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

1 Answer
Apr 8, 2016

#(x,y)=(color(cyan)(1,3))#

Explanation:

If
#color(white)("XXX")-2x+y=1#
then
#color(white)("XXX")color(red)(y)=color(blue)(1+2x)#

Since we are also told that
[3]#color(white)("XXX")-4x+color(red)(y)=-1#
we can substitute #(color(blue)(1+2x))# for #color(red)(y)# to get
#color(white)("XXX")-4x+(color(blue)(1+2x))=-1#

#color(white)("XXX")-2x+1=-1#

#color(white)("XXX")color(green)(x)=color(brown)(1)#

Then substituting #(color(brown)(1))# for #color(green)(x)#
in the original #-2color(green)(x)+y=1#

#color(white)("XXX")-2*(color(brown)(1))+y=1#

#color(white)("XXX")-2+y=1#

#color(white)("XXX")y=3#