How do you solve #x = y - 8# and #-x - y = 0# using substitution?

From the second equation -x-y=0, it is y=-x. Now substitute this in the first equation ,

x= -x-8 #-># 2x =-8 #-># x= -4

Hence y=4

#"Given "color(red)(x=y-8)# we can #color(blue)"substitute"# this directly into the other equation, and solve for y

#rArr-(color(red)(y-8))-y=0#

distributing gives.

#-y+8-y=0#

simplifying.

#-2y+8=0#

subtract 8 from both sides of the equation.

#-2ycancel(+8)cancel(-8)=0-8#

#rArr-2y=-8#

To solve for y, divide both sides by - 2

#(cancel(-2) y)/cancel(-2)=(-8)/(-2)#

#rArry=4#

To find x, substitute y = 4 into #x=y-8#

#y=4tox=4-8=-4#

#rArr(-4,4)" is the solution"#

Each of these equations represents a line in 2D-space. Solving the system of these two equations means finding all the #(x,y)# points where the lines cross.

We are given the equations #x=y-8# and #"–"x-y=0#. From the first equation, we have a value for #x# in terms of #y#. That is, for the first line, #x# is always #y-8#. To see if this line has any points in common with the other line #"–"x-y=0#, we want to check if an #x# value of #y-8# works for any point on that other line.

So, we substitute #x# out for #y-8# in the second equation:

#"      ""–"x"    "-y=0#
#"–"(y-8)-y=0#
#"  ""–"y+8" "-y=0#
#"               –"2y="–"8#
#"                   "y=4#

So yes—there is a point on the second line where #x# will be #y-8#, and that point occurs when #y=4#.

The only thing left to do is to find the #x#-value for this point. To do that, we can plug #y=4# into either of our line equations and solve for #x#. (Since #y=4# is where the lines cross, both equations will have the same #x#-value for that #y#).

Using the first equation, we get:

#x=y-8#
#x=4-8#
#color(white)x="–"4#

(Or, using the second equation, we get

#"–"x-y=0#
#"–"x-4=0#
#"–"x"       "=4#
#"         "x="–"4#

which gives the same #x#-value, as we'd expect.)

So our solution for the system is #(x,y)=("–"4, 4)#.

graph{(x-y+8)(x+y)=0 [-12.17, 7.83, -2.76, 7.24]}