Question 1103090: What is the solution of the linear-quadratic system of equations?
y=x^2+5x-3
y-x=2 Found 2 solutions by stanbon, Boreal:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! What is the solution of the linear-quadratic system of equations?
y=x^2+5x-3
y-x=2
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Substitute for "y" and solve for "x"::
x+2 = x^2 + 5x - 3
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x^2 + 4x - 5 = 0
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Factor:
(x+5)(x-1) = 0
x = -5 or x = 1
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If x = -5, y = x+2 = -3
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If x = 1, y = x+2 = 3
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Cheers,
Stan H.
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You can put this solution on YOUR website! y=x^2+5x-3
y-x=2
y=x+2
x+2=x^2+5x-3
0=x^2+4x-5
0=(x+5)(x-1)
x=-5, 1
when x=-5, y =-3 (-5, -3)
when x=1, y=3 (1, 3)
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