How do you solve #3x + y = 1# and #y + 4 = 5x # using substitution?

1 Answer
Aug 27, 2017

Answer:

See a solution process below: #(5/8, -7/8)#

Explanation:

Step 1)* Solve the second equation for #y#:

#y + 4 = 5x#

#y + 4 - color(red)(4) = 5x - color(red)(4)#

#y + 0 = 5x - 4#

#y = 5x - 4#

Step 2) Substitute #(5x - 4)# for #y# in the first equation and solve for #x#:

#3x + y = 1# becomes:

#3x + (5x - 4) = 1#

#3x + 5x - 4 = 1#

#(3 + 5)x - 4 = 1#

#8x - 4 = 1#

#8x - 4 + color(red)(4) = 1 + color(red)(4) #

#8x - 0 = 5#

#8x = 5#

#(8x)/color(red)(8) = 5/color(red)(8)#

#(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = 5/8#

#x = 5/8#

**Step 3)# Substitute #5/8# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = 5x - 4# becomes:

#y = (5 xx 5/8) - 4#

#y = 25/8 - 4#

#y = 25/8 - (8/8 xx 4)#

#y = 25/8 - 32/8#

#y = (25 - 32)/8#

#y = -7/8#

The Solution Is: #x = 5/8# and #y = -7/8# or #(5/8, -7/8)#