How do you solve by substitution #x-y=-4# and #x+3y=4#?

2 Answers
Don't Memorise
Jun 11, 2015

The solution for the system of equations is
# color(red)(x = -2 , y=2#

Explanation:

#x-y=-4#
# x =color(red)( -4+y # .....equation #(1)#
#x+3y=4#........equation #(2)#

Substituting equation #1# in #2# to find #x#
#x+3y=4#
# color(red)(-4+y) +3y=4#
# -4 +4y = 4#
# 4y = 8#
# color(red)(y=2#

Substituting #y# in equation #1# to obtain #x#:
#x =-4+y #
# x = -4+2#
#color(red)( x=-2#

Alan P.
Jun 11, 2015

#(x,y) = (-2,2)#

Explanation:

Given:
[1]#color(white)("XXXX")##x-y = -4#
[2]#color(white)("XXXX")##x+3y = 4#

We can rewrite [1] as
[3]#color(white)("XXXX")##x = y -4#
then substituting the expression #y-4# from [3] in place of #x# in [2]
[4]#color(white)("XXXX")##(y-4) +3y =4#
which simplifies as
[5]#color(white)("XXXX")##4y = 8#
or
[6]#color(white)("XXXX")##y = 2#
substituting #2# for #y# in [1], we get
[7]#color(white)("XXXX")##x-2 = -4#
[8]#color(white)("XXXX")##x = -2#

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