How do solve the following linear system?: # 3x + 2y = 12, y = 4x + 4 #?

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Answer 1 · 2018-06-19T22:59:19

See a solution process below:

Step 1) Because the second equation is already solved for #y# we can substitute #(4x + 4)# for #y# in the first equation and solve for #x#:

#3x + 2y = 12# becomes:

#3x + 2(4x + 4) = 12#

#3x + (2 xx 4x) + (2 xx 4) = 12#

#3x + 8x + 8 = 12#

#(3 + 8)x + 8 = 12#

#11x + 8 = 12#

#11x + 8 - color(red)(8) = 12 - color(red)(8)#

#11x + 0 = 4#

#11x = 4#

#(11x)/color(red)(11) = 4/color(red)(11)#

#(color(red)(cancel(color(black)(11)))x)/cancel(color(red)(11)) = 4/11#

#x = 4/11#

Step 2) Substitute #4/11# for #x# in the second equation and solve for #y#:

#y = 4x + 4# becomes:

#y = (4 xx 4/11) + 4#

#y = 16/11 + 4#

#y = 16/11 + (4 xx 11/11)#

#y = 16/11 + 44/11#

#y = (16 + 44)/11#

#y = 60/11#

The Solution Is:

#x = 4/11# and #y = 60/11#

Or

#(4/11, 60/11)#