Question 1103090
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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