How do you solve this system of equations: #2x + 8y = 12, x-2y =0#?

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Answer 1 · 2017-08-04T19:20:20

See a solution process below:

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

#x - 2y = 0#

#x - 2y + color(red)(2y) = 0 + color(red)(2y)#

#x - 0 = 2y#

#x = 2y#

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

#2x + 8y = 12# becomes:

#2(2y) + 8y = 12#

#4y + 8y = 12#

#(4 + 8)y = 12#

#12y = 12#

#(12y)/color(red)(12) = 12/color(red)(12)#

#(color(red)(cancel(color(black)(12)))y)/cancel(color(red)(12)) = 1#

#y = 1#

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

#x = 2y# become

#x = 2 * 1#

#x = 2#

The Solution Is: #x = 2# and #y = 1# or #(2, 1)#