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 $4^2+4=y\\ 16+4=y\\ 20=y$