How to solve linear equations correctly. Things to Remember When Solving Linear Equations

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.

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. Give 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 expand the parentheses, if there are 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 bring 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. Please note: we are talking 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 brackets here, but they are not multiplied by anything, they are simply preceded by different 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 help you avoid making stupid and hurtful mistakes in high school, when doing such things 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 necessarily 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 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 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 an algebraic sum is. 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!

Linear equations. Solution, examples.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

Linear equations.

Linear equations- not the most difficult topic in school mathematics. But there are some tricks there that can puzzle even a trained student. Let's figure it out?)

Typically a linear equation is 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 don’t notice the words: "where a and b are any numbers"... And if you notice and 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, This turns out to be something completely absurd:

Which is annoying and undermines confidence in mathematics, yes...) Especially during exams. But out 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 to do this. In this lesson.

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

A linear equation can be clearly recognized in some cases. Let's say, if we have an equation in which there are only unknowns to the first degree and numbers. And in the equation there is no fractions divided by unknown , it is important! And division by number, or a numerical fraction - that's welcome! For example:

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

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

It turns out that it is impossible to recognize the linear equation in some complicated example until you almost solve it. This is upsetting. But in assignments, as a rule, they don’t ask about the form of the equation, right? The assignments ask for equations decide. This makes me happy.)

Solving linear equations. Examples.

The entire solution of linear equations consists of identical transformations of the equations. By the way, these transformations (two of them!) are the basis of the solutions all equations of mathematics. In other words, the solution any the equation begins with these very transformations. In the case of linear equations, it (the solution) is based on these transformations and ends with a full answer. It makes sense to follow the link, right?) Moreover, there are also examples of solving linear equations there.

First, let's look at the simplest example. Without any pitfalls. Suppose we need to solve this equation.

x - 3 = 2 - 4x

This is a linear equation. The X's are all in the first power, there is no division by X's. But, in fact, it doesn’t matter to us what kind of equation it 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 in left side, with a change of sign, of course, and - 3 - to the right. By the way, this is the first identical transformation of equations. Surprised? This means that you didn’t follow the link, but in vain...) We get:

x + 4x = 2 + 3

Here are similar ones, we consider:

What do we need for complete happiness? Yes, so that there is a pure X on the left! Five is in the way. Getting rid of the five with the help the second identical transformation of equations. Namely, we divide both sides of the equation by 5. We get a ready answer:

An elementary example, of course. This is for warming up.) It’s not very clear why I remembered identical transformations here? OK. Let's take the bull by the horns.) Let's decide something more solid.

For example, here's the equation:

Where do we start? With X's - to the left, without X's - to the right? Could be so. Small steps along a long road. Or you can immediately, universally and in a powerful way. If, of course, you have identical transformations of equations in your arsenal.

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

95 out of 100 people will answer: fractions ! The answer is correct. So let's get rid of them. Therefore, we start immediately with second identity transformation. What do you need to multiply the fraction on the left by so that the denominator is completely reduced? That's right, at 3. And on the right? By 4. But mathematics allows us to multiply both sides by the same number. How can we get out? Let's multiply both sides by 12! Those. to a common denominator. Then both the three and the four will be reduced. Don't 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 put it in brackets! This is because when multiplying fractions, the entire numerator is multiplied! Now you can reduce fractions:

Expand the remaining brackets:

Not an example, but sheer pleasure!) Now let’s remember the spell from junior classes: with an X - to the left, without an X - to the right! And apply this transformation:

Here are some similar ones:

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

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

Please note: to bring the original confusing equation into a nice form, we used two (just two!) identity transformations– translation left-right with a change of sign and multiplication-division of an equation by the same number. This is a universal method! We will work in this way with any equations! Absolutely anyone. That’s why I tediously repeat about 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 using identical transformations until we get the answer. The main problems here are in the calculations, 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 you into a strong stupor...) Fortunately, there can only be two such surprises. Let's call them special cases.

Special cases in solving linear equations.

First surprise.

Suppose you come across a very basic equation, something like:

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

Slightly bored, we move it with an X to the left, without an X - to the right... With a change of sign, everything is perfect... We get:

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

We count, and... oops!!! We get:

This equality in itself is not objectionable. Zero really is zero. But X is missing! And we must write down in the answer, what is x equal to? Otherwise, the solution doesn't count, right...) Dead end?

