How do you solve the system #x^2 - x - y = 2#, #4x - 3y = 0#?

Question

2794 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-21T20:22:11

#(-2/3,-8/9),(3,4)#

First modify the equation #x^2-x-y=2# to solve for #y#:

#y=x^2-x-2#

Now, take this and plug it into #4x-3y=0#

#4x-3(x^2-x-2)=0#

#4x-3x^2+3x+6=0#

#-3x^2+7x+6=0#

#3x^2-7x-6=0#

#(3x^2-9x)+(2x-6)=0#

#3x(x-3)+2(x-3)=0#

#(3x+2)(x-3)=0#

Thus, we have two possible values for #x#: #{(x=-2/3),(x=3):}#

Plug these both into #4x-3y=0#

When #x=-2/3#:

#4(-2/3)-3y=0#
#-8/3=3y#
#y=-8/9#

Solution point: #(-2/3,-8/9)#

When #x=3#:

#4(3)-3y=0#

#12=3y#

#y=4#

Solution point: #(3,4)#

graph{(x^2-x-y-2)(4x-3y)=0 [-6.92, 13.08, -3.04, 6.96]}