How do you solve the following system: # 1/2x - 2y = 4 , 4x + y = 2 #?

1 Answer
May 18, 2018

Answer:

See a solution process below:

Explanation:

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

#1/2x - 2y = 4#

#color(red)(2)(1/2x - 2y) = color(red)(2) * 4#

#(color(red)(2) * 1/2x) - (color(red)(2) * 2y) = 8#

#color(red)(2)/2x - 4y = 8#

#1x - 4y = 8#

#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#:

#4x + y = 2# becomes:

#4(8 + 4y) + y = 2#

#(4 * 8) + (4 * 4y) + y = 2#

#32 + 16y + y = 2#

#32 + 16y + 1y = 2#

#32 + (16 + 1)y = 2#

#32 + 17y = 2#

#32 - color(red)(32) + 17y = 2 - color(red)(32)#

#0 + 17y = -30#

#17y = -30#

#(17y)/color(red)(17) = -30/color(red)(17)#

#(color(red)(cancel(color(black)(17)))y)/cancel(color(red)(17)) = -30/17#

#y = -30/17#

Step 3) Substitute #-30/17# 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 * -30/17)#

#x = 8 + (-120/17)#

#x = 8 - 120/17#

#x = (17/17 * 8) - 120/17#

#x = 136/17 - 120/17#

#x = 16/17#

The Solution Is:

#x = 16/17# and #y = -30/17#

Or

#(16/17, -30/17)#