Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.1 Systems of Linear Equations: Substitution and Elimination - 10.1 Assess Your Understanding - Page 735: 60


$\text{10 liters of 30% solution, and 4 liters of 65% solution.}$

Work Step by Step

Let us consider that $\text{x=liters of 30% solution, and}$ $\text{y =liters of 65% solution}$ We are given: $x+y=14~~~~(1) $ and $0.30x+0.65y=0.40(14)~~~~(2)$ Re-write equation (1) as: $y=14-x~~~~(3)$ Here, the first equation indicates the amounts of solutions in litres and the second indicates litres of pure acid in the mixture. We substitute equation (3) into equation (2): $0.30x+0.65(14−x)=0.40(14) \\ 30x+65(14−x)=40(14)\\ −35x=−350\\ x=10 \ liters$ Now, back substitute the value of $x$ into Equation (3) to solve for $y$: $y=14-10=4 \ liters$ Therefore, our desired results are: $\text{10 liters of 30% solution,, and 4 liters of 65% solution.}$
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