How do you solve the following system: #-3x + y = -4, 6y + 4x = 12 #?

1 Answer
Aug 27, 2017

Answer:

See a solution process below: #(18/11, 10/11)#

Explanation:

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

#-3x + y = -4#

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

#0 + y = 3x - 4#

#y = 3x - 4#

Step 2) Substitute #(3x - 4)# for #y# in the second equation and solve for #x#:

#6y + 4x = 12# becomes:

#6(3x - 4) + 4x = 12#

#(6 * 3x) - (6 * 4) + 4x = 12#

#18x - 24 + 4x = 12#

#18x + 4x - 24 = 12#

#(18 + 4)x - 24 = 12#

#22x - 24 = 12#

#22x - 24 + color(red)(24) = 12 + color(red)(24)#

#22x - 0 = 36#

#22x = 36#

#(22x)/color(red)(22) = 36/color(red)(22)#

#(color(red)(cancel(color(black)(22)))x)/cancel(color(red)(22)) = (2 xx 18)/color(red)(2 xx 11)#

#x = (color(red)(cancel(color(black)(2))) xx 18)/color(red)(color(black)(cancel(color(red)(2))) xx 11)#

#x = 18/11#

Step 3) Substitute #18/11# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = 3x - 4# becomes:

#y = (3 xx 18/11) - 4#

#y = 54/11 - 4#

#y = 54/11 - (11/11 xx 4)#

#y = 54/11 - 44/11#

#y = 10/11#

The Solution Is: #x = 18/11# and #y = 10/11# or #(18/11, 10/11)#