How do you use substitution to solve the system of equations #y=7x+8# and #y=x+20#?

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Answer 1 · 2017-06-04T19:10:11

See a solution process below:

Step 1) Because the first equation is already solved for #y# we can substitute #(7x + 8)# for #y# in the second equation and solve for #x#:

#y = x + 20# becomes:

#7x + 8 = x + 20#

#-color(blue)(x) + 7x + 8 - color(red)(8) = -color(blue)(x) + x + 20 - color(red)(8)#

#-color(blue)(1x) + 7x + 0 = 0 + 12#

#(-color(blue)(1) + 7)x = 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 first equation and calculate #y#:

#y = 7x + 8# becomes:

#y = (7 xx 2) + 8#

#y = 14 + 8#

#y = 22#

The solution is: #x = 2# and #y = 22# or #(2, 22)#