How do you solve #4x+y=-5# and #2x+3y=5# using substitution?

1 Answer
May 25, 2018

#x=-2#
#y=3#

Explanation:

#4x+y=-5#
#2x+3y=5#

Rearrange the first equation for #y#:

#4x+y=-5#

#y=-5-4x#

Substitute #y# into second equation and solve for #x#:

#2x+3y=5#

#2x+3(-5-4x)=5#

#2x-15-12x=5#

#2x-12x=5+15#

#-10x=20#

#x=20/-10#

#x=-2#

Then substitute the value of #x# into one of the original equations to find the solution for #y#:

#4x+y=-5#

#4(-2)+y=-5#

#-8+y=-5#

#y=-5+8#

#y=3#

Double check your solution by substituting values of #x=-2# and #y=3# into any of the original equations and see whether you get the numerical solution (#-5# or #5#).