The document provides examples of using the substitution method to solve systems of linear equations. In Example 1, the system is solved to get the solution (1, 5). Example 2 is similarly worked through, yielding the solution (-2, 2). The guided practice exercises ask to use substitution to solve three additional systems of linear equations.

1. EXAMPLE 1 Use the substitution method Solve the linear system: y = 3 x + 2 Equation 2 Equation 1 x + 2 y = 11 Solve for y . Equation 1 is already solved for y . SOLUTION STEP 1

2. EXAMPLE 1 Use the substitution method 7 x + 4 = 11 Simplify. 7 x = 7 Subtract 4 from each side. x = 1 Divide each side by 7. Substitute 3 x + 2 for y . x + 2( 3 x + 2 ) = 11 Write Equation 2 . x + 2 y = 11 Substitute 3 x + 2 for y in Equation 2 and solve for x . STEP 2

3. EXAMPLE 1 Use the substitution method Substitute 1 for x in the original Equation 1 to find the value of y . y = 3 x + 2 = 3( 1 ) + 2 = 3 + 2 = 5 STEP 3 ANSWER The solution is (1, 5).

4. GUIDED PRACTICE CHECK y = 3 x + 2 Substitute 1 for x and 5 for y in each of the original equations. x + 2 y = 11 EXAMPLE 1 Use the substitution method 5 = 3( 1 ) + 2 ? 5 = 5 1 + 2 ( 5 ) = 11 ? 11 = 11

5. EXAMPLE 2 Use the substitution method Solve the linear system : x – 2 y = –6 Equation 1 4 x + 6 y = 4 Equation 2 SOLUTION Solve Equation 1 for x . x – 2 y = –6 Write original Equation 1 . x = 2 y – 6 Revised Equation 1 STEP 1

6. EXAMPLE 2 Use the substitution method Substitute 2 y – 6 for x in Equation 2 and solve for y . 4 x + 6 y = 4 Write Equation 2 . 4( 2 y – 6 ) + 6 y = 4 Substitute 2 y – 6 for x . Distributive property 8 y – 24 + 6 y = 4 14 y – 24 = 4 Simplify. 14 y = 28 Add 24 to each side. y = 2 Divide each side by 14. STEP 2

7. EXAMPLE 2 Use the substitution method Substitute 2 for y in the revised Equation 1 to find the value of x . x = 2 y – 6 Revised Equation 1 x = 2( 2 ) – 6 Substitute 2 for y . x = –2 Simplify. STEP 3 ANSWER The solution is (–2, 2).

8. GUIDED PRACTICE CHECK Substitute –2 for x and 2 for y in each of the original equations. 4 x + 6 y = 4 Equation 1 Equation 2 x – 2 y = –6 EXAMPLE 2 Use the substitution method 4( –2 ) + 6 ( 2 ) = 4 ? – 2 – 2 ( 2 ) = –6 ? –6 = –6 4 = 4

9. EXAMPLE 1 Use the substitution method Solve the linear system using the substitution method. 3 x + y = 10 GUIDED PRACTICE for Examples 1 and 2 y = 2 x + 5 1. ANSWER (1, 7)

10. EXAMPLE 2 Use the substitution method x + 2 y = –6 GUIDED PRACTICE for Examples 1 and 2 Solve the linear system using the substitution method. x – y = 3 2. ANSWER (0, – 3 )

11. EXAMPLE 2 Use the substitution method –2 x + 4 y = 0 GUIDED PRACTICE for Examples 1 and 2 Solve the linear system using the substitution method. 3 x + y = –7 3. ANSWER (–2, –1)