Solving systems of equations using the substitution method

Let us remember what a system of equations is.

A system of two equations with two variables is two equations written below each other, joined by a curly brace. Solving a system means finding a pair of numbers that will solve both the first and second equations at the same time.

In this lesson we will get acquainted with such a method 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 construct graphs of each of the equations in one coordinate system, transforming them to the form:

Then find the coordinates of the intersection point of the graphs, which will be the solution to the system. But the graphical method is not always convenient, because differs in low accuracy, or even inaccessibility. Let's try to take a closer look at our system. Now it looks like:

You can notice that the left sides of the equations are equal, which means the right sides must also be equal. Then we get the equation:

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

Now let’s 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 = 3 - x; after substitution we get y = 2. Now let’s analyze the work done. First, in the first equation we expressed the y variable in terms of the x variable. 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 finally, 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 in only one equation of the system and use it instead of the corresponding variable in the second. Moreover, you can express any variable from any equation. Here the choice depends solely on the convenience of the account. Mathematicians called this procedure an 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 another in one of the equations of the system.

2.Substitute the resulting expression instead of the corresponding variable into 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 step one and find the value of another variable.

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

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

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

Let's write this equation separately and solve it. First, let's open the brackets. We get the equation 3x - 16 + 4x = 5. Let's collect the unknown terms on the left side of the equation, and the known ones on the right, and present similar terms. We get the equation 7x = 21, hence x = 3.

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

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

Thus, in this lesson we 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 7th grade in 2 parts, Part 1, Textbook for general education institutions / A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007.
2. Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007.
3. HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, Mnemosyne, 2008.
4. Alexandrova L.A., Algebra 7th grade. Thematic test papers in a new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011.
5. Alexandrova L.A. Algebra 7th grade. Independent works for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010.

Let's figure it out How to solve systems of equations using the substitution method?

1) Express the unknown from the first or second equation of the system X or at(whatever is more convenient for us);

2) Substitute into another equation (into the one from which the unknown was not expressed) instead of the unknown X or at(if expressed X, substitute instead X; if expressed at, substitute instead at) the resulting expression;

3) Solve the equation that we received. We find X or y;

4) Substitute the resulting value of the unknown and find the second unknown.

The rule is written. Now let's try to apply it to solve a system of equations.

Example 1.

Let's take a close look at the system of equations. We note that from the first equation it is easier to express at.

We express at:

–2у = 11 – 3х

y = (11 – 3x)/(–2)

y = –5.5 + 1.5x

Now let’s carefully substitute into the second equation instead at expression –5.5 + 1.5x.

We get: 4x – 5(–5.5 + 1.5x) = 3

Let's solve this equation:

4x + 27.5 – 7.5x = 3

–3.5x = 3 – 27.5

–3.5x = –24.5

x = –24.5/(–3.5)

We substitute y = – 5.5 + 1.5x into the expression instead X the value we found. We get:

y = – 5.5+ 1.5 7 = –5.5 + 10.5 = 5.

Answer: (7; 5)

It’s interesting, but if we express from the first equation not at, A X, will the answer change?

Let's try to express X from the first equation.

x = (11 + 2y)/3

Let's substitute instead X into the second equation the expression (11 +2у)/3, we get an equation with one unknown and solve it.

4(11 + 2у)/3 – 5у = ​​3, multiply both sides of the equation by 3, we get

4(11 + 2y) – 15y=9

44 + 8у – 15у = 9

–7у = 9 – 44

y = –35/(–7)

We find the variable x by substituting 5 into the expression x = (11 +2y)/3.

x = (11 +2 5)/3 = (11+10)/3 = 21/3 = 7

Answer: (7; 5)

As you can see, the answer was the same. If you are careful and careful, then no matter what variable you express - X or at, you will get the correct answer.

Quite often students ask: “ Are there other ways to solve systems besides addition and substitution?»

There is some modification of the substitution method - way to compare unknowns .

1) It is necessary to express the same unknown from each equation of the system through the second.

2) The resulting unknowns are compared and an equation with one unknown is obtained.

3) Find the value of one unknown.

4) Substitute the resulting value of the unknown and find the second unknown.

Example 2. Solve system of equations

From two equations we express the variable X through at.

From the first equation we obtain x = (13 – 6y) / 5, and from the second equation x = (–1 – 18y) / 7.

Comparing these expressions, we obtain an equation with one unknown and solve it:

(13 – 6y) / 5 = (–1 – 18y) / 7

7 (13 – 6y) = 5 (–1 – 18y)

91 – 42у = –5 – 90у

–42у + 90у = –5 – 91

y = – 96 / 48

Unknown X let's find by substituting the value at into one of the expressions for X.

(13 – 6(– 2)) / 5= (13+12) / 5 = 25/5 = 5

Answer: (5; –2).

I think that you will succeed too. If you have any questions, come to my lessons.

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Usually the equations of the system are written in a column one below the other and combined with a curly brace

A system of equations of this type, where a, b, c- numbers, and x, y- variables are called system of linear equations.

When solving a system of equations, properties that are valid for solving equations are used.

Solving a system of linear equations using the substitution method

Let's look at an example

1) Express the variable in one of the equations. For example, let's express y in the first equation, we get the system:

2) Substitute into the second equation of the system instead of y expression 3x-7:

3) Solve the resulting second equation:

4) We substitute the resulting solution into the first equation of the system:

A system of equations has a unique solution: a pair of numbers x=1, y=-4. Answer: (1; -4) , written in brackets, in the first position the value x, On the second - y.

Solving a system of linear equations by addition

Let's solve the system of equations from the previous example addition method.

1) Transform the system so that the coefficients for one of the variables become opposite. Let's multiply the first equation of the system by "3".

2) Add the equations of the system term by term. We rewrite the second equation of the system (any) without changes.

3) We substitute the resulting solution into the first equation of the system:

Solving a system of linear equations graphically

The graphical solution of a system of equations with two variables comes down to finding the coordinates of the common points of the graphs of the equations.

The graph of a linear function is a straight line. Two lines on a plane can intersect at one point, be parallel, or coincide. Accordingly, a system of equations can: a) have a unique solution; b) have no solutions; c) have an infinite number of solutions.

2) The solution to the system of equations is the point (if the equations are linear) of the intersection of the graphs.

Graphic solution of the system

Method for introducing new variables

Changing variables can lead to solving a simpler system of equations than the original one.

Consider the solution of the system

Let's introduce the replacement , then

Let's move on to the initial variables

Special cases

Without solving a system of linear equations, you can determine the number of its solutions from the coefficients of the corresponding variables.

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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 in two unknowns. The general view of a system of two linear equations with two unknowns is presented 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 in two unknowns is a pair of numbers (x,y) such that if we substitute these numbers 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.

Solution algorithm by substitution method

Algorithm for solving a system of linear equations using the substitution method:

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

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

3. Solve the resulting linear equation and obtain a solution.

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

5. Check 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. Solve the resulting linear equation: 24 - 4y - 3*y = -18; 24-7*y =-18; -7*y = -42; y=6;

4. 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 resulting solution; to do this, we substitute the found numbers into 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 found the solution correctly.