How do you solve the system of equations #3x + 2y = - 4# and #3x - y = 2#?

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Answer 1 · 2017-08-01T14:01:40

See a solution process below:

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

#3x - y = 2#

#-color(red)(3x) + 3x - y = -color(red)(3x) + 2#

#0 - y = -3x + 2#

#-y = -3x + 2#

#color(red)(-1) xx -y = color(red)(-1)(-3x + 2)#

#y = 3x - 2#

Step 2) Substitute #(3x - 2)# for #y# in the first equation and solve for #x#:

#3x + 2y = -4# becomes:

#3x + 2(3x - 2) = -4#

#3x + (2 xx 3x) - (2 xx 2) = -4#

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

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

#9x - 4 = -4#

#9x - 4 + color(red)(4) = -4 + color(red)(4)#

#9x - 0 = 0#

#9x = 0#

#9x/color(red)(9) = 0/color(red)(9)#

#color(red)(cancel(color(black)(9)))x/cancel(color(red)(9)) = 0#

#x = 0#

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

#y = 3x - 2# becomes:

#y = (3 xx 0) - 2#

#y = 0 - 2#

#y = -2#

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