1. 3.2 Solving Systems Algebraically
Substitution & Elimination

3. Equation 1 Equation 2 Revised Equation 2 EXAMPLE 1
Use the substitution method
Solve the system using the substitution method.
2x + 5y = –5
Equation 1
x + 3y = 3
Equation 2
SOLUTION
STEP 1
Solve Equation 2 for x.
x = –3y + 3
Revised Equation 2

4. Write revised Equation 2.
EXAMPLE 1
Use the substitution method
STEP 2
Substitute the expression for x into Equation 1 and solve for y.
2x +5y = –5
Write Equation 1.
2( ) + 5y = –5
–3y + 3
Substitute –3y + 3 for x.
y = 11 11
Solve for y.
STEP 3
Substitute the value of y into revised Equation 2 and solve for x.
x = –3y + 3
Write revised Equation 2.
x = –3( ) + 3
Substitute 11 for y.
x = –30
Simplify.

5. Substitute for x and y. Solution checks. EXAMPLE 1
Use the substitution method
The solution is (– 30, 11).
ANSWER
CHECK Check the solution by substituting into the original equations.
2(–30) + 5(11) –5 = ?
Substitute for x and y. = ?
–30 + 3(11) 3
–5 = –5
Solution checks.
3 = 3

6. Equation 1 Equation 2 EXAMPLE 2 Use the elimination method
Solve the system using the elimination method.
3x – 7y = 10
Equation 1
6x – 8y = 8
Equation 2
SOLUTION
STEP 1
Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign.
–6x + 14y = -20
3x – 7y = 10
6x – 8y = 8
6x – 8y = 8
STEP 2
6y = –12
Add the revised equations and solve for y.
y = –2

7. Write Equation 1. Substitute –2 for y. Simplify. Solve for x.
EXAMPLE 2
Use the elimination method
STEP 3
Substitute the value of y into one of the original equations. Solve for x.
3x – 7y = 10
Write Equation 1.
3x – 7(–2) = 10
Substitute –2 for y.
3x + 14 = 10
Simplify.
x = 4 3 –
Solve for x.

8. EXAMPLE 2
Use the elimination method
The solution is ( , –2) 4 3 –
ANSWER
CHECK
You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.

9. GUIDED PRACTICE
for Examples 1 and 2
Solve the system using the substitution for 1 and the elimination method for 2. 1.
4x + 3y = –2
x + 5y = –9 2.
3x + 3y = –15
5x – 9y = 3
The solution is (1,–2).
ANSWER
The solution is ( -3 , –2)
ANSWER

10. GUIDED PRACTICE
for Examples 1 and 2
Solve the system using the substitution or the elimination method. 3.
3x – 6y = 9
–4x + 7y = –16
ANSWER
The solution is (11, 4)

11. Write first equation. Solve for x. EXAMPLE 4
Solve linear systems with many or no solutions
Solve the linear system. a.
x – 2y = 4
3x – 6y = 8 b.
4x – 10y = 8
– 14x + 35y = – 28
SOLUTION a.
Because the coefficient of x in the first equation is 1, use the substitution method.
Solve the first equation for x.
x – 2y = 4
Write first equation.
x = 2y + 4
Solve for x.

12. Write second equation. Substitute 2y + 4 for x. Simplify. EXAMPLE 4
Solve linear systems with many or no solutions
Substitute the expression for x into the second equation.
3x – 6y = 8
Write second equation.
3(2y + 4) – 6y = 8
Substitute 2y + 4 for x.
12 = 8
Simplify.
Because the statement 12 = 8 is never true, there is no solution.
ANSWER

13. EXAMPLE 4
Solve linear systems with many or no solutions b.
Because no coefficient is 1 or – 1, use the elimination method.
Multiply the first equation by 7 and the second equation by 2.
4x – 10y = 8
28x – 70y = 56
– 14x + 35y = – 28
– 28x + 70y = – 56
Add the revised equations.
0 = 0
ANSWER
Because the equation 0 = 0 is always true, there are infinitely many solutions.

14. Write second equation. Solve for y. GUIDED PRACTICE for Example 4
Solve the linear system using any algebraic method. 5.
12x – 3y = – 9
– 4x + y = 3
SOLUTION
Because the coefficient of y in the second equation is y, use the substitution method.
Solve the 2nd equation for y.
– 4x + y = 3
Write second equation.
y = 4x + 3
Solve for y.

15. Write second equation. Substitute 4x + 3 for y. Simplify.
GUIDED PRACTICE
for Example 4
Substitute the expression for y into the first equation.
12x – 3y = – 9
Write second equation.
12x – 3(4y + 3) = – 9
Substitute 4x + 3 for y.
0 = 0
Simplify.
Because the equation 0 = 0 is always true, there are infinitely many solutions.
ANSWER

16. GUIDED PRACTICE
for Example 4
Solve the linear system using any algebraic method. 6.
6x + 15y = – 12
– 2x – 5y = 9
Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 3
6x + 15y = – 12
6x + 15y = – 12
– 2x – 5y = 9 3
– 6x – 15y = 27
0 = 15
Add the revised equations.
ANSWER
Because the statement 0 = 15 is never true, there are no solutions.

17. Write second equation. Solve for x. GUIDED PRACTICE for Example 4
Solve the linear system using any algebraic method. 7.
5x + 3y = 20
– x – y = – 4 3 5
Because the coefficient of x in the first equation is – 1, use the substitution method.
Solve the 2nd equation for x.
– x – y = – 4 3 5
Write second equation.
x = – y + 4 3 5
Solve for x.

18. Write first equation. Substitute for x. Simplify. GUIDED PRACTICE
for Example 4
Substitute the expression for x into the first equation.
5x + 3y = 20
Write first equation. 5
( – y + 4 ) + 3y = 20 3
Substitute for x. 5 3
– y + 4
20 = 20
Simplify.
Because the statement 20 = 20 is always true, there are infinitely many solution.
ANSWER

19. GUIDED PRACTICE
for Example 4
Solve the linear system using any algebraic method. 8.
12x – 2y = 21
3x + 12y = – 4
Because no coefficient is 1 or – 1, use the elimination method.
Multiply the second equation by 6
12x – 2y = 21 6
72x – 12y = 126
3x + 12y = – 4
3x + 12y = – 4
Add the revised equations.
75x = 122 x
122 75 =

20. Write second equation. 122 Substitute for x. 75 Simplify.
GUIDED PRACTICE
for Example 4
Substitute the expression for x into the second equation.
3x + 12y = – 4
Write second equation.
3( ) + 12y = 8
122 75
Substitute for x.
122 75
y =
– 37 50
Simplify.
The solution is ( , )
ANSWER
122 75
– 37 50

21. GUIDED PRACTICE
for Example 4
Solve the linear system using any algebraic method.
5x + 5y = 5
10.
5x + 3y = 4.2 9.
8x + 9y = 15
5x – 2y = 17
(3, –1)
ANSWER
(0.6, 0.4)
ANSWER