EXAMPLE 1

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

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

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

4. 6x – 8y = 8 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 xdiffer only in sign. –6x + 14y = 220 3x – 7y = 10 6x – 8y = 8

5. 4 – x= 3 EXAMPLE 2 Use the elimination method STEP 2 6y = –12 Add the revised equations and solve for y. y= –2 STEP 3 Substitute the value of yinto 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. Solve for x.

6. 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. 4 3 The solution is ( , –2) – EXAMPLE 2 Use the elimination method

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

8. 3x – 6y = 9 3. –4x + 7y = –16 for Examples 1 and 2 GUIDED PRACTICE Solve the system using the substitution or the elimination method. ANSWER The solution is (11, 4)