If
color(white)("XXX")-2x+y=1XXX−2x+y=1
then
color(white)("XXX")color(red)(y)=color(blue)(1+2x)XXXy=1+2x
Since we are also told that
[3]color(white)("XXX")-4x+color(red)(y)=-1XXX−4x+y=−1
we can substitute (color(blue)(1+2x))(1+2x) for color(red)(y)y to get
color(white)("XXX")-4x+(color(blue)(1+2x))=-1XXX−4x+(1+2x)=−1
color(white)("XXX")-2x+1=-1XXX−2x+1=−1
color(white)("XXX")color(green)(x)=color(brown)(1)XXXx=1
Then substituting (color(brown)(1))(1) for color(green)(x)x
in the original -2color(green)(x)+y=1−2x+y=1
color(white)("XXX")-2*(color(brown)(1))+y=1XXX−2⋅(1)+y=1
color(white)("XXX")-2+y=1XXX−2+y=1
color(white)("XXX")y=3XXXy=3