Solve by substitution -8x+4y=20 y+6=(x+4)^2

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)