How do you solve the system of equations #12x - 6y = - 6# and #- 6x + 2y = 8#?

1 Answer
Apr 19, 2017

See the entire solution process below:

Explanation:

Step 1) Solve the first equation for #y#:

#12x - 6y = -6#

#(12x - 6y)/color(red)(-6) = (-6)/color(red)(-6)#

#(12x)/color(red)(-6) - (6y)/color(red)(-6) = 1#

#-2x + y = 1#

#color(red)(2x) - 2x + y = color(red)(2x) + 1#

#0 + y = 2x + 1#

#y = 2x + 1#

Step 2) Substitute #2x + 1# for #y# in the second equation and solve for #x#:

#-6x + 2y = 8# becomes:

#-6x + 2(2x + 1) = 8#

#-6x + (2 * 2x) + (2 * 1) = 8#

#-6x + 4x + 2 = 8#

#-2x + 2 = 8#

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

#-2x + 0 = 6#

#-2x = 6#

#(-2x)/color(red)(-2) = 6/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2)) = -3#

#x = -3#

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

#y = 2x + 1# becomes:

#y = (2 * -3) + 1#

#y = -6 + 1#

#y = -5#

The solution is: #x = -3# and #y = -5# or #(-3, -5)#