How do you solve the following system: #3x-y=12, 3x+4y=-10 #?

1 Answer
May 29, 2017

Answer:

See a solution process below:

Explanation:

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

#3x - y = 12#

#3x - y + color(red)(y) - color(blue)(12) = color(red)(y) + 12 - color(blue)(12)#

#3x cancel( - y + color(red)(y)) - 12 = y + cancel(12 - color(blue)(12))#

#3x - 12 = y#

#y = 3x - 12#

Step 2) Substitute #(3x - 12)# for #y# in the second equation and solve for #x#:

#3x + 4y = -10# becomes:

#3x + 4(3x - 12) = -10#

#3x + (4 xx 3x) - (4 xx 12) = -10#

#3x + 12x - 48 = -10#

#(3 + 12)x - 48 = -10#

#15x - 48 = -10#

#15x - 48 + color(red)(48) = -10 + color(red)(48)#

#15x cancel(- 48 + color(red)(48)) = 38#

#15x = 38#

#15x/color(red)(15) = 38/color(red)(15)#

#color(red)(cancel(color(black)(15)))x/cancel(color(red)(15)) = 38/15#

#x = 38/15#

Step 3) Substitute #38/15# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = 3x - 12# becomes:

#y = (3 xx 38/15) - 12#

#y = (color(red)(cancel(color(black)(3))) xx 38/(color(red)(cancel(color(black)(15)))5)) - 12#

#y = 38/5 - 12#

#y = 38/5 - (5/5 xx 12)#

#y = 38/5 - 60/5#

#y = -22/5#

The solution is: #x = 38/15# and #y = -22/5# or #(38/15, -22/5)#