How do you solve #x-3y=4# and #-3x+5y=-14# using substitution?

1 Answer
Mar 20, 2016

#y=0.5#
#x=5.5#

Explanation:

First, we rearrange one of the formulae to give us #x# in terms of #y# (or vice versa)
#x-3y = 4" "# (Add #3y# to both sides)
#x = 4+3y#

Now we can substitute in our value for #x# into the other equation so that we only have one unknown, #y#.

Then you expand the bracket and simplify it to find #y#.

#-3x + 5y = -14#
#-3 (4+3y) + 5y = -14#

(expand the brackets)

#-12 - 9y +5y = -14#

(simplify the #y# terms)

#-12-4y = -14#

(add #12# to both sides)

#-4y=-2#

(divide both sides by #-4# to isolate the #y# term)

#y=-2/-4#
#y = 0.5#

We then substitute the #y# value into the original equation to find #x# (you can use either of the equations for this)

#x=4+3y#
#x=4 +(3xx0.5)#
#x=4 + 1.5#
#x=5.5#

Then we can use the other equation to check our answer.

#-3x+5y=-14#
#-(3xx5.5) + (5xx0.5) = -14#
#-16.5+ 2.5=-14#
#-14 = -14#
So we know our answer is right!