How do you solve the system of equations #2x + y - 4= 0# and #2x - y = 4#?

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Answer 1 · 2018-07-25T21:27:10

See a solution process below:

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

#2x + y - 4 = 0#

#2x - color(red)(2x) + y - 4 + color(blue)(4) = -color(red)(2x) + color(blue)(4)#

#0 + y - 0 = -2x + 4#

#y = -2x + 4#

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

#2x - y = 4# becomes:

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

#2x + 2x - 4 = 4#

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

#4x - 4 = 4#

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

#4x - 0 = 8#

#4x = 8#

#(4x)/color(red)(4) = 8/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 2#

#x = 2#

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

#y = -2x + 4# becomes:

#y = (-2 xx 2) + 4#

#y = -4 + 4#

#y = 0#

The Solution Is:

#x = 2# and #y = 0#

Or

#(2, 0)#