What is called the solution of a linear equation. How to solve a cubic equation? The principle of solving linear equations

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

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

\(x=[number]\)

This number will be the root.

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

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

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

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

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

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

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

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

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

Add \(2x\) left and right

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

Again, we present like terms

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

Answer : \(7\)

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

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

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

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

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

The answer is simple:

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

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

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

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

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

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

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

Now we give like terms on the left and right.

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

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

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

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

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

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

Decision :

Answer : no roots.

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

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

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

Decision :

Answer : any number.

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

More complex linear equations.

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

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

Decision :

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

Answer : \(x=5\)

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

Decision :

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

		Well, dividing by \ (-4 \) the right and left parts, we get the answer

Answer : \(x=-1.25\)

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:

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.
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:

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

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

Equality with a variable is called an equation.
Solving an equation means finding the set of its roots. An equation can have one, two, several, many roots, or none at all.
Each value of the variable at which the given equation turns into a true equality is called the root of the equation.
Equations that have the same roots are called equivalent equations.
Any term of the equation can be transferred from one part of the equality to another, while changing the sign of the term to the opposite.
If both sides of the equation are multiplied or divided by the same non-zero number, then an equation is obtained that is equivalent to this equation.

Examples. Solve the equation.

1. 1.5x+4 = 0.3x-2.

1.5x-0.3x = -2-4. We collected the terms containing the variable on the left side of the equality, and the free members on the right side of the equality. The following property was used:

1.2x = -6. We brought like terms according to the rule:

x = -6 : 1.2. Both parts of the equality were divided by the coefficient of the variable, since

x = -5. Divided according to the rule of dividing a decimal fraction by a decimal fraction:

to divide a number by a decimal, you need to move the commas in the dividend and divisor as many digits to the right as they are after the decimal point in the divisor, and then divide by a natural number:

6 : 1,2 = 60 : 12 = 5.

Answer: 5.

2. 3∙ (2x-9) = 4 ∙ (x-4).

6x-27 = 4x-16. We opened the brackets using the distributive law of multiplication with respect to subtraction: (a-b) ∙ c = a ∙ c-b ∙ c.

6x-4x = -16+27. We collected the terms containing the variable on the left side of the equality, and the free members on the right side of the equality. The following property was used: any term of the equation can be transferred from one part of the equality to another, while changing the sign of the term to the opposite.

2x \u003d 11. They brought like terms according to the rule: to bring similar terms, you need to add their coefficients and multiply the result by their common letter part (i.e., add their common letter part to the result).

x = 11 : 2. Both parts of the equality were divided by the coefficient of the variable, since if both parts of the equation are multiplied or divided by the same non-zero number, then an equation is obtained that is equivalent to this equation.

Answer: 5,5.

3. 7x-(3+2x)=x-9.

7x-3-2x = x-9. We opened the brackets according to the rule for opening brackets, which are preceded by a "-" sign: if there is a “-” sign in front of the brackets, then we remove the brackets, the “-” sign and write the terms in brackets with opposite signs.

7x-2x-x \u003d -9 + 3. We collected the terms containing the variable on the left side of the equality, and the free members on the right side of the equality. The following property was used: any term of the equation can be transferred from one part of the equality to another, while changing the sign of the term to the opposite.

4x = -6. We brought like terms according to the rule: to bring similar terms, you need to add their coefficients and multiply the result by their common letter part (i.e., add their common letter part to the result).

x = -6 : 4. Both parts of the equality were divided by the coefficient of the variable, since if both parts of the equation are multiplied or divided by the same non-zero number, then an equation is obtained that is equivalent to this equation.

Answer: -1,5.

3 ∙ (x-5) = 7 ∙ 12 — 4 ∙ (2x-11). Multiply both sides of the equation by 12 - the lowest common denominator for the denominators of these fractions.

3x-15 = 84-8x+44. We opened the brackets using the distributive law of multiplication with respect to subtraction: in order to multiply the difference of two numbers by the third number, you can multiply the separately reduced and separately subtracted by the third number, and then subtract the second result from the first result, i.e.(a-b) ∙ c = a ∙ c-b ∙ c.

3x+8x = 84+44+15. We collected the terms containing the variable on the left side of the equality, and the free members on the right side of the equality. The following property was used: any term of the equation can be transferred from one part of the equality to another, while changing the sign of the term to the opposite.

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:

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;
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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