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

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

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

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

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

Example #1:

Let's solve by 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, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y

2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1

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

The solution of 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 paragraph where we expressed we substitute y there.
x=3+10y
x=3+10*(-0.2)=1

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

Example #2:

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

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

1. Select a variable, let's say we select 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. Multiply the first equation by 2 and the second by 3 to get overall coefficient 6.

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

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

2. From the first equation, subtract the second 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 in any of the equations, let's say in 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 point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)

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A system of linear equations with two unknowns is two or more linear equations for which it is necessary to find all their common solutions. We will consider systems of two linear equations with two unknowns. A general view of a system of two linear equations with two unknowns is shown in the figure below:

( a1*x + b1*y = c1,
( a2*x + b2*y = c2

Here x and y are unknown variables, a1, a2, b1, b2, c1, c2 are some real numbers. A solution to a system of two linear equations with two unknowns is a pair of numbers (x, y) such that if these numbers are substituted into the equations of the system, then each of the equations of the system turns into a true equality. Consider one of the ways to solve a system of linear equations, namely the substitution method.

Algorithm for solving by substitution method

Algorithm for solving a system of linear equations by substitution method:

1. Choose one equation (it is better to choose the one where the numbers are smaller) and express one variable from it through another, for example, x through y. (you can also y through x).

2. Substitute the resulting expression instead of the corresponding variable in another equation. Thus, we get a linear equation with one unknown.

3. We solve the resulting linear equation and get the solution.

4. We substitute the obtained solution into the expression obtained in the first paragraph, we obtain the second unknown from the solution.

5. Verify the resulting solution.

Example

To make it more clear, let's solve a small example.

Example 1 Solve the system of equations:

(x+2*y=12
(2*x-3*y=-18

Solution:

1. From the first equation of this system, we express the variable x. We have x= (12 -2*y);

2. Substitute this expression into the second equation, we get 2*x-3*y=-18; 2*(12 -2*y) - 3*y = -18; 24 - 4y - 3*y = -18;

3. We solve the resulting linear equation: 24 - 4y - 3*y = -18; 24-7*y=-18; -7*y = -42; y=6;

4. We substitute the result obtained into the expression obtained in the first paragraph. x= (12 -2*y); x=12-2*6 = 0; x=0;

5. We check the obtained solution, for this we substitute the numbers found in the original system.

(x+2*y=12;
(2*x-3*y=-18;

{0+2*6 =12;
{2*0-3*6=-18;

{12 =12;
{-18=-18;

We got the correct equalities, therefore, we correctly found the solution.

Solving systems of equations by the substitution method

Recall what a system of equations is.

A system of two equations with two variables is two equations written one below the other, united by a curly bracket. Solving a system means finding a pair of numbers that will be a solution to both the first and second equations at the same time.

In this lesson, we will get acquainted with such a way of solving systems as the substitution method.

Let's look at the system of equations:

You can solve this system graphically. To do this, we will need to build graphs of each of the equations in one coordinate system, converting them to the form:

Then find the coordinates of the intersection point of the graphs, which will be the solution of the system. But the graphical method is far from always convenient, because. differs in low accuracy, and even inaccessibility at all. Let's take a closer look at our system. Now it looks like:

It can be seen that the left-hand sides of the equations are equal, which means that the right-hand sides must also be equal. Then we get the equation:

This is a familiar one-variable equation that we know how to solve. Let's transfer the unknown terms to the left side, and the known ones - to the right, not forgetting to change the signs +, - when transferring. We get:

Now we substitute the found value of x into any equation of the system and find the value of y. In our system, it is more convenient to use the second equation y \u003d 3 - x, after substitution we get y \u003d 2. Now let's analyze the work done. First, in the first equation, we expressed the variable y in terms of the variable x. Then the resulting expression - 2x + 4 was substituted into the second equation instead of the variable y. Then we solved the resulting equation with one variable x and found its value. And in conclusion, we used the found value of x to find another variable y. Here the question arises: was it necessary to express the variable y from both equations at once? Of course not. We could express one variable in terms of another only in one equation of the system and use it instead of the corresponding variable in the second. Moreover, any variable from any equation can be expressed. Here the choice depends solely on the convenience of the account. Mathematicians called this procedure the algorithm for solving systems of two equations with two variables using the substitution method. Here's what it looks like.

1. Express one of the variables in terms of the other in one of the equations of the system.

2. Substitute the resulting expression instead of the corresponding variable in another equation of the system.

3. Solve the resulting equation with one variable.

4. Substitute the found value of the variable into the expression obtained in the first paragraph and find the value of another variable.

5. Write down the answer as a pair of numbers that were found in the third and fourth steps.

Let's look at one more example. Solve the system of equations:

Here it is more convenient to express the variable y from the first equation. We get y \u003d 8 - 2x. The resulting expression must be substituted for y in the second equation. We get:

We write this equation separately and solve it. Let's open the parentheses first. We get the equation 3x - 16 + 4x \u003d 5. Let's collect the unknown terms on the left side of the equation, and the known ones on the right side and give similar terms. We get the equation 7x \u003d 21, hence x \u003d 3.

Now, using the found value of x, you can find:

Answer: a pair of numbers (3; 2).

Thus, in this lesson, we have learned to solve systems of equations with two unknowns in an analytical, accurate way, without resorting to dubious graphical methods.

List of used literature:

1. Mordkovich A.G., Algebra grade 7 in 2 parts, Part 1, Textbook for educational institutions / A.G. Mordkovich. - 10th ed., revised - Moscow, "Mnemosyne", 2007.
2. Mordkovich A.G., Algebra grade 7 in 2 parts, Part 2, Task book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, Mnemosyne, 2007.
3. HER. Tulchinskaya, Algebra Grade 7. Blitz survey: a guide for students of educational institutions, 4th edition, revised and supplemented, Moscow, Mnemozina, 2008.
4. Alexandrova L.A., Algebra Grade 7. Thematic verification work in new form for students of educational institutions, edited by A.G. Mordkovich, Moscow, "Mnemosyne", 2011.
5. Aleksandrova L.A. Algebra 7th grade. Independent work for students of educational institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, "Mnemosyne", 2010.

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. The substitution method makes it easy to solve systems of linear equations of any complexity. The essence of the method is that, using the first expression of the system, we express "y", and then we substitute the resulting expression into the second equation of the system instead of "y". Since the equation already contains not two unknowns, but only one, we can easily find the value of this variable, and then use it to determine the value of the second.

Suppose we are given a system of linear equations of the following form:

\[\left\(\begin(matrix) 3x-y-10=0\\ x+4y-12=0 \end(matrix)\right.\]

Express \

\[\left\(\begin(matrix) 3x-10=y\\ x+4y-12=0 \end(matrix)\right.\]

Substitute the resulting expression into the 2nd equation:

\[\left\(\begin(matrix) y=3x-10\\ x+4(3x-10)-12=0 \end(matrix)\right.\]

Find the value \

Simplify and solve the equation by opening the brackets and taking into account the rules for transferring terms:

Now we know the value of \ Let's use this to find the value of \

Answer: \[(4;2).\]

Where can I solve a system of equations online using the substitution method?

You can solve the system of equations on our website. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver. You can also learn how to solve the equation on our website. And if you have any questions, you can ask them in our Vkontakte group.

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