How do you solve by substitution #y = 3x + 5# and #4x – y = 5#?

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Answer 1 · 2015-06-07T07:22:19

We can substitute the expression for #y# given to us by the first equation into the second equation:

#5 = 4x-y = 4x-(3x+5) = 4x-3x-5#

#= (4-3)x-5 = x-5#

Then add #5# to both ends to get:

#x = 10#

Then substitute this value of #x# for #x# in the first equation:

#y = 3x+5 = 3*10+5 = 30+5 = 35#