Precalculus (10th Edition)

$\{(1,1)\}$
We are given the system of equations: $\begin{cases} 2x-3y=-1\\ 10x+y=11 \end{cases}$ Use the elimination method. Multiply the second equation by $3$ and add it to the first equation to eliminate $y$ and determine $x$: $\begin{cases} 2x-3y=-1\\ 3(10x+y)=3(11) \end{cases}$ $\begin{cases} 2x-3y=-1\\ 30x+3y=33 \end{cases}$ $2x-3y+30x+3y=-1+33$ $32x=32$ $x=1$ Determine $y$ using the second equation: $10x+y=11$ $10(1)+y=11$ $10+y=11$ $y=1$ The solution set of the system is: $\{(1,1)\}$