We can set the 2nd equation to y that way we have something to solve for
\(\left \{ {{3x-1=11} \atop {x^2+x=y}} \right.\)
First we solve our first equation by moving the one to the other side
\(3x=12\)
Then we divide by 3 on each side
\(x=4\)
So now we can plug 4 into the x's of the 2nd equation
\(4^2+4=y\\
16+4=y\\
20=y\)
