Solving the equation x 1. How is a system of equations solved? Methods for solving systems of equations

An equation with one unknown, which, after opening the brackets and bringing similar terms, takes the form

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

For example, all equations:

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

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

For example, if in the equation 3x + 7 = 13 instead of the unknown x we ​​substitute the number 2, we obtain the correct equality 3 2 +7 = 13. This means that the value x = 2 is the solution or root of the equation.

And the value x = 3 does not turn the equation 3x + 7 = 13 into a true equality, since 3 2 +7 ≠ 13. This means that the value x = 3 is not a solution or a root of the equation.

Solving any linear equations reduces to solving equations of the form

ax + b = 0.

Let's move the free term from the left side of the equation to the right, 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.

Let's move 2 from the left side of the equation to the right, changing the sign in front of 2 to the opposite, we get
3x = 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.

This means that the value x = 3 is the solution or root of the equation.

Answer: x = 3.

If a = 0 and b = 0, then we get the equation 0x = 0. This equation has infinitely many solutions, since when we multiply any number by 0 we get 0, but b is also equal to 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 = 3x – 12 + 2x ‒ 1.

5x – 3x ‒ 2x = – 12 ‒ 1 + 15 ‒ 2.

Here are some similar terms:
0x = 0.

Answer: x - any number.

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

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

Let’s group terms containing unknowns on the left side, and free terms on the right side:
x – x = 5 – 8.

Here are some similar terms:
0х = ‒ 3.

Answer: no solutions.

On Figure 1 shows a diagram for solving a linear equation

Let's draw up a general scheme for solving equations with one variable. Let's consider the solution to Example 4.

Example 4. Suppose we need to 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 terms containing unknown and free terms, open the brackets:
4x – 16 + 6x + 6 – 12 = 30x – 90 + 24x – 22x – 86.

4) Let us group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x – 30x – 24x + 22x = ‒ 90 – 86 + 16 – 6 + 12.

5) Let us present similar terms:
- 22x = - 154.

6) Divide by – 22, We get
x = 7.

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

Generally such equations can be solved using the following scheme:

a) bring the equation to its integer form;

b) open the 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 similar terms.

However, this scheme is not necessary for every equation. When solving many simpler equations, you have 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.

Find the unknown x = 1/4: 2,
x = 1/8 .

Let's look at solving some linear equations found in the main state exam.

Example 6. Solve the 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) = 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.

Let's decide linear equation x + 2 = 6,
we get x = 6 – 2, x = 4.

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

Answer: 27.

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At the stage of preparation for the final test, high school students need to improve their knowledge on the topic “Exponential Equations.” The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to thoroughly master the theory, remember the formulas and understand the principle of solving such equations. Having learned to cope with this type of problem, graduates can count on high scores when passing the Unified State Exam in mathematics.

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When reviewing the materials they have covered, many students are faced with the problem of finding the formulas needed to solve equations. A school textbook is not always at hand, and selecting the necessary information on a topic on the Internet takes a long time.

The Shkolkovo educational portal invites students to use our knowledge base. We implement completely new method preparation for the final test. By studying on our website, you will be able to identify gaps in knowledge and pay attention to those tasks that cause the most difficulty.

Shkolkovo teachers collected, systematized and presented all the material necessary for successfully passing the Unified State Exam in the simplest and most accessible form.

Basic definitions and formulas are presented in the “Theoretical background” section.

To better understand the material, we recommend that you practice completing the assignments. Carefully review the examples presented on this page. exponential equations with the solution to understand the calculation algorithm. After that, proceed to perform tasks in the “Directories” section. You can start with the easiest tasks or move straight to solving complex exponential equations with several unknowns or . The database of exercises on our website is constantly supplemented and updated.

Those examples with indicators that caused you difficulties can be added to “Favorites”. This way you can quickly find them and discuss the solution with your teacher.

To successfully pass the Unified State Exam, study on the Shkolkovo portal every day!

The online fraction calculator allows you to perform simple arithmetic operations with fractions: adding fractions, subtracting fractions, multiplying fractions, dividing fractions. To make calculations, fill in the fields corresponding to the numerators and denominators of the two fractions.

Fractions in mathematics is a number representing a part of a unit or several parts of it.

A common fraction is written as two numbers, usually separated by a horizontal line indicating the division sign. The number above the line is called the numerator. The number below the line is called the denominator. The denominator of a fraction shows the number of equal parts into which the whole is divided, and the numerator of the fraction shows the number of these parts of the whole taken.

