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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]
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