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