How do you solve the system of equations #x= 6y - 7# and #x + 6y = 9#?

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Answer 1 · 2018-01-26T22:37:24

See a solution process below:

Step 1) Because the first equation is already solved for #x# we can substitute #(6y - 7)# for #x# in the second equation and solve for #y#:

#x + 6y = 9# becomes:

#(6y - 7) + 6y = 9#

#6y - 7 + 6y = 9#

#6y + 6y - 7 = 9#

#(6 + 6)y - 7 = 9#

#12y - 7 = 9#

#12y - 7 + color(red)(7) = 9 + color(red)(7)#

#12y - 0 = 16#

#12y = 16#

#(12y)/color(red)(12) = 16/color(red)(12)#

#y = 4/3#

Step 2) Substitute #4/3# for #y# in the first equation and calculate #x#:

#x = 6y - 7# becomes

#x = (6 xx 4/3) - 7#

#x = 24/3 - 7#

#x = 8 - 7#

#x = 1#

The Solution Is:

#x = 1# and #y = 4/3#

Or

#(1, 4/3)#

Answer 2 · 2018-01-26T22:38:19

X=1, y= #8/6#

Rearrange equations to be in the form of x=...
Ie.
#x=9-6y#

Make these equations equal to each other
#6y-7=9-6y#

#16=12y#

#y=16/12# #=8/6#

Find x
#x=9-6(8/6) =9-8= 1#