1. Chapter 3 Section 2

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

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

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

7. EXAMPLE 2 Use the elimination method Solve the system using the elimination method. 3x – 7y = 10 6x – 8y = 8 Equation 1 Equation 2 SOLUTION Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign. STEP 1 3x – 7y = 10 6x – 8y = 8 –6x + 14y = 220 6x – 8y = 8

8. EXAMPLE 2 Use the elimination method STEP 2 Add the revised equations and solve for y. 6y = –12 y = –2 STEP 3 Substitute the value of y into one of the original equations. Solve for x. 3x – 7y = 10 3x – 7(–2) = 10 3x + 14 = 10 x =x = 4 3 – Solve for x. Simplify. Substitute –2 for y. Write Equation 1.

9. 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.

10. 1.4x + 3y = –2 x + 5y = –9 Solve the system using the substitution or the elimination method. GUIDED PRACTICE for Examples 1 and 2 The solution is (1,–2). ANSWER 2. 3x + 3y = –15 5x – 9y = 3 The solution is (, –2) 3 – ANSWER

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

12. EXAMPLE 3 Standardized Test Practice SOLUTION Write verbal models for this situation.

13. Equation 1 EXAMPLE 3 Standardized Test Practice Equation 2

14. EXAMPLE 3 Standardized Test Practice STEP 2 Write a system of equations. Equation 1 Equation 2 5x + 7y = 2500 8x + 12y = 4200 Total cost for all T-shirts Total revenue from all T-shirts sold STEP 3 Solve the system using the elimination method. Multiply Equation 1 by –8 and Equation 2 by 5 so that the coefficients of x differ only in sign. 5x + 7y = 2500 8x + 12y = 4200 – 40x – 56y = – 20,000 40x + 60y = 21,000 Add the revised equations and solve for y. 4y = 1000 y = 250

15. EXAMPLE 3 Standardized Test Practice Substitute the value of y into one of the original equations and solve for x. 5x + 7y = 2500 5x + 7(250) = 2500 5x + 1750 = 2500 x = 150 Write Equation 1. Substitute 250 for y. Simplify. Solve for x. The school sold 150 short sleeve T-shirts and 250 long sleeve T-shirts. ANSWER The correct answer is C.

16. GUIDED PRACTICE for Example 3 4. WHAT IF? In Example 3, suppose the school spends a total of $3715 on T-shirts and sells all of them for $6160. How many of each type of T-shirt are sold? The school sold 365 short sleeve T-shirts and 270 long sleeve T-shirts. ANSWER

18. 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 x = 2y + 4 Write first equation. Solve for x.

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

20. 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 – 14x + 35y = – 28 28x – 70y = 56 – 28x + 70y = – 56 Add the revised equations. 0 = 0 ANSWER Because the equation 0 = 0 is always true, there are infinitely many solutions.

21. GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 5.12x – 3y = – 9 –4x + y = 3 infinitely many solutions ANSWER 6.6x + 15y = – 12 – 2x – 5y = 9 ANSWER no solutions

22. GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 8.12x – 2y = 21 3x + 12y = – 4 The solution is (, ) ANSWER 122 75 – 37 50 7. 5x + 3y = 20 – x – y = – 4 3 5 infinitely many solutions ANSWER

23. GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 9.8x + 9y = 15 5x – 2y = 17 5x + 5y = 510. 5x + 3y = 4.2 (3, –1) ANSWER (0.6, 0.4) ANSWER