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

Equation systems are used not only in the field of mathematics, but also in physics, chemistry and biology, when solving problems of finding the population size.

A system of linear equations is a term for 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, the value of which must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving the equation by plotting its graph will look like a straight line, all points of which are the solution of the polynomial.

Types of systems of linear equations

The simplest are examples of 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 a system of equations - it means to find such values ​​(x, y) for which the system becomes a true equality, or to establish that there are no suitable values ​​of x and y.

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

If the systems have one common solution or there is no solution, they are called equivalent.

Homogeneous systems of linear equations are systems whose right side 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 not homogeneous.

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 variables or more.

Faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not so. The number of equations in the system does not depend on the variables, there can be an arbitrarily large number of them.

Simple and complex methods for solving systems of equations

There is no general analytical way to solve such systems, all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as the graphical and matrix method, the solution by the Gauss method.

The main task in teaching methods of solving is to teach how to correctly analyze the system and find the optimal solution algorithm 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 applying a particular method.

The solution of examples of systems of linear equations of the 7th grade of the general education school program is quite simple and is explained in great detail. In any textbook on mathematics, this section is given enough attention. The solution of examples of systems of linear equations by the method of Gauss and Cramer is studied in more detail in the first courses of higher educational institutions.

Solution of systems by the substitution method

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

Let's give an example of a system of linear equations of the 7th class by 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. The solution of this example does not cause difficulties and allows you to get 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 the expression of 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, the substitution solution is also impractical.

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

Solution using algebraic addition

When searching for a solution to systems by the addition method, term-by-term addition and multiplication of equations by various numbers are performed. The ultimate goal of mathematical operations is an equation with one variable.

The application of this method requires practice and observation. It is not easy to solve a system of linear equations using the addition method with the number of variables 3 or more. Algebraic addition is useful when the equations contain fractions and decimal numbers.

Solution action algorithm:

1. Multiply both sides of the equation by some number. As a result of the arithmetic operation, 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.

Solution method by introducing a new variable

A new variable can be introduced if the system needs to find 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 with respect to the entered unknown, and the resulting value is used to determine the original variable.

It can be seen from the example that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard square trinomial. You can solve a polynomial by finding the discriminant.

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

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

A visual method for solving systems

Suitable for systems with 3 equations. The method consists in plotting graphs of each equation included in the system on the coordinate axis. The coordinates of the points of intersection of the curves will be the general solution of the system.

The graphic method has a number of nuances. Consider several examples of solving systems of linear equations in a visual way.

As can be seen from the example, two points were constructed for each line, 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.

In the following example, it is required to find a graphical solution to the system 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 the system has a solution or not, it is always necessary to build a graph.

Matrix and its varieties

Matrices are used to briefly write down a system of linear equations. A matrix is ​​a special type of table filled with numbers. n*m has n - rows and m - columns.

A matrix is ​​square when the number of columns and rows is equal. A matrix-vector is a single-column matrix with an infinitely possible number of rows. A matrix with units along one of the diagonals and other zero elements is called identity.

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

Rules for transforming a system of equations into a matrix

With regard to systems of equations, the coefficients and free members of the equations are written as numbers of the matrix, one equation is one row of the matrix.

A matrix row is called non-zero if at least one element of the row is not equal to 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 columns of the matrix must strictly correspond to the variables. This means that the coefficients of the variable x can only be written in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all matrix elements 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 is the inverse matrix and |K| - matrix determinant. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix, it is only necessary to multiply the elements diagonally 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 column and row numbers of the elements do not repeat in the product.

Solution of examples of systems of linear equations by the matrix method

The matrix method of finding a solution makes it possible to reduce cumbersome entries when solving systems with a large number of variables and equations.

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

Solution of systems by the Gauss method

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

The Gaussian method is very similar to substitution and algebraic addition solutions, but is more systematic. In the school course, the Gaussian solution is used for systems of 3 and 4 equations. The purpose of the method is to bring the system to the form of an inverted trapezoid. By 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, and 3 and 4 - with 3 and 4 variables, respectively.

