Algebra 1 Section 7 7 Mixture Problems Mixture

Algebra 1 Section 7. 7

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

Example 1 Number of Price per Total Pounds Pound Cost Cashews Almonds Mix Cost: Pounds: c a 80 8. 15 6. 95 7. 25 8. 15 c 6. 95 a 580 8. 15 c + 6. 95 a = 580 c + a = 80

Example 1 8. 15 c + 6. 95 a = 580 c + a = 80 a = 60; c = 20 Mix 60 pounds of almonds and 20 pounds of cashews.

Solving Word Problems 1. Read the problem carefully. 2. Plan by making a table. 3. Solve a system of equations obtained from information in the table. 4. Check your answer.

Example 2 Pieces of mail Price per Item Total Cost x y 150 60 84 --- 60 x 84 y 10, 848 $0. 60 $0. 84 Total Pieces: Cost: x + y = 150 60 x + 84 y = 10, 848

Example 2 x + y = 150 60 x + 84 y = 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 Number of Percent Gallons of fat 3% Fat-free 2% Gallons: Fat: x y 5 0. 03 0 0. 02 0. 03 x 0 0. 1 x+y=5 0. 03 x = 0. 1

Example 2 x+y=5 0. 03 x = 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