How do you solve the following system: #y = 3x – 12, 3x + 8y = -2#?

1 Answer
Mar 1, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Because the first equation is already solved for #y#, substitute #3x - 12# for #y# in the second equation and solve for #x#: xx
)# #3x + 8y = -2# becomes:

#3x + 8(3x - 12) = -2#

#3x + 24x - 96 = -2#

#27x - 96 = -2#

#27x - 96 + color(red)(96) = -2 + color(red)(96)#

#27x - 0 = 94#

#27x = 94#

#(27x)/color(red)(27) = 94/color(red)(27)#

#(color(red)(cancel(color(black)(27)))x)/cancel(color(red)(27)) = 94/27#

#x = 94/27#

Step 2) Substitute #94/27# for #x# in the first equation and calculate #y#:

#y = 3x - 12# becomes:

#y = 3xx94/27 - 12#

#y = 94/9 - 12#

#y = 94/9 - (9/9 xx 12)#

#y = 94/9 - 108/9#

#y = -14/9#

The solution is #x = 94/27# and #y = -14/9# or #(94/27, -14/9)#