Fractions can be regular or improper.

• A fraction whose numerator is less than its denominator is called a proper fraction.
• An improper fraction is when the numerator of a fraction is greater than the denominator.

A mixed fraction is a fraction written as an integer and a proper fraction, and is understood as the sum of this number and the fractional part. Accordingly, a fraction that does not have an integer part is called a simple fraction. Any mixed fraction can be converted to an improper fraction.

In order to convert a mixed fraction into a common fraction, you need to add the product of the whole part and the denominator to the numerator of the fraction:

How to convert a common fraction to a mixed fraction

In order to convert an ordinary fraction to a mixed fraction, you must:

1. Divide the numerator of a fraction by its denominator
2. The result of division will be the whole part
3. The balance of the department will be the numerator

How to convert a fraction to a decimal

In order to convert a fraction to a decimal, you need to divide its numerator by its denominator.

In order to translate decimal as usual, you need:

How to convert a fraction to a percentage

In order to convert a common or mixed fraction to a percentage, you need to convert it to a decimal fraction and multiply by 100.

How to convert percentages to fractions

In order to convert percentages into fractions, you need to obtain a decimal fraction from the percentage (dividing by 100), then convert the resulting decimal fraction into an ordinary fraction.

Adding Fractions

The algorithm for adding two fractions is as follows:

1. Perform addition of fractions by adding their numerators.

Subtracting Fractions

Algorithm for subtracting two fractions:

1. Convert mixed fractions to ordinary fractions (get rid of the whole part).
2. Reduce fractions to a common denominator. To do this, you need to multiply the numerator and denominator of the first fraction by the denominator of the second fraction, and multiply the numerator and denominator of the second fraction by the denominator of the first fraction.
3. Subtract one fraction from another by subtracting the numerator of the second fraction from the numerator of the first.
4. Find the greatest common divisor (GCD) of the numerator and denominator and reduce the fraction by dividing the numerator and denominator by GCD.
5. If the numerator of the final fraction is greater than the denominator, then select the whole part.

Multiplying fractions

Algorithm for multiplying two fractions:

1. Convert mixed fractions to ordinary fractions (get rid of the whole part).
2. Find the greatest common divisor (GCD) of the numerator and denominator and reduce the fraction by dividing the numerator and denominator by GCD.
3. If the numerator of the final fraction is greater than the denominator, then select the whole part.

Division of fractions

Algorithm for dividing two fractions:

1. Convert mixed fractions to ordinary fractions (get rid of the whole part).
2. To divide fractions, you need to transform the second fraction by swapping its numerator and denominator, and then multiply the fractions.
3. Multiply the numerator of the first fraction by the numerator of the second fraction and the denominator of the first fraction by the denominator of the second.
4. Find the greatest common divisor (GCD) of the numerator and denominator and reduce the fraction by dividing the numerator and denominator by GCD.
5. If the numerator of the final fraction is greater than the denominator, then select the whole part.

Online calculators and converters:

Let us analyze two types of solutions to systems of equations:

1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.

In order to solve the system of equations by substitution method you need to follow a simple algorithm:
1. Express. From any equation we express one variable.
2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
3. Solve the resulting equation with one variable. We find a solution to the system.

To solve system by term-by-term addition (subtraction) method need to:
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.

The solution to the system is the intersection points of the function graphs.

Let us consider in detail the solution of systems using examples.

Example #1:

Let's solve by substitution method

Solving a system of equations using the substitution method

2x+5y=1 (1 equation)
x-10y=3 (2nd equation)

1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, which means that it is easiest to express the variable x from the second equation.
x=3+10y

2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1

3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2

The solution to the equation system is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first point where we expressed it we substitute y.
x=3+10y
x=3+10*(-0.2)=1

It is customary to write points in the first place by writing the variable x, and in the second place by the variable y.
Answer: (1; -0.2)

Example #2:

Let's solve using the term-by-term addition (subtraction) method.

Solving a system of equations using the addition method

3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)

1. We choose a variable, let’s say we choose x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get overall coefficient 6.

3x-2y=1 |*2
6x-4y=2

2x-3y=-10 |*3
6x-9y=-30

2. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y=2

5y=32 | :5
y=6.4

3. Find x. We substitute the found y into any of the equations, let’s say into the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6

The intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)

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