How do you solve the system #x^2+4x-4y-16=0# and #-2x+y+1=0#?

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Answer 1 · 2016-08-30T21:13:09

The solutions are #(6, 11)# and #(-2, -5)#

Solve for one of the variables. Let's solve for #y# in the second equation.

#y = -1 + 2x#

Insert into the first equation:

#x^2 + 4x - 4(-1 + 2x) - 16 = 0#

#x^2 + 4x + 4 - 8x - 16 = 0#

#x^2 - 4x - 12 = 0#

#(x - 6)(x + 2) = 0#

#x = 6 and -2#

We can now substitute and solve for #y#:

#y = 2x - 1#

#y = 2(6) - 1" AND "2(-2) - 1#

#y = 11" AND "-5#

Hopefully this helps!