15.2 solving systems of equations by substitution

Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Systems of Linear
Equations and
Inequalities
15

1. Solve linear systems by substitution.
2. Solve special systems by substitution.
3. Solve linear systems with fractions and
decimals.
Objectives
15.2 Solving Systems of Linear Equations by
Substitution

Example Solve the system by the substitution method.
Solve Linear Systems by Substitution
5x + 2y = 2
y = – 3x
5x + 2y = 2
5x + 2(– 3x) = 2
5x + – 6x = 2
– 1x = 2
x = – 2
Let y = – 3x.
Multiply.
Combine like terms.
Multiply by – 1.
The second equation is already solved for y. This equation says that
y = – 3x. Substituting – 3x for y in the first equation gives

y = – 3x = – 3(– 2) = 6
Check that the solution of the given system is (– 2, 6) by
substituting – 2 for x and 6 for y in both equations.
5x + 2y = 2
Let x = – 2.
y = – 3x
Example (cont) Because x = – 2, we find y from the equation
y = – 3x by substituting – 2 for x.
5(– 2) + 2(6) = 2
2 = 2
6 = – 3(– 2)
6 = 6
The solution set of the system is {(– 2, 6)}.
? ?
True True

Example Solve the system
by the substitution method.
3x + 4y = 4
x = 2y + 18
3x + 4y = 4
3(2y + 18) + 4y = 4
6y + 54 + 4y = 4
10y + 54 = 4
Let x = 2y + 18.
Distributive property
Combine like terms.
The second equation gives x in terms of y. Substitute 2y + 18 for x
in the first equation.
10y = – 50 Subtract 54.
y = – 5 Divide by 10.

x = 2y + 18 = 2(– 5) + 18 = 8
Check that the solution of the given system is (8, – 5) by
substituting 8 for x and – 5 for y in both equations.
3x + 4y = 4
Let y = – 5.
x = 2y + 18
Example 2 (cont) Because y = – 5, we find x from the equation
x = 2y + 18 by substituting – 5 for y.
3(8) + 4(– 5) = 4
4 = 4
8 = 2(– 5) + 18
8 = 8
The solution set of the system is {(8, – 5)}.
? ?
True True

Solving a Linear System by Substitution
Step 1 Solve one equation for either variable. If one of
the variables has a coefficient of 1 or –1, choose it because
the substitution method is usually easier.
Step 2 Substitute for that variable in the other equation.
The result should be an equation with just one variable.
Step 3 Solve the equation from Step 2.
Step 4 Find the other value. Substitute the result from Step 3
into the equation from Step 1 and solve for the other
variable.
Step 5 Check the solution in both of the original equations.
Then write the solution set as a set containing an ordered
pair.

3x + 2y = 26
5x – y = 13
5x – y = 13
(1)
(2)
(2)
Step 1 For the substitution method, we must solve one of the
equations for either x or y. Because the coefficient of y in
equation (2) is – 1, we choose equation (2) and solve for y.
5x – y – 5x = 13 – 5x Subtract 5x.
– y = 13 – 5x Combine like terms.
y = – 13 + 5x Multiply by – 1.

Example (cont) Use substitution to solve the system.
3x + 2y = 26
5x – y = 13
3x + 2y = 26
(1)
(2)
(1)
Step 2 Now substitute – 13 + 5x for y in equation (1).
3x + 2(–13 + 5x) = 26 Let y = – 13 + 5x.

3x + 2y = 26
5x – y = 13
(1)
(2)
Step 3 Now solve the equation from Step 2.
3x + 2(–13 + 5x) = 26 From Step 2.
3x – 26 + 10x = 26 Distributive property
13x – 26 = 26 Combine like terms.
13x – 26 + 26 = 26 + 26 Add 26.
13x = 52 Combine like terms.
x = 4 Divide by 13.