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 Gaussian solution 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. The solution of any of the equations will allow you to find out one of the variables x n.

Theorem 5, which is mentioned in the text, says 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 Gauss method is difficult for middle school students to understand, but is one of the most interesting ways to develop the ingenuity of children studying in the advanced study program in math and physics classes.

For ease of recording calculations, it is customary to do the following:

Equation coefficients 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 side. Roman numerals denote the numbers of equations in the system.

First, they write down the matrix with which to work, then all the actions carried out with one of the rows. The resulting matrix is ​​written after the "arrow" sign and continue to perform the necessary algebraic operations until the result is achieved.

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

This notation is less cumbersome and allows you not to be distracted by the enumeration of numerous unknowns.

The free application of any method of solution will require care and a certain amount of experience. Not all methods are applied. Some ways of finding solutions are more preferable in a particular area of ​​human activity, while others exist for the purpose of learning.

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 in linear equations also occur! All around. A little lower, we will also analyze such evil equations.)

Well, let's not pull the cat by the tail and start to figure it out, shall we? Then we read and understand.)

What is a linear equation? Examples.

Typically, a linear equation has the following form:

ax + b = 0,

Where a and b are any numbers. Anything: integer, fractional, negative, irrational - everyone can be!

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 we take a closer look at the common notation ax+b=0, and think a little? Because a and b 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 fun! And if, say, a = 0, b = -10? Then it turns out quite some nonsense:

0 = 10.

Which is very, very annoying and undermines the trust in mathematics won by sweat and blood ... Especially in tests and exams. But of these incomprehensible and strange equalities, you also need to find X! Which does not exist at all! And here even well-prepared students, at times, can fall, as they say, into a stupor ... But don't worry! In this lesson, we will also consider all such surprises. And x from such equalities will also be sure to find.) Moreover, this very x is searched for very, very simply. Yes Yes! Surprising but true.)

Okay, that's understandable. But how can you know by the appearance of the task that we have a linear equation, and not some other one? Unfortunately, it is far from always possible to recognize the type of equation only by appearance. The thing is that not only equations of the form ax + b = 0 are called linear, but also any other equations that, by identical transformations, one way or another, are reduced to this form. How do you know if it fits or not? Until you almost solve the example - almost nothing. It's upsetting. But for some types of equations, it is possible, with one quick glance, to immediately say with certainty whether it is linear or not.

To do this, we once again turn to the general structure of any linear equation:

ax + b = 0

Note that in a linear equation always there is only variable x in the first degree and some numbers! And that's it! Nothing else. At the same time, there are no x squared, cubed, under the root, under the logarithm and other exotics. And (most importantly!) 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 xes in higher powers - squared, cubed, and so on. Yes, there are fractions here, but at the same time they sit in the denominators of fractions 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 degree. For, 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, and square - anyone.

How to solve linear equations? Examples.

So how do you solve linear equations? Read on and be surprised.) The whole 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.

This is the use of brackets, opening brackets, working with fractions, working with negative numbers, the multiplication table, and so on. This knowledge and skills are necessary not only for solving linear equations, but for all mathematics in general. And, if this is a problem, remember the lower grades. Otherwise, you will have a hard time ...

2)

There are only two of them. Yes Yes! Moreover, these very basic identical transformations underlie the solution of not only linear, but in general any equations of mathematics! In a word, the solution of any other equation - quadratic, logarithmic, trigonometric, irrational, etc. - as a rule, begins with these very basic transformations. But the solution of precisely linear equations, in fact, ends on them (transformations). Ready answer.) So do not be lazy and take a walk through the link.) Moreover, linear equations are also analyzed in detail there.

Well, I think it's time to start the analysis of examples.

