How do you solve #4x - y = -8# and #x + 3y = -17#?

1 Answer
Jul 8, 2017

See a solution process below:

Explanation:

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

#x + 3y = -17#

#x + 3y - color(red)(3y) = -17 - color(red)(3y)#

#x + 0 = -17 - 3y#

#x = -17 - 3y#

Step 2) Substitute #(-17 - 3y)# for #x# in the first equation and solve for #y#:

#4x - y = -8# becomes:

#4(-17 - 3y) - y = -8#

#(4 xx -17) - (4 xx 3y) - y = -8#

#-68 - 12y - 1y = -8#

#-68 + (-12 - 1)y = -8#

#-68 + (-13)y = -8#

#-68 - 13y = -8#

#color(red)(68) - 68 - 13y = color(red)(68) - 8#

#0 - 13y = 60#

#-13y = 60#

#(-13y)/color(red)(-13) = 60/color(red)(-13)#

#(color(red)(cancel(color(black)(-13)))y)/cancel(color(red)(-13)) = -60/13#

#y = -60/13#

Step 3) Substitute #-60/13# for #y# in the solution to the second equation at the end of Step 1 and calculate #y#:

#x = -17 - 3y# becomes:

#x = -17 - (3 * -60/13)#

#x = -17 - (-180/13)#

#x = -17 + 180/13#

#x = (13/13 * -17) + 180/13#

#x = -221/13 + 180/13#

#x = -41/13#

The solution is: #x = -41/13# and #y = -60/13# or #(-41/13, -60/13)#