How do you solve by substitution #x - 4y = -1 # and #3x + 5y = 31#?

1 Answer
Jul 13, 2015

#x=7#
#y=2#

Explanation:

We are given:

#x-4y=-1#

and

#3x+5y=31#

We can do this a number of ways, but let's start with #x-4y=-1#

Let's multiply #x-4y=-1# by #-3# to get its #x# to look like the #x# in #3x+5y=31# but with an opposite sign:

#x-4y=-1#

#(-3)(x-4y)=(-3)(-1)#

#-3x+12y=3#

Now, let's add #-3x+12y=3# to #3x+5y=31# to get rid of those #x#'s:

#-3x+12y=3#
#3x+5y=31#

#17y=34#

Now, divide both sides by #17# to get:

#y=2#

Now, just plug in the value for #y# into #3x+5y=31#

#3x+5y=31#
#3x+5(2)=31#
#3x+10=31#

subtract both sides by #10#:

#3x+10=31#
#3x=21#

divide both sides by #3#:

#3x=21#
#x=7#

So, #x=7# and #y=2#

To check if our answers are correct, you can take either of the two given equations and plug in the values of #x# and #y#.

Let's try this with #x-4y=-1#

#x-4y=-1#

#(7)-4(2)=-1#

#7-8=-1#

#-1=-1#