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

But we have true equality already happened! 0=0, how much more accurate?! It remains to figure out at what x's this happens. What values ​​of X can be substituted into original equation if these x's will they still be reduced to zero? Come on?)

Yes!!! X's can be substituted any! Which ones 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 values ​​of X into original equation and calculate. All the time you will get the pure truth: 0=0, 2=2, -7.1=-7.1 and so on.

Here's your answer: x - any number.

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

Second surprise.

Let's take the same elementary linear equation and change just 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. We solved a linear equation and got a strange equality. Speaking mathematical language, we got false equality. And speaking in simple language, this is not true. Rave. But nevertheless, this nonsense is a very good reason for the correct solution of the equation.)

Again we think based on general rules. What x's, when substituted into the original equation, will give us true equality? Yes, none! There are no such X's. No matter what you put in, everything will be reduced, only nonsense will remain.)

Here's your answer: there are no solutions.

This is also a completely complete answer. In mathematics, such answers are often found.

Like this. Now, I hope, the disappearance of X's in the process of solving any (not just linear) equation will not confuse you at all. This is already a familiar matter.)

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. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

In this lesson we will look at methods for solving a system of linear equations. In a course of higher mathematics, systems of linear equations are required to be solved both in the form of separate tasks, for example, “Solve the system using Cramer’s formulas,” and in the course of solving other problems. Systems of linear equations have to be dealt with in almost all branches of higher mathematics.

First, a little theory. What in in this case stands for the mathematical word "linear"? This means that the equations of the system All variables included in the first degree: without any fancy stuff like etc., which only participants in mathematical Olympiads are delighted with.

In higher mathematics, not only letters familiar from childhood are used to denote variables.
A fairly popular option is variables with indexes: .
Or initial letters Latin alphabet, small and large:
It is not so rare to find Greek letters: – known to many as “alpha, beta, gamma”. And also a set with indices, say, with the letter “mu”:

The use of one or another set of letters depends on the section of higher mathematics in which we are faced with a system of linear equations. So, for example, in systems of linear equations encountered when solving integrals, differential equations It is traditional to use the notation

But no matter how the variables are designated, the principles, methods and methods for solving a system of linear equations do not change. Thus, if you come across something scary like , do not rush to close the problem book in fear, after all, you can draw the sun instead, a bird instead, and a face (the teacher) instead. And, funny as it may seem, a system of linear equations with these notations can also be solved.

I have a feeling that the article will turn out to be quite long, so a small table of contents. So, the sequential “debriefing” will be like this:

– Solving a system of linear equations using the substitution method (“ school method») ;
– Solving the system by term-by-term addition (subtraction) of the system equations;
– Solution of the system using Cramer’s formulas;
– Solving the system using an inverse matrix;
– Solving the system using the Gaussian method.

Everyone is familiar with systems of linear equations from school mathematics courses. Essentially, we start with repetition.

Solving a system of linear equations using the substitution method

This method can also be called the "school method" or the method of eliminating unknowns. Figuratively speaking, it can also be called “an unfinished Gaussian method.”

Example 1

Here we are given a system of two equations with two unknowns. Note that the free terms (numbers 5 and 7) are located on the left side of the equation. Generally speaking, it doesn’t matter where they are, on the left or on the right, it’s just that in problems in higher mathematics they are often located that way. And such a recording should not lead to confusion; if necessary, the system can always be written “as usual”: . Don’t forget that when moving a term from part to part, it needs to change its sign.

What does it mean to solve a system of linear equations? Solving a system of equations means finding many of its solutions. The solution of a system is a set of values ​​of all variables included in it, which turns EVERY equation of the system into a true equality. In addition, the system can be non-joint (have no solutions).Don't worry, it's general definition=) We will have only one value “x” and one value “y”, which satisfy each equation c-we.

Exists graphic method solution of the system, which can be found in class The simplest problems with a line. There I talked about geometrically systems of two linear equations with two unknowns. But now this is the era of algebra, and numbers-numbers, actions-actions.

Let's decide: from the first equation we express:
We substitute the resulting expression into the second equation:

We open the brackets, add similar terms and find the value:

Next, we remember what we danced for:
We already know the value, all that remains is to find:

Answer:

After ANY system of equations has been solved in ANY way, I strongly recommend checking (orally, on a draft or on a calculator). Fortunately, this is done easily and quickly.

