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What is the solution of the linear-quadratic system of equations?
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\[x^{2}+5x-3 \] y-x=2
@agent0smith
I think the first one should be equal to something, is that all there is?
my bad its supposed to be y=x^2+5x-3
K. Well we can probably solve it using substitution... since we have y-x = 2 add x to both sides to get y = 2+x got it so far?
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yes you are just putting it into standard form to make it easier to solve right?
well we're putting it that way because... y=x^2+5x-3 we know y = x+2 from above, so let's replace that y there with x+2... x+2=x^2+5x-3 now we can try finding x...
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