How do you solve the system of equations #x=-10+y# and #-6x+8y=22#?

1 Answer
Feb 11, 2017

See the entire solution process below:

Explanation:

Step 1) Because the first equation is already solved for #x#, substitute #-10 + y# for #x# in the second equation and solve for #y#:

#-6x + 8y = 22# becomes:

#-6(-10 + y) + 8y = 22#

#60 - 6y + 8y = 22#

#60 + 2y = 22#

#60 + 2y - color(red)(60) = 22 - color(red)(60)#

#60 - color(red)(60) + 2y = -38#

#0 + 2y = -38#

#2y = -38#

#(2y)/color(red)(2) = -38/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = -19#

#y = -19#

Step 2) Substitute #-19# for #y# in the first equation and calculate #x#:

#x = -10 + y# becomes:

#x = -10 + -19#

#x = -29#

The solution is: #x = -29# and #y = -19# or (-29, -19)