How do you solve the system #x+4y=8# and #2x-5y=29# using substitution?

1 Answer
Mar 5, 2017

Answer:

See the entire solution process below:

Explanation:

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

#x + 4y - color(red)(4y) = 8 - color(red)(4y)#

#x + 0 = 8 - 4y#

#x = 8 - 4y#

Step 2) Substitute #8 - 4y# for #x# in the second equation and solve for #y#:

#2x - 5y = 29# becomes:

#2(8 - 4y) - 5y = 29#

#(2 xx 8) - (2xx4y) - 5y = 29#

#16 - 8y - 5y = 29#

#16 - 13y = 29#

#-color(red)(16) + 16 - 13y = -color(red)(16) + 29#

#0 - 13y = 13#

#-13y = 13#

#(-13y)/color(red)(-13) = 13/color(red)(-13)#

#(color(red)(cancel(color(black)(-13)))y)/cancel(color(red)(-13)) = -1#

#y = -1#

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

#x = 8 - 4y# becomes:

#x = 8 - (4 xx -1)#

#x = 8 - (-4)#

#x = 8 + 4#

#x = 12#

The solution is #x = 12# and #y = -1# or #(12, -1)#