How do you solve #x=y+4# and #x=2y+8# using substitution?

Step 1) Because both equations are solve solve for #x#, we substitute #y + 4# from the first equation for #x# in the second equation and solve for #y#:

#y + 4 = 2y + 8#

#-color(red)(y) + y + 4 - color(blue)(8) = -color(red)(y) + 2y + 8 - color(blue)(8)#

#0 - 4 = -color(red)(1y) + 2y + 0#

#-4 = (-color(red)(1) + 2)y#

#-4 = 1y#

#-4 = y#

#y = -4#

Step 2) Substitute #-4# for #y# in either of the original equations and calculate #x#. I will substitute it into the first equation:

#x = y + 4# becomes:

#x = -4 + 4#

#x = 0#

The solution is: #x = 0# and #y = -4# or #(0, -4)#

# x = y + 4# so putting this into the second equation gives.

# y + 4 = 2y +8# solve for y by subtracting y and 8 from both sides.

# y -y + 4 -8 = 2y -y + 8 -8# Which gives

# -4 = y # now put -4 into the first equation for solve for x

# x = -4 +4 # so

# x = 0#

#x=y+4----(1)#

#x=2y+8----(2)#

substitute #color(magenta)(x=y+4 # in (1)

#:.(color(magenta)(y+4))=2y+8#

#:.y-2y=8-4#

#:.-y=4#

#:.color(magenta)(y=-4#

substitute #color(magenta)(y=-4# in (1)

#:.x=(color(magenta)(-4))+4#

#:.color(magenta)(x=0#

substitute #y=-4#and#x=0# in (2)

#:.0=2(-4)+8#

#:.0=-8+8#

#:.color(magenta)(0=0#