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

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Answer 1 · 2018-06-19T21:38:17

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#x = y+3" " color(red)((1) #

#x - y^2 = 3y " " color(blue)((2) #

Substitute #color(red)((1) # into #color(blue)((2) #

#=> (y+3 ) - y^2 = 3y #

#=> y^2 +2y -3 = 0 #

#=> ( y-1)(y+3) = 0 #

#=> y = 1 , y =-3 #

Using #color(red)((1) #

#y=1 => x = 1+3 = 4 #

#y =-3 => x = -3 + 3 = 0 #

Hence the solutions are #x = 4 , y=1 # and #x = 0 , y=-3#

In coordinate form:

#(4,1) # and #(0,-3) #