To begin with, as a warm-up, consider some elementary. Without any fractions and other bells and whistles. For example, this equation:

x - 2 \u003d 4 - 5x

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

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

x + 5x = 4 + 2

Well. Half the battle is done: the x's are gathered in a pile, the numbers, too. Now we give similar ones on the left, and we count on the right. We get:

6x = 6

What do we lack now for complete happiness? Yes, so that a clean X remains on the left! And the six interferes. How to get rid of it? Now we start the second identical transformation - we divide both sides of the equation by 6. And - voila! Answer ready.)

x = 1

Of course, the example is quite primitive. To get the general idea. Well, let's do something more substantial. For example, consider the following equation:

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

What will we do first? With X - to the left, without X - to the right? In principle, it is possible and so. Fly to Sochi via Vladivostok.) Or you can take the shortest path, immediately using the universal and powerful method. If you know the identical transformations, of course.)

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

Here we multiply. The entire left side and the entire right side entirely! Therefore, we use brackets. This is what the procedure looks like:

Now let's open these parentheses:

Now, representing 6 as 6/1, multiply the six by each of the fractions on the left and right. This is the usual multiplication of fractions, but, so be it, I will write in detail:

And here - attention! I took the numerator (x-3) in brackets! This is all because when multiplying fractions, the numerator is multiplied in its entirety, entirely and completely! And with the expression x-3 it is necessary to work as with one solid construction. But if you write the numerator like this:

6x - 3,

But we have everything right and we need to finish it. What to do next? Open brackets in the numerator on the left? In no case! You and I multiplied both parts by 6 in order to get rid of fractions, and not to take a steam bath with opening brackets. At this stage, we need reduce our fractions. With a feeling of deep satisfaction, we reduce all the denominators and get the 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 just keeps getting better and better! Now we recall again the first identical transformation. With a stone face, we repeat the spell from the lower grades: with x - to the left, without x - to the right. And apply this transformation:

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

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

13x = 39

It remains to divide both parts by 13. That is, re-apply the second transformation. 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 (transfer of terms) once and the second one 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 before x. And the solution of 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. Somewhere it is more profitable to start with a transfer, and somewhere (as in this example) - with multiplication (or division).

We work from simple to complex. Consider now frank tin. With a bunch of fractions and brackets. And I'll tell you how not to overstrain.)

For example, here's an equation:

We look at the equation for a minute, we are horrified, but still we 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 share ... 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 do not violate the identity of your transformations and, thereby, do not make mistakes ...

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

Here we are guessing. On the left are the sixes in the denominators. Personally, I don't like them, but they are very easy to remove. Let me multiply both sides of the equation by 6! Then the sixes on the left will be safely reduced, the fractions in brackets will not go anywhere yet. Well, no big deal. We will deal with them a bit later.) But on the right, the denominators 2 and 3 will decrease. It is with this action (multiplication 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 turned out, then you didn’t understand the analysis of the previous example well. And I tried, by the way ...

So let's open it:

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 give similar ones on the right side. We get:

Already much better. Now the left side has prepared itself for multiplication. What should be multiplied by the left side so that both the five and the four are immediately reduced? At 20! But we also have downsides 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, the minuses will disappear, and the fractions.

Here we multiply:

For those who still do not understand this step, it means that the problems are not in the equations. Problems are at the core! Again, remember the golden rule of opening parentheses:

If the number is multiplied by some expression in brackets, then this number must be successively multiplied by each term of this very expression. Moreover, if the number is positive, then the signs of the expressions after expansion are preserved. If negative, they are reversed:

a(b+c) = ab+ac

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

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

And now you can reduce fractions:

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

Expand the remaining parentheses. Again, we open correctly. The first brackets are multiplied by a 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 empty spaces left. With x to the left, without x to the right:

-20x - 15x = 20 - 10 - 12

-35x = -2

That's almost all. On the left, you need a clean X, and the number -35 gets in the way. So we divide both parts by (-35). I remind you that the second identity transformation allows us to multiply and divide both parts by whatever number. Including the negative.) If only not to zero! Feel free to share and get the answer:

X=2/35

This time 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 twisted ones) is quite simple: we take the original equation and, by identical transformations, we sequentially simplify it right up to the answer. With the basics, of course! The main problems here are precisely in non-compliance with the basics (say, there is a minus before the brackets, and they forgot to change the signs when opening), as well as in banal arithmetic. So don't neglect the basics! They are the foundation of all the rest of mathematics!

