How do you solve the system #y - 2x = 3#, #x² + y² = 18#?

1 Answer
Nov 16, 2015

#(x,y) = (3/5, 4 1/5)#
or
#(x,y)=(-3,-3)#

Explanation:

Given
[1]#color(white)("XXX")y-2x=3#
[2]#color(white)("XXX")x^2+y^2 =18#

Re-writing [1]
[3]#color(white)("XXX")y=2x+3#

Substituting #(2x+3)# from [3] for #y# in [2]
[4]#color(white)("XXX")x^2+(2x+3)^2=18#

Simplifying
[5]#color(white)("XXX")x^2+4x^2+12x+9=18#

[6]#color(white)("XXX")5x^2+12x-9=0#

Factoring
[7]#color(white)("XXX")(5x-3)(x+3)=0#

#rArr x= 3/5# or #x=-3#

case 1: #color(black)(x=3/5)#
Substituting #3/5# for #x# in [3]
#color(white)("XXX")y=2(3/5)+3= 4 1/5#

case 2: #color(black)(x=-3)#
Substituting #(-3)# for #x# in [3]
#color(white)("XXX")y = 2(-3) +3 = -3#