How do solve the following linear system?: # -4x-14y=28 , x-2y=8 #?

1 Answer
Mar 12, 2017

Answer:

See the entire solution process below:

Explanation:

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

#x - 2y = 8#

#x - 2y + color(red)(2y) = 8 + color(red)(2y)#

#x - 0 = 8 + 2y#

#x = 8 + 2y#

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

#-4x - 14y = 28# becomes:

#-4(8 + 2y) - 14y = 28#

#(-4 xx 8) + (-4 xx 2y) - 14y = 28#

#-32 - 8y - 14y = 28#

#-32 - 22y = 28#

#color(red)(32) - 32 - 22y = color(red)(32) + 28#

#0 - 22y = 60#

#-22y = 60#

#(-22y)/color(red)(-22) = 60/color(red)(-22)#

#(color(red)(cancel(color(black)(-22)))y)/cancel(color(red)(-22)) = (2 xx 30)/color(red)(2 xx -11)#

#y = (color(red)(cancel(color(black)(2))) xx 30)/color(red)(cancel(2) xx -11)#

#y = -30/11#

Step 3) Substitute #-30/11# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 8 + 2y# becomes:

#x = 8 + (2 xx -30/11)#

#x = (11/11 xx 8) - 60/11#

#x = 88/11 - 60/11#

#x = (88 - 60)/11#

#x = 28/11#

The solution is: #x = 28/11# and #y = -30/11# or #(28/11, -30/11)#