How do you solve the system of equations #3x - \frac { 4} { 3} y = - 8# and #- 4y = x + 16#?

1 Answer
Aug 23, 2017

Answer:

See a solution process below:

Explanation:

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

#-4y = x + 16#

#-4y - color(red)(16) = x + 16 - color(red)(16)#

#-4y - 16 = x + 0#

#-4y - 16 = x#

#x = -4y - 16#

Step 2) Substitute #(-4y - 16)# for #x# in the first equation and solve for #y#:

#3x - 4/3y = -8# becomes:

#3(-4y - 16) - 4/3y = -8#

#(3 * -4y) - (3 * 16) - 4/3y = -8#

#-12y - 48 - 4/3y = -8#

#-12y - 4/3y - 48 = -8#

#(3/3 * -12)y - 4/3y - 48 = -8#

#-36/3y - 4/3y - 48 = -8#

#-40/3y - 48 = -8#

#-40/3y - 48 + color(red)(48) = -8 + color(red)(48)#

#-40/3y - 0 = 40#

#-40/3y = 40#

#-color(red)(3)/color(blue)(40) * -(40)/3y = -color(red)(3)/color(blue)(40) * 40#

#-cancel(color(red)(3))/cancel(color(blue)(40)) * -color(blue)(cancel(color(black)(40)))/color(red)(cancel(color(black)(3)))y = -color(red)(3)/cancel(color(blue)(40)) * color(blue)(cancel(color(black)(40)))#

#- -y = -3#

#y = -3#

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

#x = -4y - 16# becomes:

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

#x = 12 - 16#

#x = -4#

The Solution Is: #x = -4# and #y = -3# or #(-4, -3)#