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: 32

Answer

$6,667$

Work Step by Step

Let $s_1(x)$ be the monthly salary of option 1 and $x$ be the sale amount. Given that the commission rate of this option is $ 4\% $ and the at the base salary is $200$ dollars, the monthly salary can be modelled as a linear function with a slope of $m= \frac{4}{100} = 0.04$ and a y-intercept of $200$. Hence, we have: $$ s_1(x) = 0.04x+200. $$ In a similar way, we can model the second option as $$ s_2(x) = 0.055x+100.$$ The sale amount can be found by setting the two functions equal and solving for $x$. $$\begin{aligned} s_2(x)& = s_1(x) \\\\ 0.055x+100 & = 0.04x+200\\ 0.055x-0.04x& = 200-100\\ 0.015x& = 100 \\ x&= \frac{100}{0.015}\\ &\approx 6667. \end{aligned}$$ Check $$\begin{aligned} s_2(6667) &= 0.055\cdot 6667+100 \\ & = \$466.68\\ s_1(6667) &= 0.04\cdot 6667+200 \\ & = \$466.68. \end{aligned}$$ Hence, the sale amount is $6,667$ units.
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