How do you solve the following system?: #2x +y = 6 , x + 3y = -8#

1 Answer
Feb 27, 2017

See the entire solution process below:

Explanation:

Step 1) Solve the second equation for #x#:

#x + 3y = -8#

#x + 3y - color(red)(3y) = -8 - color(red)(3y)#

#x + 0 = -8 - 3y#

#x = -8 - 3y#

Step 2) Substitute #-8 - 3y# for #x# in the first equation and solve for #y#:

#2x + y = 6# becomes:

#2(-8 - 3y) + y = 6#

#-16 - 6y + y = 6#

#-16 - 5y = 6#

#color(red)(16) - 16 - 5y = color(red)(16) + 6#

#0 - 5y = 22#

#-5y = 22#

#(-5y)/color(red)(-5) = 22/color(red)(-5)#

#(color(red)(cancel(color(black)(-5)))y)/cancel(color(red)(-5)) = -22/5#

#y = -22/5#

Step 3) Substitute #-22/5# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = -8 - 3y# becomes:

#x = -8 - (3 xx -22/5)#

#x = -8 - (-66/5)#

#x = -8 + 66/5#

#x = (5/5 xx -8) + 66/5#

#x = -40/5 + 66/5#

#x = 26/5#

The solution is: #x = 26/5# and #y = -22/5# or #(26/5, -22/5)#