How do you solve the following system?: #-x -4y =31, x -y = -42#

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

#x - y = -42#

#x - y + color(red)(y) = -42 + color(red)(y)#

#x - 0 = -42 + y#

#x = -42 + y#

Step 2) Substitute #-42 + y# for #x# in the first equation and solve for #y#:

#-x - 4y = 31# becomes:

#-(-42 + y) - 4y = 31#

#42 - y - 4y = 31#

#42 - 5y = 31#

#-color(red)(42) + 42 - 5y = -color(red)(42) + 31#

#0 - 5y = -11#

#-5y = -11#

#(-5y)/color(red)(-5) = -11/color(red)(-5)#

#(color(red)(cancel(color(black)(-5)))y)/cancel(color(red)(-5)) = 11/5#

#y = 11/5#

Step 3) Substitute #11/5# for #y# into the solution for the second equation at the end of Step 1 and calculate #x#:

#x = -42 + y# becomes:

#x = -42 + 11/5#

#x = (5/5 xx -42) + 11/5#

#x = -210/5 + 11/5#

#x = -199/5#

The solution is: #x = -199/5# and #y = 11/5# or #(-199/5, 11/5)#

#-x -4y =31 " "and " " x -y = -42#

This system of equations is a perfect scenario for eliminating
the #x#-terms because they are additive inverses.
Additive inverses give 0 when added together.

#(-x) + (+x) = 0#

#color(white)(........)-x -4y =31...................................A#
#color(white)(..............)x -y = -42...............................B#

#A+B:color(white)(..)-5y = -11#

#color(white)(.....)y = (-11)/-5 = 11/5 = 2 1/5#

Substitute #2 1/5# for #y# in #B#

#color(white)(..............)x -2 1/5 = -42#

#color(white)(..................)x = -42 +2 1/5#

#color(white)(..................)x = -39 4/5#

NOte that in this case the solutions are easier to work with as mixed numbers rather than as improper fractions which tend to involve large numbers.