How do you solve #x+4y=-1, -3x-14=y# by substitution?

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Answer 1 · 2016-12-07T02:15:46

#x=-5# and #y=1#

#x+4y=-1#
#-3x-14=y#

From the second equation, we know the value of #y# as #color(red)((-3x-14))#. So in the first equation, substitute #y# with #color(red)((-3x-14))#.

#x+4y=-1#

#x+ 4color(red)((-3x-14))=-1#

Open the brackets and simplify. The product of positive and negative is a negative.

#x-12x-56=-1#

#-11x-56=-1#

Add #56# to both sides.

#-11x=55#

Divide both sides by #11#.

#-x=5# or #x=-5#

In the first equation, substitute #x# with #color(blue)(-5)#.

#x+4y=-1#

#color(blue)(-5)+4y=-1#

Add #5# to both sides.

#4y=4#

Divide both sides by #4#.

#y=1#