Warm-Up Exercises Solve the system by substitution. 2x2x y = – 1. 3x3x – y – = 1 4x+y = 2. 72x2x+3y3y = – ANSWER () 1, 2 – ANSWER () 1, 3

Warm-Up Exercises Three T-shirts plus 4 sweatshirts cost $96. A T-shirt costs $3 less than a sweatshirt. How much does a T-shirt cost? 3. ANSWER $12

Example 1 Multiply One Equation Solve the linear system using the linear combination method. Equation 1 63y3y2x2x = – Equation 2 85y5y4x4x = – SOLUTION STEP 1Multiply the first equation by 2 so that the coefficients of x differ only in sign. – 63y3y2x2x = – 85y5y4x4x = –85y5y4x4x = – 126y6y4x4x = +–– 4y = –

Example 1 Multiply One Equation STEP 2Add the revised equations and solve for y. STEP 3Substitute 4 for y in one of the original equations and solve for x. – Write Equation 1. 63y3y2x2x = – Substitute 4 for y. 62x2x = – () 4 – 3 – Simplify. 6122x2x = + Subtract 12 from each side. 62x2x = – Solve for x. 3x = – 4y = –

Example 1 Multiply One Equation STEP 4Check by substituting 3 for x and 4 for y in the original equations. –– ANSWER The solution is. () 3,3, –– 4

Example 2 Multiply Both Equations Solve the system using the linear combination method. Equation 1 Equation 2 SOLUTION STEP 1Multiply the first equation by 2 and the second equation by y7x7x = –– 148y8y5x5x = +– 2212y7x7x = –– 148y8y5x5x = + – 4424y14x = –– 4224y15x = + – 2x = ––

Example 2 Multiply Both Equations STEP 2Add the revised equations and solve for x. 2x = 2x = –– STEP 3Substitute 2 for x in one of the original equations and solve for y. 148y8y5x5x = + – Write Equation y8y = + Substitute 2 for x. – () y8y10 = + – Multiply. 3y = Solve for y.

Example 2 Multiply Both Equations STEP 4Check by substituting 2 for x and 3 for y in the original equations. ANSWER The solution is (2, 3).

Example 3 A Linear System with No Solution Solve the system using the linear combination method. Equation 1 Equation 2 74y4y2x2x = – 128y8y4x4x = +–– SOLUTION Multiply the second equation by 2 so that the coefficients of y differ only in sign. 8y8y – 124x4x = + 148y8y4x4x = –– 74y4y2x2x = – 128y8y4x4x = +–– 20 = Add the revised equations.

Example 3 A Linear System with No Solution ANSWER Because the statement 0 2 is false, there is no solution. =

Checkpoint ANSWER infinitely many solutions 1. Solve the system using the linear combination method. Solve a Linear System 54y4yx = – 1y2x2x = + ANSWER (1, 1) – 2. 4y2x2x = – 82y2y4x4x = – 3. 22y2y3x3x = – 13y3y4x4x = – ANSWER (4, 5 )

Checkpoint ANSWER if you get a false equation; if you get a true equation 4. How can you tell when a system has no solution? infinitely many solutions? Solve a Linear System

Use a Linear System as a Model Example 4 Catering A customer hires a caterer to prepare food for a party of 30 people. The customer has $80 to spend on food and would like there to be a choice of sandwiches and pasta. A $40 pan of pasta contains 10 servings, and a $10 sandwich tray contains 5 servings. The caterer must prepare enough food so that each person receives one serving of either food. How many pans of pasta and how many sandwich trays should the caterer prepare?

Use a Linear System as a Model Example 4 SOLUTION VERBAL MODEL Servings per pan Pans of pasta Sandwich trays = + Servings per sandwich tray Servings needed Price per pan Pans of pasta Sandwich trays = + Price per tray Money to spend on food

Use a Linear System as a Model Example 4 LABELS Servings per pan of pasta 10 = (servings) Pans of pasta p = (pans) Servings per sandwich tray 5 = (trays) Sandwich trays s = Servings needed 30 = Price per pan of pasta 40 = (dollars) Price per sandwich tray 10 = (dollars) Money to spend on food 80 = (dollars) (servings)

Use a Linear System as a Model Example 4 ALGEBRAIC MODEL Equation 1 (servings needed) 30 = 10p + 5s5s Equation 2 (money to spend on food) 80 = 40p + 10s Multiply Equation 1 by 2 so that the coefficients of s differ only in sign. – 30 = 10p + 5s5s 80 = 40p + 10s = 20p10s –– 60 – 80 = 40p + 10s 20 = 20p Add the revised equations and solve for p. 1 = p

Use a Linear System as a Model Example 4 ANSWER The caterer should make 1 pan of pasta and 4 sandwich trays. Substitute 1 for p in one of the original equations and solve for s. Write Equation = 10p + 5s5s Substitute 1 for p. 30 = s5s () 1 Subtract 10 from each side. 20 = 5s5s 4 = s Solve for s.

Checkpoint Solve a Linear System 5. Another customer asks the caterer in Example 4 to plan a party for 40 people. This customer also wants both sandwiches and pasta and has $120 to spend. How many pans of pasta and how many sandwich trays should the caterer prepare? ANSWER 2 pans of pasta and 4 sandwich trays