Beginning Algebra With Applications, Chapter 6, Test, Section Test, Problem 20

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

5x-3y =& 29 \\
4x+7y =& -5

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

7(5x-3y) =& (29)7
&& \text{Eliminate $y$: Multiply each side of equation 1 by 7}
\\
3(4x+7y) =& (-5)3
&& \text{and Multiply each side of equation 2 by 3}
\\
\\
35x-21y =& 203
&&
\\
12x + 21y =& -15
&&
\\
\hline
\\
47x =& 188
&& \text{Add the equations}
\\
x =& 4
&& \text{Solve for } x

\end{aligned}
\end{equation}
$


Substitute the value of $x$ in equation 1


$
\begin{equation}
\begin{aligned}

5(4)-3y =& 29
\qquad \text{Solve for } y
\\
20-3y =& 29
\\
-3y =& 9
\\
y =& -3

\end{aligned}
\end{equation}
$


The solution is $(4,-3)$.