Some tricks in solving linear equations. Or special occasions.

Everything would be nothing. However ... Among the linear equations, there are also such funny pearls that, in the process of solving them, can drive them into a strong stupor. Even an excellent student.)

For example, here is a harmless-looking equation:

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

Yawning wide 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 give similar ones, consider and get:

0 = 0

That's it! Issued primerchik focus! In itself, this equality raises no objections: zero is indeed equal to zero. But X is gone! Without a trace! And we must write in the answer, what is x equal to. Otherwise, the decision is not considered, yes.) What to do?

No panic! In such non-standard cases, the most general concepts and principles of mathematics save. 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 the correct equality already done! 0=0, or rather nowhere!) It remains to be guessed at which x's we get this equality. What kind of x's can be substituted into original equation if, when substituting, they all still shrink to zero? Haven't you figured it out yet?

Yes, of course! Xs can be substituted any!!! Absolutely any. Whatever you want, put them in. 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 is your answer:

x is any number.

In scientific notation, this equality is written like this:

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

Or in another form, at intervals:

As you like, arrange it. This is the correct and completely complete answer!

And now I'm going to change just one number in our original equation. Let's solve this equation now:

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

We again transfer the terms, count and get:

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

0 = 1

And how do you like this joke? There was an ordinary linear equation, but there was an incomprehensible equality

0 = 1…

In scientific terms, we have wrong equality. But in Russian it's not true. Bullshit. Nonsense.) For zero is not equal to one!

And now again we think what kind of x when substituting into the original equation will give us correct equality? Which? But none! Whatever X you substitute, everything will still be reduced and there will be crap.)

Here is the answer: no solutions.

In mathematical notation, such an answer is drawn up like this:

It reads: "X belongs to the empty set."

Such answers in mathematics are also quite common: not always any equation has 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 Xs in the equation will not confuse you forever. The case is quite familiar.)

And then I hear a logical question: will they be in the OGE or the USE? On the exam, by themselves as a task - no. Too simple. But in the OGE or in text problems - easily! So now - we train and decide:

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

Everything worked out? Fine! You have good chances in the exam.

Something doesn't fit? Hm ... Sadness, of course. So there are gaps somewhere. Either in bases or in identical transformations. Or it's a matter of banal inattention. Reread the lesson again. For this is not a topic that one can do without so easily in mathematics ...

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

Equations. In other words, the solution of all equations begins with these transformations. When solving linear equations, it (solution) on identical transformations and ends with the final answer.

The case of a non-zero coefficient for an unknown variable.

ax+b=0, a ≠ 0

We transfer members with x to one side, and numbers to the other side. Be sure to remember that when transferring the terms to the opposite side of the equation, you need to change the sign:

ax:(a)=-b:(a)

We reduce a at X and we get:

x=-b:(a)

This is the answer. If you want to check if a number is -b:(a) root of our equation, then we need to substitute in the initial equation instead of X this is the same number:

a(-b:(a))+b=0 ( those. 0=0)

Because this equality is true, then -b:(a) and truth is the root of the equation.

Answer: x=-b:(a), a ≠ 0.

First example:

5x+2=7x-6

We transfer to one side the terms from X, and on the other side of the number:

5x-7x=-6-2

-2x:(-2)=-8:(-2)

With an unknown coefficient, they reduced it and got the answer:

This is the answer. If you need to check whether the number 4 is really the root of our equation, we substitute this number instead of x in the original equation:

5*4+2=7*4-6 ( those. 22=22)

Because this equality is true, then 4 is the root of the equation.

Second example:

Solve the equation:

5x+14=x-49

Transferring the unknowns and the numbers in different directions, we got:

We divide the parts of the equation by the coefficient at x(on 4) and get:

Third example:

Solve the equation:

First, we get rid of the irrationality in the coefficient of the unknown by multiplying all the terms by:

This form is considered simplified, because the number has the root of the number in the denominator. We need to simplify the answer by multiplying the numerator and denominator by the same number, we have this:

The case of no solutions.

