Algebra 1 Section 3.8

Mixture Problems Mixture problems involve the combination of different substances. Making a table can be quite helpful.

Example 1 The number of bushels, multiplied by the price per bushel, will equal the total cost. The costs of the corn and barley, added together, will equal the total cost of the mix.

Example 1 Let c = number of bushels of corn. Then 150 – c = number of bushels of barley. Now it’s time to fill in our chart!

Example 1 3.8c + 2.6(150 – c) = 450 Number of Bushels Price per Bushel Total Cost Corn Barley Mix c 150 – c 150 3.80 2.60 3.00 3.8c 2.6(150 – c) 450 3.8c + 2.6(150 – c) = 450

Example 1 3.8c + 2.6(150 – c) = 450 3.8c + 390 – 2.6c = 450 1.2c = 60 50 bushels of corn [c] 100 bushels of barley [150 – c]

Example 2 Pure nickel, by definition, is 100% nickel. Pounds of metal × % nickel = pounds of nickel The alloy is a combination of nickel silver and pure nickel. p = number of lb of pure nickel

Example 2 0.15(70) + 1p = 0.2(70 + p) Nickel Silver Pure Nickel Alloy Percent of Nickel Pounds of Nickel Pounds of Metal Nickel Silver Pure Nickel Alloy 70 p 70 + p 0.15 1 0.2 15% 100% 20% 0.15(70) 1p 0.2(70 + p) 0.15(70) + 1p = 0.2(70 + p)

Example 2 0.15(70) + 1p = 0.2(70 + p) 10.50 + p = 14 + 0.2p p = 4.375 lb of pure nickel

Solving Mixture Problems Read the problem carefully. Plan a strategy. Solve an equation. Check.

Homework: pp. 137-139