115 1.2 Applications and Modeling with Linear Equations EXAMPLE 3 Solving a Mixture Problem A chemist needs a 20% solution of alcohol. She has a 15% solution on hand, as well as a 30% solution. How many liters of the 15% solution should she add to 3 L of the 30% solution to obtain the 20% solution? SOLUTION Step 1 Read the problem. We must find the required number of liters of 15% alcohol solution. Step 2 Assign a variable. Let x = the number of liters of 15% solution to be added. Figure 2 and the table show what is happening in the problem. The numbers in the last column were found by multiplying the strengths and the numbers of liters. 30% 20% x L + = 3 L (x + 3) L 15% Figure 2 Step 3 Write an equation. The number of liters of pure alcohol in the 15% solution plus the number of liters in the 30% solution must equal the number of liters in the final 20% solution. Liters of pure alcohol in 15% solution + Liters of pure alcohol in 30% solution = Liters of pure alcohol in 20% solution (111111111)111111111* (1111+11)1111+11* (111111111)11111111* 0.15x + 0.30132 = 0.201x + 32 Step 4 Solve. 0.15x + 0.90 = 0.20x + 0.60 Distributive property 0.30 = 0.05x Subtract 0.60 and 0.15x. 6 = x Divide by 0.05. Step 5 State the answer. Thus, 6 L of 15% solution should be mixed with 3 L of 30% solution, giving 6 + 3 = 9 L of 20% solution. Step 6 Check. The answer checks because the amount of alcohol in the two solutions is equal to the amount of alcohol in the mixture. 0.15162 + 0.9 = 0.9 + 0.9 = 1.8 Solutions 0.2016 + 32 = 0.20192 = 1.8 Mixture S Now Try Exercise 29. PROBLEM-SOLVING HINT In mixed investment problems, multiply the principal amount P by the interest rate r, expressed as a decimal, and the time t, in years, to find the amount of interest earned I. I =Prt Simple interest formula Strength Liters of Solution Liters of Pure Alcohol 15% x 0.15x 30% 3 0.30132 20% x + 3 0.201x + 32 Sum must equal
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