How do you solve the system of equations #-4x + 4y = 4# and #y = 9x + 73#?

1 Answer
Mar 20, 2018

Answer:

See a solution process below:

Explanation:

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

#-4x + 4y = 4#

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

#-4x + 0 = 4 - 4y#

#-4x = 4 - 4y#

#(-4x)/color(red)(-4) = (4 - 4y)/color(red)(-4)#

#x = 4/color(red)(-4) - (4y)/color(red)(-4)#

#x = -1 + y#

Step 2) Substitute #(-1 + y)# for #x# in the second equation and solve for #y#:

#y = 9x + 73# becomes:

#y = 9(-1 + y) + 73#

#y = (9 * -1) + (9 * y) + 73#

#y = -9 + 9y + 73#

#y - color(red)(9y) = -9 + 9y - color(red)(9y) + 73#

#1y - color(red)(9y) = -9 + 0 + 73#

#(1 - color(red)(9))y = -9 + 73#

#-8y = 64#

#(-8y)/color(red)(-8) = 64/color(red)(-8)#

#y = -8#

Step 3) Substitute #-8# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = -1 + y# becomes:

#x = -1 + (-8)#

#x = -1 - 8#

#x = -9#

The Solution Is:

#x = -9# and #y = -8#

Or

#(-9, -8)#