How do you solve the following system: #-6x + y = -8, 2x+3y =4 #?

1 Answer
Mar 17, 2018

See a solution process below:

Explanation:

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

#-6x + y = -8#

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

#0 + y = -8 + 6x#

#y = -8 + 6x#

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

#2x + 3y = 4# becomes:

#2x + 3(-8 + 6x) = 4#

#2x + (3 xx -8) + (3 xx 6x) = 4#

#2x - 24 + 18x = 4#

#2x + 18x - 24 = 4#

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

#20x - 24 = 4#

#20x - 24 + color(red)(24) = 4 + color(red)(24)#

#20x - 0 = 28#

#20x = 28#

#(20x)/color(red)(20) = 28/color(red)(20)#

#(color(red)(cancel(color(black)(20)))x)/cancel(color(red)(20)) = 7/5#

#x = 7/5#

Step 3) Substitute #7/5# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = -8 + 6x# becomes:

#y = -8 + (6 xx 7/5)#

#y = (5/5 xx -8) + 42/5#

#y = -40/5 + 42/5#

#y = 2/5#

The Solution Is:

#x = 7/5# and #y = 2/5#

Or

#(7/5, 2/5)#