3x + 2y = 26
5x – y = 13
(1)
(2)
Step 4 Since y = – 13 + 5x and x = 4, y = – 13 + 5(4) = 7.
Step 5 Check that (4, 7) is the solution.
3x + 2y = 26 5x – y = 13
3(4) + 2(7) = 26
12 + 14 = 26
26 = 26
5(4) – 7 = 13
20 – 7 = 13
13 = 13
Since both results are true, the solution set of the system is {(4, 7)}.
? ?
True True
? ?

Example Use substitution to solve the system.
Solve Special Systems by Substitution
y = 2x – 7
3y – 6x = –6
3y – 6x = –6
(1)
(2)
(2)
Substitute 2x – 7 for y in equation (2).
3(2x – 7) – 6x = –6 Let y = 2x – 7.
6x – 21 – 6x = –6 Distributive property
– 21 = –6 False.
This false result means that the equations in the system have graphs
that are parallel lines. The system is inconsistent and the solution set
is 0 .

x
y
y = 2x – 7
3y – 6x = – 6.
(1)
(2)
(2)
(1)Example (cont)

3 = 5x – y
– 4y – 12 = – 20x
– 4y – 12 = – 20x
(1)
(2)
(2)
Begin by solving equation (1) for y to get y = 5x – 3. Substitute
5x – 3 for y in equation (2) and solve the resulting equation.
– 4(5x – 3) – 12 = – 20x Let y = 5x – 3.
– 20x + 12 – 12 = – 20x Distributive property
0 = 0 Add 20x; combine terms.
This true result means that every solution of one equation is also a
solution of the other, so the system has an infinite number of solutions.

x
y
Example (cont)
(1)
(2)
3 = 5x – y
– 4y – 12 = – 20x
The system has an infinite number of solutions – all the ordered pairs
corresponding to points that lie on the common graph.

Solve Linear Systems with Fractions and Decimals
(1)
(2)
Clear equation (1) of fractions by multiplying each side by 3.
x + 5y = 12
3
x – y = –1
4
1
8
13
8
Multiply by 3.x + 5y = 12
3
3 3
x + 5y = 12
3
3 33 Distributive property
2x + 15y = 3 (3)

Example (cont) Solve the system by the substitution method.
(1)
(2)
Now, clear equation (2) of fractions by multiplying each side by 8.
x + 5y = 12
3
x – y = –1
4
1
8
13
8
Multiply by 8.
Distributive property
2x – y = – 13 (4)
x – y = –1
4
1
8
13
8
8 8
x – y = –1
4
1
8
13
8
8 8 8

Example (cont) The given system of equations has been
simplified to the equivalent system.
To solve this system by substitution, equation (4) can be solved for y.
(1)
(2)
x + 5y = 12
3
x – y = –1
4
1
8
13
8
Subtract 2x.
2x – y = – 13
2x + 15y = 3 (3)
(4)
2x – y = – 13 (4)
2x – y – 2x = – 13 – 2x
Combine like terms.– y = – 13 – 2x
Multiply by – 1.y = 13 + 2x

Now substitute 13 + 2x for y in equation (3).
(1)
(2)
x + 5y = 12
3
x – y = –1
4
1
8
13
8
Let y = 13 + 2x.
2x – y = – 13
2x + 15y = 3 (3)
(4)
2x + 15y = 3 (3)
2x + 15(13 + 2x) = 3
Distributive property2x + 195 + 30x = 3
Combine like terms.32x + 195 = 3

Now substitute 13 + 2x for y in equation (3).
(1)
(2)
x + 5y = 12
3
x – y = –1
4
1
8
13
8
2x – y = – 13
2x + 15y = 3 (3)
(4)
Combine like terms.32x + 195 = 3
Subtract 195.32x + 195 – 195 = 3 – 195
Combine like terms.32x = – 192
Divide by 32.x = – 6

Substitute – 6 for x in y = 13 + 2x to get
(1)
(2)
x + 5y = 12
3
x – y = –1
4
1
8
13
8
2x – y = – 13
2x + 15y = 3 (3)
(4)
Check by substituting – 6 for x and 1 for y in both of the original
equations. The solution set is {(– 6, 1)}.
y = 13 + 2(– 6) = 1.

15.2 solving systems of equations by substitution

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