Solve the following system of equations:
5x + 3y = 5(4 - x) AND 2x - 5y = -(y + 11)/2
$\displaystyle \left\{
\begin{array}{rcl}
5x + 3y & = & 5(4 - x)\\
2x - 5y & = & -\frac12(y + 11)
\end{array}
\right.$
$\displaystyle \Rightarrow\left\{
\begin{array}{rcll}
5x + 3y & = & 20 - 5x & \text{\color{red}\quad(distribution)}\\
-4x + 10y & = & y + 11 & \text{\color{red}\quad(multiplication by -2)}
\end{array}
\right.$
$\displaystyle \Rightarrow\left\{
\begin{array}{rcl}
10x + 3y - 20 & = & 0\\
-4x + 9y - 11 & = & 0
\end{array}
\right.\text{\color{red}\quad(moving terms over to left-hand side)}$
$\displaystyle \Rightarrow\left\{
\begin{array}{rcll}
30x + 9y - 60 & = & 0 & \text{\color{red}\quad(multiplication by 3)}\\
-4x + 9y - 11 & = & 0 &
\end{array}
\right.$
$\displaystyle \Rightarrow\left\{
\begin{array}{rcl}
34x + 0y - 49 & = & 0\\
-4x + 9y - 11 & = & 0
\end{array}
\right.\text{\color{red}\quad(subtracting equation 2 from equation 1)}$
$\displaystyle \Rightarrow 34x -49 = 0\Rightarrow x = \frac{49}{34}$
Now, substitute back into equation 2 and solve for $\displaystyle y$.