How do you solve the system of equations #-4x - 8y = 16# and #y = - 3x - 22#?

1 Answer
Jul 31, 2017

See a solution process below:

Explanation:

Step 1) Because the second equation is already solve for #y# we can substitute #(-3x - 22)# for #y# in the first equation and solve for #x#:

#-4x - 8y = 16# becomes:

#-4x - 8(-3x - 22) = 16#

#-4x - (8 * -3x) - (8 * -22) = 16#

#-4x - (-24x) - (-176) = 16#

#-4x + 24x + 176 = 16#

#(-4 + 24)x + 176 = 16#

#20x + 176 = 16#

#20x + 176 - color(red)(176) = 16 - color(red)(176)#

#20x + 0 = -160#

#20x = -160#

#(20x)/color(red)(20) = -160/color(red)(20)#

#(color(red)(cancel(color(black)(20)))x)/cancel(color(red)(20)) = -8#

#x = -8#

Step 2) Substitute #-8# for #x# in the second equation and calculate #y#:

#y = -3x - 22# becomes:

#y = (-3 * -8) - 22#

#y = 24 - 22#

#y = 2#

The Solution Is: #x = -8# and #y = 2# or #(-8, 2)#