Solve the equation:

2x+3=2x+7

For all x our equation will not become a true equality. That is, our equation has no roots.

Answer: There are no solutions.

A special case is an infinite number of solutions.

Solve the equation:

2x+3=2x+3

Transferring x's and numbers in different directions and bringing like terms, we get the equation:

Here, too, it is not possible to divide both parts by 0, because it is forbidden. However, putting in place X any number, we get the correct equality. That is, every number is a solution to such an equation. Thus, there is an infinite number of solutions.

Answer: an infinite number of solutions.

The case of equality of two complete forms.

ax+b=cx+d

ax-cx=d-b

(a-c)x=d-b

x=(d-b):(a-c)

Answer: x=(d-b):(a-c), if d≠b and a≠c, otherwise there are infinitely many solutions, but if a=c, a d≠b, then there are no solutions.

A linear equation is an algebraic equation whose full degree of polynomials is equal to one. Solving linear equations is part of the school curriculum, and not the most difficult. However, some still experience difficulties in the passage of this topic. We hope that after reading this material, all the difficulties for you will remain in 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.

Even though 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 an infinite number of solutions;
• 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 have non-zero values, the equation has to be solved in order to derive the final expression for the variable.

How to decide?

Solving a linear equation means finding what a variable is equal to. How to do it? Yes, it's very simple - using simple algebraic operations and following the rules of transfer. If the equation appeared before you in a 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 ax = -b should be obtained.
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, an expression of the form should be obtained: x \u003d -b / a.

That's all - the solution is found!

Now let's look at a specific example:

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

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

As you can see, finding a solution to a linear equation with one variable is quite simple, but everything is so simple if we are lucky to meet the equation in a general form. In most cases, before solving the equation in the two steps described above, it is also necessary to bring the existing expression to a general form. However, this is also not a daunting task. Let's look at some special cases with examples.

Solving special cases

First, let's take a 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 will always get zero! Right! Therefore, they say that the equation has an infinite number of solutions - whatever number you take, the equality will be true, 0x \u003d 0 or 0 \u003d 0.
• If a=0, b≠0, the equation will look like: 0x + 3 = 0. We perform the first step, we get 0x = -3. Nonsense again! It is obvious that this equality will never be true! That is why they say that the equation has no solutions.
• If a≠0, b=0, the equation will look like: 3x + 0 = 0. Taking the first step, we get: 3x = 0. What is the solution? It's easy, x = 0.

Difficulties in translation

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

• 12x - 14 = 2x + 6

Is this a linear equation? But what about the zero on the right side? We will not rush to conclusions, we will act - we will transfer all the components of our equation to the left side. We get:

• 12x - 2x - 14 - 6 = 0

Now subtracting like from like, we get:

• 10x - 20 = 0

Learned? The most linear equation ever! Whose solution: 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 done. Let's expand the brackets first:

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

As you can see, everything is solved, the main thing is not to worry, but to act. Remember, if your equation contains only variables of the first degree and numbers, this is 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!

In this article, we consider the principle of solving such equations as linear equations. Let us write down the definition of these equations and set the general form. We will analyze all the conditions for finding solutions to linear equations, using, among other things, practical examples.

Please note that the material below contains information on linear equations with one variable. Linear equations with two variables are considered in a separate article.

Yandex.RTB R-A-339285-1

What is a linear equation

Definition 1

Linear Equation is an equation written like this:
a x = b, where x- variable, a and b- some numbers.

This formulation is used in the algebra textbook (grade 7) by Yu.N. Makarychev.

Example 1

Examples of linear equations would be:

3x=11(one variable equation x at a = 5 and b = 10);

− 3 , 1 y = 0 ( linear equation with variable y, where a \u003d - 3, 1 and b = 0);

x = -4 and − x = 5 , 37(linear equations, where the number a written explicitly and equal to 1 and - 1, respectively. For the first equation b = - 4 ; for the second - b = 5, 37) etc.

