How do you solve the system of equations #x+ 3y = 8# and #- x + 9y = 16#?

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Answer 1 · 2017-05-10T22:26:56

See a solution process below:

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

#x + 3y = 8#

#x + 3y - color(red)(3y) = 8 - color(red)(3y)#

#x + 0 = 8 - 3y#

#x = 8 - 3y#

Step 2) Substitute #8 - 3y# for #x# in the second equation and solve for #y#:

#-x + 9y = 16# becomes:

#-(8 - 3y) + 9y = 16#

#-8 + 3y + 9y = 16#

#-8 + (3 + 9)y = 16#

#-8 + 12y = 16#

#color(red)(8) - 8 + 12y = color(red)(8) + 16#

#0 + 12y = 24#

#12y = 24#

#(12y)/color(red)(12) = 24/color(red)(12)#

#(color(red)(cancel(color(black)(12)))y)/cancel(color(red)(12)) = 2#

#y = 2#

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

#x = 8 - 3y# becomes:

#x = 8 - (3 * 2)#

#x = 8 - 6#

#x = 2#

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