1. Algebra 1
Section 7.7

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

3. 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 a = 580
c + a = 80

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

5. 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.

6. 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

7. 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.

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

9. 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.

10. 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

11. 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.

12. Homework: pp