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Applications of Linear Equations. Use percent in solving problems involving rates. Recall that percent means “per hundred.” Thus, percents are ratios.

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Presentation on theme: "Applications of Linear Equations. Use percent in solving problems involving rates. Recall that percent means “per hundred.” Thus, percents are ratios."— Presentation transcript:

1 Applications of Linear Equations

2 Use percent in solving problems involving rates. Recall that percent means “per hundred.” Thus, percents are ratios in which the second number is always 100. For example, 50% represents the ratio 50 to 100 and 27% represents the ratio 27 to 100. PROBLEM-SOLVING HINT Percents are often used in problems involving mixing different concentrations of a substance or different interest rates. In each case, to get the amount of pure substance or the interest, we multiply. In an equation, percent is always written as a decimal. For example, 35% is written 0.35, not 35, and 7 % is written 0.07 not 7. Interest Problems (annual) principle × rate (%) = interest p × r = I Mixture Problems base × rate (%) = percentage b × r = p

3 EXAMPLE 1 What is the amount of pure acid in 40 L of a 16% acid solution? Find the annual interest if $5000 is invested at 4%. Using Percents to Find Percentages Solution:

4 PROBLEM-SOLVING HINT In the examples that follow, we use tables to organize the information in the problems. A table enables us to more easily set up an equation, which is usually the most difficult step.

5 EXAMPLE 2 Kg ofPercentageKg of Metal(as a decimal)Copper x0.40.4x 800.780(0.7)=56 x+800.50.5(x + 80) Solution: Let x = kg of 40% copper metal. Solving a Mixture Problem 160 kg of the 40% copper metal is needed. A certain metal is 40% copper. How many kilograms of this metal must be mixed with 80 kg of a metal that is 70% copper to get a metal that is 50% copper?

6 EXAMPLE 3 Amount InvestedRate ofInterest for in DollarsInterestOne Year x0.050.05x 2x + 30000.080.08(2x+3000) Solution: Let x = amount invested at 5%. $7000 was invested at 5% interest. With income earned by selling a patent, an engineer invests some money at 5% and $3000 more than twice as much at 8%. The total annual income from the investment is $1710. Find the amount invested at 5%.

7 Textbook Homework 2.1 1-71 ODD 2.2 1-75 ODD 2.3 5 - 53 ODD 2.4  5 – 65 ODD 2.5 5 – 85 EOO 2.6 3 – 67 EOO 2.7 7 – 53 EOO OR MYMATHLAB!

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