How do you solve #y=x^2-x-20, y=3x+12# using substitution?

1 Answer
Jan 6, 2017

#(x,y)=color(green)(""(8,36))# or #color(green)(""(-4,0))#

Explanation:

Given
[1]#color(white)("XXX")y=x^2-x-20#
[2]#color(white)("XXX")y=3x+12#

Using [2] we can substitute #3x+12# for #y# in [1] to get
[3]#color(white)("XXX")3x+12=x^2-x-20#

[4]#color(white)("XXX")rarr x^2-4x-32=0#

[5]#color(white)("XXX")rarr (x-8)(x+4)=0#

[6]#color(white)("XXX")rarr x=8#
or
[7]#color(white)("XXX")rarr x=-4#

Using [6] in [2]
[8]#color(white)("XXX")y=3 * 8 +12 =36# (when #x=8#)
or
Using [7] in [2]
[9]#color(white)("XXX")y=3 * (-4) +12 =0# (when #x=-4#)