How do you solve #5x + y = 8# and #–4x – 4y = – 16# using substitution?

1 Answer
William Woods
Apr 18, 2018

#x=1# and #y=3#

Explanation:

So always when you try to solve systems, if you find either a single y or x then use the substitution, just something I always do. Anyway so here we have

#5x+y=8#
#-4x-4y=-16#

And in the first one we have a single y ( shown below)

#5x+color(red)(y)=8#
#-4x-4y=-16#

Now we re-write it to get:

#color(red)(y) =8-5x#
#-4x-4y=-16#

Put this equation :
#color(red)(y) =8-5x#

into the other to get:

#-4x-4(8-5x)=-16#

Simplify:

#-4x-32+20x=-16#

Simplify more:

#16x=16#

Divide 16 on both sides to get:

#x=1#

Now put that into any equation to get:

#5+y=8#

Simplify to get

y=3

Now check you work by putting in the numbers into the other equation to get:

#-4(1)-4(3)=-16#
#-4-12=-16#

So it's right with

#x=1#
#y=3#

Hope this helps!

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