How do you solve #6x + 2y = -2# and #4x + y = 1# using substitution?

1 Answer
Jun 20, 2017

Answer:

See a solution process below:

Explanation:

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

#4x + y = 1#

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

#0 + y = -4x + 1#

#y = -4x + 1#

Step 2) Substitute #(-4x + 1)# for #y# in the first equation and solve for #x#:

#6x + 2y = -2# becomes:

#6x + 2(-4x + 1) = -2#

#6x + (2 * -4x) + (2 * 1) = -2#

#6x + (-8x) + 2 = -2#

#6x - 8x + 2 = -2#

#(6 - 8)x + 2 = -2#

#-2x + 2 = -2#

#-2x + 2 - color(red)(2) = -2 - color(red)(2)#

#-2x + 0 = -4#

#-2x = -4#

#(-2x)/color(red)(-2) = (-4)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2)) = 2#

#x = 2#

Step 3) Substitute #2# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = -4x + 1# becomes:

#y = (-4 * 2) + 1#

#y = -8 + 1#

#y = -7#

The solution is: #x = 2# and #y = -7# or #(2, -7)#