How do you solve this system of equations: #-8x - 6y = - 12; 8x + 8y = 0#?

2 Answers
Jan 30, 2018

See a solution process below:

Explanation:

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

#8x + 8y = 0#

#8x + 8y - color(red)(8y) = 0 - color(red)(8y)#

#8x + 0 = -8y#

#8x = -8y#

#(8x)/color(red)(8) = -8y/color(red)(8)#

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

#x = -y#

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

#-8x - 6y = -12# becomes:

#-8(-y) - 6y = -12#

#8y - 6y = -12#

#(8 - 6)y = -12#

#2y = -12#

#(2y)/color(red)(2) = -12/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = -6#

#y = -6#

Step 3)

Substitute #-6# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = -y# becomes:

#x = -(-6)#

#x = 6#

The Solution Is:

#x = 6# and #y = -6#

Or

#(6, -6)#

Jan 30, 2018

#x = 6 and y =-6#

Explanation:

This is the BEST scenario you could hope for!.

Notice that the #x# terms are additive inverses. #-8x+8x =0#

Simply add the two equations together and the #x# terms will be eliminated.

#color(white)(xxxxx)-8x-6y =-12" "A#
#color(white)(xxxxx)+8x+8y =" "0" "B#

#A+B:rarr" "2y = -12#
#color(white)(xxxxxxxxxxx)y=-6" "#

Substitute #-6# for #y# in either equation. Let's use #B#

#color(white)(xxxxx)8x+8y" " =0" "B#
#color(white)(xxxxx)8x+8(-6) =0#

#color(white)(xxxxx)8x-48 =" "0#
#color(white)(xxxxxxxx)8x=48#
#color(white)(xxxxxxxx)x=6#

Check in equation #A#

#color(white)(xxxxx)-8(6)-6(-6) =-12" "A#
#color(white)(xxxxxx.xxx)-48+36 =-12#
#color(white)(xxxxxxx.xx.xxx)-12 =-12" "larr# correct