Beginning Algebra With Applications, Chapter 6, Review Exercises, Section Review Exercises, Problem 14

Solve by addition method: $
\begin{equation}
\begin{aligned}

6x+4y =& -3 \\
12x-10y =& -15

\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}

-2(6x+4y) =& (-3)(-2)
&& \text{Eliminate $x$: Multiply each side of equation 1 by -2}
\\
\\
12x-10y =& -15
&&
\\
\\
-12x-8y =& 6
&&
\\
\\
12x-10y =& -15
&&
\\
\\
\hline
\\
\\
-18y =& -9
&& \text{Add the equations}
\\
\\
y =& \frac{-9}{-18}
&& \text{Solve for } y
\\
\\
y =& \frac{1}{2}
&&

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

6x + 4 \left( \frac{1}{2} \right) =& -3
&& \text{Substitute the value of $y$ in equation 1}
\\
\\
6x + 2 =& -3
&& \text{Solve for } x
\\
\\
6x =& -5
&&
\\
\\
x =& \frac{-5}{6}
&&

\end{aligned}
\end{equation}
$



The solution is $\displaystyle \left( \frac{-5}{6}, \frac{1}{2}
\right)$.