How do you solve #y=3x^2-x-2, y=-2x+2# using substitution?

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Answer 1 · 2016-12-23T02:29:51

You can simply substitute the second equation into the first or vice versa.

#-2x + 2 = 3x^2 - x - 2#

#0 = 3x^2 + x - 4#

#0 = 3x^2 - 3x + 4x - 4#

#0 = 3x(x -1) + 4(x - 1)#

#0 = (3x + 4)(x - 1)#

#x = -4/3 and 1#

Case 1: #x= -4/3#

#y = -2(-4/3) + 2#

#y = 8/3 + 2#

#y = 14/3#

Case 2: #x= 1#:

#y = -2x + 2#

#y = -2 + 2#

#y = 0#

The solution set is therefore #(1, 0)# and #(-4/3, 14/3)#.

Hopefully this helps!