1) Substitute the found answer into the first equation:

– the correct equality is obtained.

2) Substitute the found answer into the second equation:

– the correct equality is obtained.

Or, to put it more simply, “everything came together”

The considered method of solution is not the only one; from the first equation it was possible to express , and not .
You can do the opposite - express something from the second equation and substitute it into the first equation. By the way, note that the most disadvantageous of the four methods is to express from the second equation:

The result is fractions, but why? There is a more rational solution.

However, in some cases you still can’t do without fractions. In this regard, I would like to draw your attention to HOW I wrote down the expression. Not like this: and in no case like this: .

If in higher mathematics you are dealing with fractional numbers, then try to carry out all calculations in ordinary improper fractions.

Exactly, and not or!

A comma can be used only sometimes, in particular if it is the final answer to some problem, and no further actions need to be performed with this number.

Many readers probably thought “why do this? detailed explanation, as for a correction class, and so everything is clear.” Nothing like that, it seems so simple school example, and how many VERY important conclusions! Here's another one:

You should strive to complete any task in the most rational way. If only because it saves time and nerves, and also reduces the likelihood of making a mistake.

If in a problem in higher mathematics you come across a system of two linear equations with two unknowns, then you can always use the substitution method (unless it is indicated that the system needs to be solved by another method). Not a single teacher will think that you are a sucker and will reduce your grade for using the “school method” "
Moreover, in some cases it is advisable to use the substitution method with a larger number of variables.

Example 2

Solve a system of linear equations with three unknowns

A similar system of equations often arises when using the so-called method uncertain coefficients when we find the integral of a fractional rational function. The system in question was taken from there by me.

When finding the integral, the goal is fast find the values ​​of the coefficients, and not resort to Cramer’s formulas, the method inverse matrix etc. Therefore, in this case, the substitution method is appropriate.

When any system of equations is given, first of all it is desirable to find out whether it is possible to somehow simplify it IMMEDIATELY? Analyzing the equations of the system, we notice that the second equation of the system can be divided by 2, which is what we do:

Reference: the mathematical sign means “from this follows that” and is often used in problem solving.

Now let's analyze the equations; we need to express some variable in terms of the others. Which equation should I choose? You probably already guessed that the easiest way for this purpose is to take the first equation of the system:

Here, no matter what variable to express, one could just as easily express or .

Next, we substitute the expression for into the second and third equations of the system:

We open the brackets and present similar terms:

Divide the third equation by 2:

From the second equation we express and substitute into the third equation:

Almost everything is ready, from the third equation we find:
From the second equation:
From the first equation:

Check: Substitute the found values ​​of the variables into the left side of each equation of the system:

1)
2)
3)

The corresponding right-hand sides of the equations are obtained, thus the solution is found correctly.

Example 3

Solve a system of linear equations with 4 unknowns

This is an example for independent decision(answer at the end of the lesson).

Solving the system by term-by-term addition (subtraction) of the system equations

When solving systems of linear equations, you should try to use not the “school method”, but the method of term-by-term addition (subtraction) of the equations of the system. Why? This saves time and simplifies calculations, however, now everything will become clearer.

Example 4

Solve a system of linear equations:

I took the same system as in the first example.
Analyzing the system of equations, we notice that the coefficients of the variable are identical in magnitude and opposite in sign (–1 and 1). In such a situation, the equations can be added term by term:

Actions circled in red are performed MENTALLY.
As you can see, as a result of term-by-term addition, we lost the variable. This, in fact, is what the essence of the method is to get rid of one of the variables.

Linear equations are a fairly harmless and understandable topic in 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 coefficients. 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 - they can be anything!

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 take a closer look at the general notation ax+b=0, 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 look at all such surprises. And we will 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’s look once again at the general structure of 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 X's in higher powers - 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 the lower grades. 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 now lack for 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 get the general idea. 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.)

First, I ask a key question: what stands out to you most 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. A right side? For a 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. In order to satisfy our desires (to get rid of fractions) and not to offend mathematics.) Let’s multiply both parts by six!) That is, by the common denominator of 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 the parentheses. 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 this equation we had to apply the first transformation once (transferring 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, create errors...

And, in order not to make mistakes, in such sophisticated examples as this one, it is always most useful to evaluate its 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, what is x equal to. 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 record this equality is written like this:

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…

Scientifically speaking, 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 such a response is formatted 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 in identical 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!)