How do you solve #y=|x|# and #y=1/2x+4# using substitution?

1 Answer
Feb 12, 2018

Answer:

#(x,y)=(8,8)# or #(x,y)=(-8/3,8/3)#

Explanation:

Given
[1]#color(white)("XXX")y=abs(x)#
[2]#color(white)("XXX")y=1/2x+4#

Substituting #abs(x)# for #y# (based on [1])
in [2], we get:
[3]#color(white)("XXX")abs(x)=1/2x+4#

which implies:
#{: ("either ",[4]color(white)("xx")-x=1/2x+4," or ",[8]color(white)("xx")x=1/2x+4), (,rarr[5]color(white)("xx")-3/2x=4,,rarr[9]color(white)("xx")1/2x=4), (,rarr[6]color(white)("xx")x=-8/3,,rarr[10]color(white)("xx")x=8), ("then based on [1]",,,), (,rarr[7]color(white)("xx")y=8," or ",rarr[11]color(white)("xx")y=8/3) :}#