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solve the following system: x2 - y = 3 x - y = -3
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rearrange the first eq. to get: y=x^2-3 and the second to get: y=x+3 then equate these two to get x (hint you'll get a quadratic)
Subtract equation#2 from equation#1, and you should get a quadratic in 'x'. Solve it to find out 'x', and then substitute the values to find out 'y'.
x-y= -3 , x = -3+y (x)^2 - y = 3 (-3+y)^2 - y = 3 y^2 -6y + 9 - y =3 y^2 -7y+6 = 0 (y-1)(y-6) = 0 y=1, y=6
thanks everybody
Don't forget also to solve for x, by substituting y into the original equation
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