How do you solve the following linear system: #y = 4x - 13 , 31x - 6y = 29 #?

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Answer 1 · 2017-07-19T02:57:34

See a solution process below:

Step 1) Because the first equation is already solved for #y# we can substitute #(4x - 13)# for #y# in the second equation and solve for #x#:

#31x - 6y = 29# becomes:

#31x - 6(4x - 13) = 29#

#31x - (6 * 4x) + (6 * 13) = 29#

#31x - 24x + 78 = 29#

#(31 - 24)x + 78 = 29#

#7x + 78 = 29#

#7x + 78 - color(red)(78) = 29 - color(red)(78)#

#7x - 0 = -49#

#7x = -49#

#(7x)/color(red)(7) = -49/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = -7#

#x = -7#

Step 2) Substitute #-7# for #x# in the first equation and calculate #y#:

#y = 4x - 13# becomes:

#y = (4 xx -7) - 13#

#y = -28 - 13#

#y = -41#

The Solution Is: #x = -7# and #y = -41# or #(-7, -41)#