Different teaching materials may contain different definitions. For example, Vilenkin N.Ya. linear also includes those equations that can be transformed into the form a x = b by transferring terms from one part to another with a sign change and bringing similar terms. If we follow this interpretation, the equation 5 x = 2 x + 6 – also linear.

And here is the textbook of algebra (Grade 7) Mordkovich A.G. specifies the following description:

Definition 2

A linear equation with one variable x is an equation of the form a x + b = 0, where a and b are some numbers, called the coefficients of the linear equation.

Example 2

An example of linear equations of this kind can be:

3 x - 7 = 0 (a = 3 , b = - 7) ;

1 , 8 y + 7 , 9 = 0 (a = 1 , 8 , b = 7 , 9) .

But there are also examples of linear equations that we have already used above: a x = b, For example, 6 x = 35.

We will immediately agree that in this article, under a linear equation with one variable, we will understand the equation of writing a x + b = 0, where x– variable; a , b are coefficients. We see this form of a linear equation as the most justified, since linear equations are algebraic equations of the first degree. And the other equations indicated above, and the equations given by equivalent transformations into the form a x + b = 0, we define as equations reducing to linear equations.

With this approach, the equation 5 x + 8 = 0 is linear, and 5 x = −8- an equation that reduces to a linear one.

The principle of solving linear equations

Consider how to determine whether a given linear equation will have roots and, if so, how many and how to determine them.

Definition 3

The fact of the presence of the roots of a linear equation is determined by the values ​​of the coefficients a and b. Let's write these conditions:

• at a ≠ 0 the linear equation has a single root x = - b a ;
• at a = 0 and b ≠ 0 a linear equation has no roots;
• at a = 0 and b = 0 a linear equation has infinitely many roots. In fact, in this case, any number can become the root of a linear equation.

Let's give an explanation. We know that in the process of solving an equation, it is possible to transform a given equation into an equivalent one, which means it has the same roots as the original equation, or also has no roots. We can make the following equivalent transformations:

• move the term from one part to another, changing the sign to the opposite;
• multiply or divide both sides of an equation by the same non-zero number.

Thus, we transform the linear equation a x + b = 0, moving the term b from the left side to the right side with a sign change. We get: a · x = - b .

So, we divide both parts of the equation by a non-zero number a, resulting in an equality of the form x = - b a . That is, when a ≠ 0 original equation a x + b = 0 is equivalent to the equality x = - b a , in which the root - b a is obvious.

By contradiction, it is possible to demonstrate that the found root is the only one. We set the designation of the found root - b a as x 1 . Let us assume that there is one more root of the linear equation with the notation x 2 . And of course: x 2 ≠ x 1, and this, in turn, based on the definition of equal numbers through the difference, is equivalent to the condition x 1 - x 2 ≠ 0. In view of the above, we can compose the following equalities by substituting the roots:
a x 1 + b = 0 and a · x 2 + b = 0 .
The property of numerical equalities makes it possible to perform a term-by-term subtraction of parts of equalities:

a x 1 + b - (a x 2 + b) = 0 - 0, from here: a (x 1 - x 2) + (b - b) = 0 and beyond a (x 1 - x 2) = 0 . Equality a (x 1 − x 2) = 0 is false, since the condition was previously given that a ≠ 0 and x 1 - x 2 ≠ 0. The obtained contradiction serves as a proof that at a ≠ 0 linear equation a x + b = 0 has only one root.

Let us substantiate two more clauses of the conditions containing a = 0 .

When a = 0 linear equation a x + b = 0 will be written as 0 x + b = 0. The property of multiplying a number by zero gives us the right to assert that no matter what number is taken as x, substituting it into the equality 0 x + b = 0, we get b = 0 . Equality is valid for b = 0; in other cases when b ≠ 0 equality becomes invalid.

Thus, when a = 0 and b = 0 , any number can be the root of a linear equation a x + b = 0, since under these conditions, substituting instead of x any number, we get the correct numerical equality 0 = 0 . When a = 0 and b ≠ 0 linear equation a x + b = 0 will not have roots at all, since under the specified conditions, substituting instead of x any number, we get an incorrect numerical equality b = 0.

