How do you solve this system of equations: #2x + 4y = 4; y = x - 2#?

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Answer 1 · 2018-01-30T15:26:56

See a solution process below:

Step 1) Because the second equation is already solved for #y# we can substitute #(x - 2)# for #y# in the first equation and solve for #x#:

#2x + 4y = 4# becomes:

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

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

#2x + 4x - 8 = 4#

#(2 + 4)x - 8 = 4#

#6x - 8 = 4#

#6x - 8 + color(red)(8) = 4 + color(red)(8)#

#6x - 0 = 12#

#6x = 12#

#(6x)/color(red)(6) = 12/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 2#

#x = 2#

Step 2) Substitute #2# for #x# in the second equation and calculate #y#:

#y = x - 2# becomes:

#y = 2 - 2#

#y = 0#

The Solution Is:

#x = 2# and #y = 0#

Or

#(2, 0)#