Q. 4.30 - Intermediate Algebra

Question

Kenneth currently sells suits for company $A$ at a salary of \($22,000$ plus a \)$10$ commission for each suit sold. Company $B$ offers him a position with a salary of \($28,000$ plus a \)$4$ commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Step-by-Step Solution

Answer

Kenneth needs to sell $1000$ suits for the options to be equal.

Step 1. Given information

The salary for Company $A$ is $$22,000$ plus a $$10$ commission for each suit sold and the salary for company $B$ is $\$28,000$ plus a $$4$commission for each suit sold.

Step 2. Explanation and Assumptions.

Let the number of suits sold be $x$ and the salary for company $A$ and $B$ be $y_1$ and $y_2$ respectively.

The equation representing the salary for company $A$ will be $y_1=10x+22000$.

The equation representing the salary for company $B$ will be $y_2=4x+28000$.

The salary must be same, so $y_1=y_2$.

Step 3. Solving the equations.

Putting the values and solving the equation,

$10x+22000=4x+28000$

Subtracting $22000$ and $4x$ from both sides,

$6x=6000$

Dividing by $6$, we get

$x=1000$

Step 4. Checking the solution

Putting $x=1000$ in both equations, we get

$$\begin{gathered} y_1=10x+22000 \\ {y}_1=10\left(1000\right)+22000 \\ {y}_1=10000+22000 \\ {y}_1=32,000 \\ {y}_2=4x+28000 \\ {y}_2=4\left(1000\right)+28000 \\ {y}_2=4000+28000 \\ {y}_2=32,000 \end{gathered}$$

Hence we get that $y_1=y_2$, the salaries are equal for each option.