20, but by -20. Then in one fell swoop both the minuses and the fractions will disappear.

So we multiply:

Anyone who still doesn’t understand this step means that the problem is not in the equations. The problems are in the basics! Let's remember again Golden Rule opening brackets:

If a number is multiplied by some expression in brackets, then this number must be sequentially multiplied by each term of this very expression. Moreover, if the number is positive, then the signs of the expressions are preserved after expansion. If negative, change to the opposite:

a(b+c) = ab+ac

-a(b+c) = -ab-ac

Our cons disappeared after multiplying both sides by -20. And now we multiply the brackets with fractions on the left by quite positive number 20. Therefore, when these brackets are opened, all the signs that were inside them are preserved. But where the brackets in the numerators of fractions come from, I already explained in detail in the previous example.

Now you can reduce fractions:

4(3-5x)-5(3x-2) = 20

Open the remaining brackets. Again, we reveal it correctly. The first brackets are multiplied by the positive number 4 and, therefore, all signs are preserved when they are opened. But the second brackets are multiplied by negative the number is -5 and, therefore, all signs are reversed:

12 - 20x - 15x + 10 = 20

There are mere trifles left. With X's to the left, without X's to the right:

-20x – 15x = 20 – 10 – 12

-35x = -2

That's almost all. On the left you need a pure X, but the number -35 is in the way. So we divide both sides by (-35). Let me remind you that the second identity transformation allows us to multiply and divide both sides by whatever number. Including negative ones.) As long as it’s not zero! Feel free to divide and get the answer:

X = 2/35

This time the X turned out to be fractional. It's OK. Such an example.)

As we can see, the principle of solving linear equations (even the most complicated ones) is quite simple: we take the original equation and, using identical transformations, successively simplify it until we get the answer. With the basics, of course! The main problems here are precisely the failure to follow the basics (for example, there is a minus in front of the brackets, and they forgot to change the signs when expanding), as well as in banal arithmetic. So don't neglect the basics! They are the foundation of all other mathematics!

Some fun things to do when solving linear equations. Or special occasions.

Everything would be fine. However... Among the linear equations there are also such funny pearls that in the process of solving them can drive you into a strong stupor. Even an excellent student.)

For example, here’s an innocuous-looking equation:

7x + 3 = 4x + 5 + 3x - 2

Yawning widely and slightly bored, we collect all the X’s on the left and all the numbers on the right:

7x-4x-3x = 5-2-3

We present similar ones, count and get:

0 = 0

That's it! I gave a sample trick! This equality in itself does not raise any objections: zero really is equal to zero. But X is missing! Without a trace! And we must write down in the answer, why equal to x . Otherwise, the decision does not count, yes.) What to do?

Don't panic! In such non-standard cases, the most general concepts and principles of mathematics. What is an equation? How to solve equations? What does it mean to solve an equation?

Solving an equation means finding All values ​​of the variable x, which, when substituted into original equation will give us the correct equality (identity)!

But we have true equality it's already happened! 0=0, or rather, nowhere!) We can only guess at what X's we get this equality. What kind of X's can be substituted in original equation, if upon substitution all of them will they still be reduced to zero? Haven't you figured it out yet?

Surely! X's can be substituted any!!! Absolutely any. Submit whatever you want. At least 1, at least -23, at least 2.7 - whatever! They will still be reduced and as a result, the pure truth will remain. Try it, substitute it and see for yourself.)

Here's your answer:

x – any number.

In scientific notation this equality is written as follows:

This entry reads like this: “X is any real number.”

Or in another form, at intervals:

Design it the way you like best. This is a correct and completely complete answer!

Now I'm going to change just one number in our original equation. Now let’s solve this equation:

7x + 2 = 4x + 5 + 3x – 2

Again we transfer the terms, count and get:

7x – 4x – 3x = 5 – 2 – 2

0 = 1

And what do you think of this joke? There was an ordinary linear equation, but it became an incomprehensible equality

0 = 1…

Speaking scientific language, we got false equality. But in Russian this is not true. Bullshit. Nonsense.) Because zero is in no way equal to one!

And now let’s figure out again what kind of X’s, when substituted into the original equation, will give us true equality? Which? But none! No matter what X you substitute, everything will still be shortened and everything will remain crap.)

Here is the answer: no solutions.

In mathematical notation, this answer is written like this:

It reads: “X belongs to the empty set.”

Such answers in mathematics also occur quite often: not always do any equations have roots in principle. Some equations may not have roots at all. At all.

Here are two surprises. I hope that now the sudden disappearance of X's from the equation will not leave you perplexed forever. This is quite familiar.)

And then I hear a logical question: will they be in the OGE or the Unified State Exam? On the Unified State Examination in itself as a task - no. Too simple. But in the OGE or in word problems - easily! So now let’s train and decide:

Answers (in disarray): -2; -1; any number; 2; no solutions; 7/13.

