How do you solve the following system of equations?: #2x + 15y = -4 , -x + 3y=12#?

1 Answer
Mar 7, 2017

Answer:

See the entire solution process below:

Explanation:

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

#-x + 3y = 12#

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

#-x + 0 = 12 - 3y#

#-x = 12 - 3y#

#-1 xx -x = -1(12 - 3y)#

#x = (-1 xx 12) - (-1 xx 3y)#

#x = -12 + 3y#

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

#2x + 15y = -4# becomes:

#2(-12 + 3y) + 15y = -4#

#(2 xx -12) + (2 xx 3y) + 15y = -4#

#-24 + 6y + 15y = -4#

#-24 + 21y = -4#

#color(red)(24) - 24 + 21y = color(red)(24) - 4#

#0 + 21y = 20#

#21y = 20#

#(21y)/color(red)(21) = 20/color(red)(21)#

#(color(red)(cancel(color(black)(21)))y)/cancel(color(red)(21)) = 20/21#

#y = 20/21#

Step 3) Substitute #20/21# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = -12 + 3y# becomes:

#x = -12 + (3 xx 20/21)#

#x = -12 + 20/7#

#x = (7/7 xx -12) + 20/7#

#x = -84/7 + 20/7#

#x = -64/7#

The solution is: #x = -64/7# and #y = 20/21# or #(-64/7, 20/21)#