It is called a linear equation with one variable. Equations with one variable

First you need to understand what it is.

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

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

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

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

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

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

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

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

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

• 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.

A linear equation with one variable has the general form
ax + b = 0.
Here x is a variable, a and b are coefficients. In another way, a is called the “coefficient of the unknown”, b is the “free term”.

The coefficients are some numbers, and solving the equation means finding the value x for which the expression ax + b = 0 is true. For example, we have a linear equation 3x - 6 \u003d 0. Solving it means finding what x must be equal to so that 3x - 6 is equal to 0. Performing transformations, we get:
3x=6
x=2

Thus the expression 3x - 6 = 0 is true for x = 2:
3 * 2 – 6 = 0
2 is the root of this equation. When you solve an equation, you find its roots.

The coefficients a and b can be any numbers, however, there are such values ​​when there is more than one root of a linear equation with one variable.

If a = 0, then ax + b = 0 turns into b = 0. Here x is "destroyed". The expression b = 0 itself can be true only if the knowledge of b is 0. That is, the equation 0*x + 3 = 0 is false, because 3 = 0 is a false statement. However, 0*x + 0 = 0 is the correct expression. From here it is concluded that if a \u003d 0 and b ≠ 0, a linear equation with one variable has no roots at all, but if a \u003d 0 and b \u003d 0, then the equation has an infinite number of roots.

If b \u003d 0, and a ≠ 0, then the equation will take the form ax \u003d 0. It is clear that if a ≠ 0, but the result of multiplication is 0, then x \u003d 0. That is, the root of this equation is 0.

If neither a nor b are equal to zero, then the equation ax + b = 0 is transformed to the form
x \u003d -b / a.
The value of x in this case will depend on the values ​​of a and b . However, it will be the only one. That is, it is impossible to obtain two or more different x values ​​for the same coefficients. For example,
-8.5x - 17 = 0
x = 17 / -8.5
x = -2
No number other than -2 can be obtained by dividing 17 by -8.5.

There are equations that at first glance do not look like the general form of a linear equation with one variable, but are easily converted to it. For example,
-4.8 + 1.3x = 1.5x + 12

If we move everything to the left side, then 0 will remain on the right:
–4.8 + 1.3x – 1.5x – 12 = 0

Now the equation is reduced to the standard form and you can solve it:
x = 16.8 / 0.2
x=84

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

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

\(x=[number]\)

This number will be the root.

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

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

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

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

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

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

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

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

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

Add \(2x\) left and right

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

Again, we present like terms

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

Answer : \(7\)

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

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

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

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

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

The answer is simple:

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

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

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

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

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

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 top five on the left. What is she doing"? multiplied on x. So we remove it division.

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

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

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

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

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

Solution :

Answer : no roots.

In fact, the fact that we will come to such a result was seen earlier, even when we got \(3x-1=3x+6\). Think about it: how can \(3x\) be equal, from which \(1\) was subtracted, and \(3x\) to which \(6\) was added? Obviously, no way, because they did 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 \(8(x+2)-4=12x-4(x-3)\)

Solution :

Answer : any number.

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

More complex linear equations.

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

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

Solution :

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

Answer : \(x=5\)

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

Solution :

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

Answer : \(x=-1.25\)

An equation with one unknown, which, after opening the brackets and reducing like terms, takes the form

ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.

For example, all equations:

2x + 3 \u003d 7 - 0.5x; 0.3x = 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.

The value of the unknown that turns the equation into a true equality is called decision or the root of the equation .

For example, if in the equation 3x + 7 \u003d 13 we substitute the number 2 instead of the unknown x, then we get the correct equality 3 2 + 7 \u003d 13. Hence, the value x \u003d 2 is the solution or the root of the equation.

And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 2 + 7 ≠ 13. Therefore, the value x \u003d 3 is not a solution or a root of the equation.

The solution of any linear equations is reduced to the solution of equations of the form

ax + b = 0.

We transfer the free term from the left side of the equation to the right, while changing the sign in front of b to the opposite, we get

If a ≠ 0, then x = – b/a .

Example 1 Solve the equation 3x + 2 =11.

We transfer 2 from the left side of the equation to the right, while changing the sign in front of 2 to the opposite, we get
3x \u003d 11 - 2.

Let's do the subtraction, then
3x = 9.

To find x, you need to divide the product by a known factor, that is,
x = 9: 3.

So the value x = 3 is the solution or the root of the equation.

Answer: x = 3.

If a = 0 and b = 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0, we get 0, but b is also 0. The solution to this equation is any number.

Example 2 Solve the equation 5(x - 3) + 2 = 3 (x - 4) + 2x - 1.

Let's expand the brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.

5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.

Here are similar members:
0x = 0.

Answer: x is any number.

If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when multiplying any number by 0, we get 0, but b ≠ 0.

Example 3 Solve the equation x + 8 = x + 5.

Let us group the terms containing unknowns on the left side, and the free terms on the right side:
x - x \u003d 5 - 8.

Here are similar members:
0x = - 3.

Answer: no solutions.

On the figure 1 the scheme for solving the linear equation is shown

Let us compose a general scheme for solving equations with one variable. Consider the solution of example 4.

Example 4 Let's solve the equation

1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.

2) After reduction we get
4 (x - 4) + 3 2 (x + 1) - 12 = 6 5 (x - 3) + 24x - 2 (11x + 43)

3) To separate members containing unknown and free members, open the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.

4) We group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.

5) Here are similar members:
- 22x = - 154.

6) Divide by - 22 , We get
x = 7.

As you can see, the root of the equation is seven.

In general, such equations can be solved as follows:

a) bring the equation to an integer form;

b) open brackets;

c) group the terms containing the unknown in one part of the equation, and the free terms in the other;

d) bring similar members;

e) solve an equation of the form aх = b, which was obtained after bringing like terms.

However, this scheme is not required for every equation. When solving many simpler equations, one has to start not from the first, but from the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.

Example 5 Solve the equation 2x = 1/4.

We find the unknown x \u003d 1/4: 2,
x = 1/8 .

Consider the solution of some linear equations encountered in the main state exam.

Example 6 Solve equation 2 (x + 3) = 5 - 6x.

2x + 6 = 5 - 6x

2x + 6x = 5 - 6

Answer: - 0.125

Example 7 Solve the equation - 6 (5 - 3x) \u003d 8x - 7.

– 30 + 18x = 8x – 7

18x - 8x = - 7 +30

Answer: 2.3

Example 8 Solve the Equation

3(3x - 4) = 4 7x + 24

9x - 12 = 28x + 24

9x - 28x = 24 + 12

Example 9 Find f(6) if f (x + 2) = 3 7's

Solution

Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.

We solve the linear equation x + 2 = 6,
we get x \u003d 6 - 2, x \u003d 4.

If x = 4 then
f(6) = 3 7-4 = 3 3 = 27

Answer: 27.

If you still have questions, there is a desire to deal with the solution of equations more thoroughly, sign up for my lessons in the SCHEDULE. I will be glad to help you!

TutorOnline also recommends watching a new video tutorial from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.

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