Solve by substitution -8x+4y=20 y+6=(x+4)^2
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OK...in the first equation, let's isolate y...we have
4y = 8x + 20 divide by 4 on both sides
y = 2x + 5 so, in the second equation, we have
(2x + 5) + 6 = (x + 4)^2 let's expand the right side and simplify the left
2x + 11 = x^2 + 8x + 16 now subtract the 2x + 11 from both sides
0 = x^2 + 6x +5 And this factors as
0 = (x+1) (x+5) and setting each factor to 0, we get that x = -1 and x = -5
Using y = 2x + 5 to find the y values, we have
y = 2(-1) + 5 = 3 and y = 2(-5) + 5 = -5
So the solutions are (-1, 3) and (-5, 5)
BTW......this is the intersection in two points of a parabola (the second equation) and a line (the first equation)