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

1 Answer
Sep 8, 2015

Answer:

The problem has two solutions:
#x=-1,y=2#.
#x=-5,y=-6#.

Explanation:

To solve by substitution, we isolate one of the variables in one equation, and use that to solve the other equation.

#x+-y =1# shows two cases: #x+y=1# or #x-y=1#.

Case 1:
Our system is:

#x+y=1#
#-2x+y=4#

Since the first equation is easier to work with, we can isolate a variable there.
#x+y=1#
#x=1-y#

Now, we substitute #1-y# into #x# in the other equation, and simplify.

#-2x+y=4#
#-2(1-y)+y=4#
#-2+2y+y=4#
#3y=6#
#y=2#

We now know the value of #y#, which we can use to solve for #x# using either equation:

#x+y=1#
#x+2=1#
#x=-1#

Therefore, the solution to Case 1 is #x=-1,y=2#.

Case 2:
Our system is:

#x-y=1#
#-2x+y=4#

Like before, we isolate one of the variables in one equation and solve using the other equation.

#x-y=1#
#x=1+y#

#-2x+y=4#
#-2(1+y)+y=4#
#-2-2y+y=4#
#-y=6#
#y=-6#

At this point, we know the #y=-6#, so we can solve for #x# using either equation:

#x-y=1#
#x+6=1#
#x=-5#

Therefore the solution to case 2 is #x=-5,y=-6#.

The problem has two solutions:
#x=-1,y=2#.
#x=-5,y=-6#.