Everything worked out? Great! You have a good chance in the exam.

Does something not add up? Hm... Sadness, of course. This means there are still gaps somewhere. Either in the basics or identity transformations. Or it’s just a matter of simple inattention. Read the lesson again. Because this is not a topic that can be so easily dispensed with in mathematics...

Good luck! She will definitely smile at you, believe me!)

Linear equation is algebraic equation, the total degree of which polynomials is equal to one. Solving linear equations - part school curriculum, and not the most difficult. However, some still have difficulty completing this topic. We hope after reading this material, all difficulties for you will be a thing of the past. So, let's figure it out. how to solve linear equations.

General form

The linear equation is represented as:

• ax + b = 0, where a and b are any numbers.

Although a and b can be any number, their values ​​affect the number of solutions to the equation. There are several special cases of solution:

• If a=b=0, the equation has infinite set decisions;
• If a=0, b≠0, the equation has no solution;
• If a≠0, b=0, the equation has a solution: x = 0.

In the event that both numbers do not have zero values, the equation must be solved to derive the final expression for the variable.

How to decide?

Solving a linear equation means finding what the variable is equal to. How to do this? Yes, it’s very simple - using simple algebraic operations and following the rules of transfer. If the equation appears in front of you in general form, you are in luck; all you need to do is:

1. Move b to the right side of the equation, not forgetting to change the sign (transfer rule!), thus, from an expression of the form ax + b = 0, an expression of the form should be obtained: ax = -b.
2. Apply the rule: to find one of the factors (x - in our case), you need to divide the product (-b in our case) by another factor (a - in our case). Thus, you should get an expression of the form: x = -b/a.

That's it - a solution has been found!

Now let's look at a specific example:

1. 2x + 4 = 0 - move b equal to in this case 4, to the right
2. 2x = -4 - divide b by a (don’t forget about the minus sign)
3. x = -4/2 = -2

That's all! Our solution: x = -2.

As you can see, the solution to a linear equation with one variable is quite simple to find, but everything is so simple if we are lucky enough to come across the equation in its general form. In most cases, before solving an equation in the two steps described above, you still need to bring the existing expression to a general form. However, this is also not an extremely difficult task. Let's look at some special cases using examples.

Solving special cases

First, let's look at the cases that we described at the beginning of the article and explain what it means to have an infinite number of solutions and no solution.

• If a=b=0, the equation will look like: 0x + 0 = 0. Performing the first step, we get: 0x = 0. What does this nonsense mean, you exclaim! After all, no matter what number you multiply by zero, you always get zero! Right! That's why they say that the equation has an infinite number of solutions - no matter what number you take, the equality will be true, 0x = 0 or 0 = 0.
• If a=0, b≠0, the equation will look like: 0x + 3 = 0. Perform the first step, we get 0x = -3. Nonsense again! It is obvious that this equality will never be true! That's why they say that the equation has no solutions.
• If a≠0, b=0, the equation will look like: 3x + 0 = 0. Performing the first step, we get: 3x = 0. What is the solution? It's easy, x = 0.

Lost in translation

The described special cases are not all that linear equations can surprise us with. Sometimes the equation is difficult to identify at first glance. Let's look at an example:

• 12x - 14 = 2x + 6

Is this a linear equation? What about the zero on the right side? Let's not rush to conclusions, let's act - let's transfer all the components of our equation into left side. We get:

• 12x - 2x - 14 - 6 = 0

Now subtract like from like, we get:

• 10x - 20 = 0

Learned? The most linear equation ever! The solution to which is: x = 20/10 = 2.

What if we have this example:

• 12((x + 2)/3) + x) = 12 (1 - 3x/4)

Yes, this is also a linear equation, only more transformations need to be carried out. First, let's open the brackets:

1. (12(x+2)/3) + 12x = 12 - 36x/4
2. 4(x+2) + 12x = 12 - 36x/4
3. 4x + 8 + 12x = 12 - 9x - now we carry out the transfer:
4. 25x - 4 = 0 - it remains to find a solution using the already known scheme:
5. 25x = 4,
6. x = 4/25 = 0.16

As you can see, everything can be solved, the main thing is not to worry, but to act. Remember, if your equation contains only variables of the first degree and numbers, you have a linear equation, which, no matter how it looks initially, can be reduced to a general form and solved. We hope everything works out for you! Good luck!

Systems of equations are widely used in economic sector at mathematical modeling various processes. For example, when solving problems of production management and planning, logistics routes ( transport problem) or equipment placement.

Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.

A system of linear equations is two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.

Linear equation

Equations of the form ax+by=c are called linear. The designations x, y are the unknowns whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.

Types of systems of linear equations

The simplest examples are considered to be systems of linear equations with two variables X and Y.

F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.

