How do you solve the following system: #-6x + y = -8, -3x + y = -4 #?

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Answer 1 · 2017-01-31T23:27:58

See the entire solution process below:

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

#-6x + y = -8#

#-6x + color(red)(6x) + y = color(red)(6x) - 8#

#0 + y = 6x - 8#

#y = 6x - 8#

Step 2) Substitute #6x - 8# for #y# in the second equation and solve for #x#:

#-3x + (6x - 8) = -4#

#-3x + 6x - 8 = -4#

#(-3 + 6)x - 8 = -4#

#3x - 8 = -4#

#3x - 8 + color(red)(8) = -4 + color(red)(8)#

#3x - 0 = 4#

#3x = 4#

#(3x)/color(red)(3) = 4/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 4/3#

#x = 4/3#

Step 3) Substitute #4/3# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = (6 xx 4/3) - 8#

#y = 24/3 - 8#

#y = 8 - 8#

#y = 0#

The solution is: #x = 4/3# and #y = 0#