How do you solve #y=x^2+6x+2# and #y=-x^2+2x+8# using substitution?

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Answer 1 · 2016-07-15T00:18:50

#-x^2 + 2x + 8 = x^2 + 6x + 2#

#0 = x^2 + x^2 + 6x - 2x + 2 - 8#

#0 = 2x^2 + 4x - 6#

#0 = 2(x^2 + 2x - 3)#

#0 = 2(x + 3)(x - 1)#

#x = -3 and 1#

#y = (-3)^2 + 6(-3) + 2" and "y = (1)^2 + 6(1) + 2#

#y = 9 - 18 + 2" and "y = 9#

#y = -7 and y = 9#

Hence, the solution set is #{-3, -7} and #{1, 9}#.

Hopefully this helps!