Solve system of equations - this means finding values ​​(x, y) at which the system turns into a true equality or establishing that suitable values ​​of x and y do not exist.

A pair of values ​​(x, y), written as the coordinates of a point, is called a solution to a system of linear equations.

If systems have one common solution or no solution exists, they are called equivalent.

Homogeneous systems of linear equations are systems right part which is equal to zero. If the right part after the equal sign has a value or is expressed by a function, such a system is heterogeneous.

The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three or more variables.

When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.

Simple and complex methods for solving systems of equations

There is no general analytical way to solve similar systems, all methods are based on numerical solutions. IN school course mathematics, such methods as permutation, algebraic addition, substitution, as well as graphical and matrix method, solution by Gaussian method.

The main task when teaching solution methods is to teach how to correctly analyze the system and find optimal algorithm solutions for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of using a particular method

Solving examples of systems of linear equations of the 7th grade program secondary school quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.

Solving systems using the substitution method

The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system

Let us give a solution to an example of a system of linear equations of class 7 using the substitution method:

As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Solution this example does not cause difficulties and allows you to obtain the Y value. The last step is to check the obtained values.

It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, solving by substitution is also inappropriate.

Solution of an example of a system of linear inhomogeneous equations:

Solution using algebraic addition

When searching for solutions to systems using the addition method, they perform term-by-term addition and multiplication of equations by different numbers. The ultimate goal of mathematical operations is an equation in one variable.

For Applications this method practice and observation are required. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.

Solution algorithm:

1. Multiply both sides of the equation by a certain number. As a result arithmetic action one of the coefficients of the variable must become equal to 1.
2. Add the resulting expression term by term and find one of the unknowns.
3. Substitute the resulting value into the 2nd equation of the system to find the remaining variable.

Method of solution by introducing a new variable

A new variable can be introduced if the system requires finding a solution for no more than two equations; the number of unknowns should also be no more than two.

The method is used to simplify one of the equations by introducing a new variable. The new equation is solved for the introduced unknown, and the resulting value is used to determine the original variable.

The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to the standard one quadratic trinomial. You can solve a polynomial by finding the discriminant.

It is necessary to find the discriminant value by well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. IN given example a=1, b=16, c=39, therefore D=100. If the discriminant Above zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant is less than zero, then there is one solution: x = -b / 2*a.

The solution for the resulting systems is found by the addition method.

Visual method for solving systems

Suitable for 3 equation systems. The method is to build on coordinate axis graphs of each equation included in the system. The coordinates of the points of intersection of the curves and will be general decision systems.

The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.

As can be seen from the example, for each line two points were constructed, the values ​​of the variable x were chosen arbitrarily: 0 and 3. Based on the values ​​of x, the values ​​for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.

The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.

The following example requires finding graphic solution systems of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.

As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.

The systems from examples 2 and 3 are similar, but when constructed it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether a system has a solution or not; it is always necessary to construct a graph.

The matrix and its varieties

Matrices are used for short note systems of linear equations. A matrix is ​​a table special type filled with numbers. n*m has n - rows and m - columns.

A matrix is ​​square when the number of columns and rows are equal. A matrix-vector is a matrix of one column with infinite possible number lines. A matrix with ones along one of the diagonals and other zero elements is called identity.

An inverse matrix is ​​a matrix when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.

Rules for converting a system of equations into a matrix

In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.

A matrix row is said to be nonzero if at least one element of the row is not zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.

The matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all elements of the matrix are sequentially multiplied by a number.

Options for finding the inverse matrix

The formula for finding the inverse matrix is ​​quite simple: K -1 = 1 / |K|, where K -1 - inverse matrix, and |K| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements by each other. For the “three by three” option, there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the numbers of columns and rows of elements are not repeated in the work.

Solving examples of systems of linear equations using the matrix method

The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with big amount variables and equations.

In the example, a nm are the coefficients of the equations, the matrix is ​​a vector x n are variables, and b n are free terms.

Solving systems using the Gaussian method

IN higher mathematics The Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variable systems with a large number of linear equations.

Gauss's method is very similar to solutions using substitutions and algebraic addition, but more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce the system to the form of an inverted trapezoid. By means of algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, while 3 and 4 are, respectively, with 3 and 4 variables.

After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.

In school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:

As can be seen from the example, at step (3) two equations were obtained: 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. Solving any of the equations will allow you to find out one of the variables x n.

Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.

The Gaussian method is difficult for students to understand high school, but is one of the most interesting ways to develop the ingenuity of children studying under the program in-depth study in math and physics classes.

For ease of recording, calculations are usually done as follows:

The coefficients of the equations and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right. Roman numerals indicate the numbers of equations in the system.

First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is ​​written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.

The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is ​​reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.

This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.

The free use of any solution method will require care and some experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of ​​human activity, while others exist for educational purposes.