Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 2 - Systems of Linear Equations and Inequalities - 2.2 Solving Systems of Equations Using the Substitution Method - 2.2 Exercises - Page 153: 38

Answer

$\$360,000$ at $5\%$ $\$240,000$ at $6.5\%$

Work Step by Step

We know that Nikki has $\$600,000 $ to invest in two different accounts, say account A and account B. She also wants to earn a total amount of $\$33,600$ per year from her investments at an interest rate of $5\%$ and $6.5\%$ respectively. Let $a$ be the amount invested in account A and $b$ be the amount invested in account B. Then, the sum of the money invested must satisfy the following equation: $$ a+b = 600000.$$ Similarly, the sum of the interest earned each year must satisfy the following equation: $$ 0.05a+0.065b = 33600.$$ We can now solve the system of equations to figure out the amount of money that she should invest in each account. $$\begin{aligned} a+b& = 600000 \\ 0.05a+0.065b &= 33600. \end{aligned}$$ Solve the above equations for $a$ and $b$. Multiply the first equation by $-0.05$ and add the result to the second to eliminate $a$. $$\begin{aligned} -0.05\cdot (a+b)& = -0.05\cdot 600000 \\ -0.05a-0.05b &= -30000. \end{aligned}$$ Now add this to the second equation and solve for $b$. $$\begin{aligned} -0.05a-0.05b &= -30000\\ 0.05a+0.065b &= 33600. \end{aligned}$$ Rearrange: $$\begin{aligned} (0.05a-0.05a)+(0.065b-0.05b)&=33600-30000)\\ 0.015b&= 3600\\ b & = \frac{3600}{0.015}\\ &= \$240,000. \end{aligned}$$ Finally, use the first equation to find the value of $a$. $$a= 600000-b = 600000-240000= \$360,000.$$ She needs to invest $\$360,000$ at the $5\%$ interest rate and $\$240,000$ at the $6.5\%$ respectively.
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