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Dinesh Maurya · Answer 1 · 2015-01-12T18:52:42+0000

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

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

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