How do you solve # 8x+7y=-5# and #x=35-8y# using substitution?

Question

1548 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-22T02:09:49

See the entire solution process below:

Step 1) Because the second equation is already solved for #x# we can substitute #35 - 8y# for #x# in the first equation and solve for #y#:

#8x + 7y = -5# becomes:

#8(35 - 8y) + 7y = -5#

#(8 xx 35) - (8 xx 8y) + 7y = -5#

#280 - 64y + 7y = -5#

#280 - 57y = -5#

#-color(red)(280) + 280 - 57y = -color(red)(280) - 5#

#0 - 57y = -285#

#-57y = -285#

#(-57y)/color(red)(-57) = (-285)//color(red)(-57)#

#(color(red)(cancel(color(black)(-57)))y)/cancel(color(red)(-57)) = 5#

#y = 5#

Step 2) Now, substitute #5# for #y# in the second equation and calculate #x#:

#x = 35 - 8y# becomes:

#x = 35 - (8 xx 5)#

#x = 35 - 40#

#x = -5#

The solution is: #x = -5# and #y = 5# or #(-5, 5)#