Beginning Algebra With Applications, Chapter 4, 4.1, Section 4.1, Problem 34

A real estate agent sold two homes and received commissions totaling $\$ 6000$. The agent's commission on one home was one and one half times the commission on the second home. Find the agent's commission on each home.

If we let $x$ and $y$ be the commission of the agent on first and second home respectively, then we have

$x+y = 6000 \qquad$ Equation 1

And,


$
\begin{equation}
\begin{aligned}

x =& \left( 1 + \frac{1}{2} \right)y
\\
\\
x =& \left( \frac{2+1}{2} \right) y
\\
\\
x =& \frac{3}{2} y \qquad
\text{Equation 2}

\end{aligned}
\end{equation}
$


By substituting Equation 2 in Equation 1, we get


$
\begin{equation}
\begin{aligned}

\frac{3}{2} y + y =& 6000
\\
\\
\frac{5}{2} y =& 6000
\\
\\
y =& 6000 \left( \frac{2}{5} \right)
\\
\\
y =& 2400

\end{aligned}
\end{equation}
$


Then, by applying back substitution, we get

$\displaystyle x = \frac{3}{2} y = \frac{3}{2} (2400) = 3600$

Thus, the agent's commission on the first house is $\$3600$ while at the second house, the agent obtained a commission of $\$ 2400$.