All the above reasoning gives us the opportunity to write an algorithm that makes it possible to find a solution to any linear equation:

• by the type of record we determine the values ​​of the coefficients a and b and analyze them;
• at a = 0 and b = 0 the equation will have infinitely many roots, i.e. any number will become the root of the given equation;
• at a = 0 and b ≠ 0
• at a, different from zero, we start searching for the only root of the original linear equation:

1. transfer coefficient b to the right side with a change of sign to the opposite, bringing the linear equation to the form a x = −b;
2. divide both parts of the resulting equality by the number a, which will give us the desired root of the given equation: x = - b a .

Actually, the described sequence of actions is the answer to the question of how to find a solution to a linear equation.

Finally, we clarify that equations of the form a x = b are solved by a similar algorithm with the only difference that the number b in such a notation has already been transferred to the desired part of the equation, and when a ≠ 0 you can immediately divide the parts of the equation by a number a.

Thus, to find a solution to the equation a x = b, we use the following algorithm:

• at a = 0 and b = 0 the equation will have infinitely many roots, i.e. any number can become its root;
• at a = 0 and b ≠ 0 the given equation will not have roots;
• at a, not equal to zero, both sides of the equation are divisible by the number a, which makes it possible to find a single root that is equal to b a.

Examples of solving linear equations

Example 3

It is necessary to solve a linear equation 0 x - 0 = 0.

Decision

By writing the given equation, we see that a = 0 and b = -0(or b = 0 which is the same). Thus, a given equation can have infinitely many roots or any number.

Answer: x- any number.

Example 4

It is necessary to determine whether the equation has roots 0 x + 2, 7 = 0.

Decision

From the record, we determine that a \u003d 0, b \u003d 2, 7. Thus, the given equation will not have roots.

Answer: the original linear equation has no roots.

Example 5

Given a linear equation 0 , 3 x − 0 , 027 = 0 . It needs to be resolved.

Decision

By writing the equation, we determine that a \u003d 0, 3; b = - 0 , 027 , which allows us to assert that the given equation has a single root.

Following the algorithm, we transfer b to the right side of the equation, changing the sign, we get: 0.3 x = 0.027. Next, we divide both parts of the resulting equality by a \u003d 0, 3, then: x \u003d 0, 027 0, 3.

Let's divide decimals:

0.027 0.3 = 27300 = 3 9 3 100 = 9 100 = 0.09

The result obtained is the root of the given equation.

Briefly write the solution as follows:

0, 3 x - 0, 027 = 0, 0, 3 x = 0, 027, x = 0, 027 0, 3, x = 0, 09.

Answer: x = 0 , 09 .

For clarity, we present the solution of the equation of record a x = b.

Example N

Equations are given: 1) 0 x = 0 ; 2) 0 x = − 9 ; 3) - 3 8 x = - 3 3 4 . It is necessary to solve them.

Decision

All given equations correspond to the record a x = b. Let's consider it in turn.

In the equation 0 x = 0 , a = 0 and b = 0, which means: any number can be the root of this equation.

In the second equation 0 x = − 9: a = 0 and b = − 9 , thus, this equation will not have roots.

By the form of the last equation - 3 8 x = - 3 3 4 we write the coefficients: a = - 3 8 , b = - 3 3 4 , i.e. the equation has a single root. Let's find him. Let's divide both sides of the equation by a , we get as a result: x = - 3 3 4 - 3 8 . Let's simplify the fraction by applying the rule for dividing negative numbers, then converting the mixed number to an ordinary fraction and dividing ordinary fractions:

3 3 4 - 3 8 = 3 3 4 3 8 = 15 4 3 8 = 15 4 8 3 = 15 8 4 3 = 10

Briefly write the solution as follows:

3 8 x = - 3 3 4 , x = - 3 3 4 - 3 8 , x = 10 .

Answer: 1) x- any number, 2) the equation has no roots, 3) x = 10 .

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