How do you solve #y² = x + 3# and #x - 2y = 12# using substitution?

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Answer 1 · 2016-05-04T00:26:39

See solution below.

#x = 12 + 2y -> y^2 = 12 + 2y + 3#

#y^2 - 2y - 15 = 0#

#(y - 5)(y + 3) = 0#

#y = 5 and -3#

#x - 2(5) = 12 or X - 2(-3) = 12#

#x = 22 or x = 6#

Your solution set is #{22, 5} and {6, -3}#

Hopefully this helps!