How do you solve the system #y=4x-1# and #y=2x-5# using substitution?

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Answer 1 · 2017-08-01T22:04:07

See a solution process below:

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

#4x - 1 = 2x - 5#

#-color(blue)(2x) + 4x - 1 + color(red)(1) = -color(blue)(2x) + 2x - 5 + color(red)(1)#

#(-color(blue)(2) + 4)x - 0 = 0 - 4#

#2x = -4#

#(2x)/color(red)(2) = -4/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -2#

#x = -2#

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

#y = 4x - 1# becomes:

#y = (4 xx -2) - 1#

#y = -8 - 1#

#y = -9#

The Solution Is: #x = -2# and #y = -9# or #(-2, -9)#