Chapter 2 - Systems of Linear Equations and Inequalities - 2.2 Solving Systems of Equations Using the Substitution Method - 2.2 Exercises - Page 153: 31

$40,000$

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 $ 5\% $ and the at the base salary is $700$ dollars, the monthly salary can be modelled as a linear function with a slope of $m= \frac{5}{100} = 0.05$ and a $y$-intercept of $700$. Hence, we have:$$ s_1(t) = 0.05x+700.$$ In a similar way, we can model the second option as $$ s_2(t) = 0.035x+1300$$ The sale amount can be found by setting the two functions equal and solving for $x$. $$\begin{aligned} s_1(x)& = s_2(x) \\\\ 0.05x+700 & = 0.035x+1300\\ 0.05x-0.035x& = 1300-700\\ 0.015x& = 600 \\ x&= \frac{600}{0.015}\\ &=40000. \end{aligned}$$ Check $$\begin{aligned} s_1(40000) &= 0.05\cdot 40000+700\\ & = \$2,700\\ s_2(40000) &= 0.035\cdot 40000+1300 \\ & = \$2,700. \end{aligned}$$ Hence, the sale amount is $40000$ units.