How do you solve the system of equations #4x - 6y = - 8# and #2x - 2y = - 4#?

1 Answer
Aug 8, 2017

See a solution process below:

Explanation:

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

#2x - 2y = -4#

#2x - 2y + color(red)(2y) = -4 + color(red)(2y)#

#2x - 0 = -4 + 2y#

#2x = -4 + 2y#

#(2x)/color(red)(2) = (-4 + 2y)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -4/color(red)(2) + (2y)/color(red)(2)#

#x = -2 + y#

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

#4x - 6y = -8# becomes:

#4(-2 + y) - 6y = -8#

#(4 xx -2) + (4 xx y) - 6y = -8#

#-8 + 4y - 6y = -8#

#-8 + (4 - 6)y = -8#

#-8 + (-2)y = -8#

#-8 - 2y = -8#

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

#-2y = 0#

#(-2y)/color(red)(-2) = 0/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = 0#

#y = 0#

Step 3) Substitute #0# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = -2 + y# becomes:

#x = -2 + 0#

#x = -2#

The Solution Is: #x = -2# and #y = 0# or #(-2, 0)#