Algebra 1 Section 7.7

Mixture Problems Mixture problems can also be solved with two variables. quantity × cost per item = total cost

Example 1 c 8.15 8.15c a 6.95 6.95a 80 7.25 580 Cost: Pounds: Number of Pounds Price per Pound Total Cost Cashews c 8.15 8.15c Almonds a 6.95 6.95a Mix 80 7.25 580 Cost: Pounds: 8.15c + 6.95a = 580 c + a = 80

Mix 60 pounds of almonds and 20 pounds of cashews. Example 1 8.15c + 6.95a = 580 c + a = 80 a = 60; c = 20 Mix 60 pounds of almonds and 20 pounds of cashews.

Solving Word Problems Read the problem carefully. Plan by making a table. Solve a system of equations obtained from information in the table. Check your answer.

Example 2 x 60 60x y 84 84y 150 --- 10,848 Pieces: Cost: x + y = 150 Pieces of mail Price per Item Total Cost $0.60 x 60 60x $0.84 y 84 84y Total 150 --- 10,848 Pieces: Cost: x + y = 150 60x + 84y = 10,848

Example 2 x + y = 150 60x + 84y = 10,848 x = 73; y = 77 The school mailed 73 pieces of mail that cost $0.60 and 77 that cost $0.84.

Mixture Problems Some problems involve combining two mixtures of different strengths in order to get a final mixture of a desired strength.

Mixture Problems Adding pure water is like adding a 0% salt solution. Adding pure salt can be thought of as adding a 100% salt solution.

Example 3 x 0.03 0.03x y 5 0.02 0.1 Gallons: Fat: x + y = 5 Number of Gallons Percent of fat Gallons of fat 3% x 0.03 0.03x Fat-free y 2% 5 0.02 0.1 Gallons: Fat: x + y = 5 0.03x = 0.1

Example 2 x + y = 5 0.03x = 0.1 x = 3⅓; y = 1⅔ 3⅓ gal of 3% milk should be mixed with 1⅔ gal of fat-free milk to produce 5 gal of 2% milk.

Homework: pp. 315-318