How do you solve the system #2x+5y=38# and #x-3y=-3# using substitution?

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Answer 1 · 2017-08-23T17:29:27

#x=9#

#y=4#

First, let's take the second equation and rearrange it to make #x# the subject:

#x-3y=-3 -> x=3y-3#

Insert the rearranged equation in the first equation, in place of #x#:

#2(3y-3)+5y=38#

Now calculate #y#:

#6y-6+5y=38#

#11y=44#

#y=4#

Now substitute the value of #y# into the rearranged equation to find #x#:

#x=3(4)-3#

#x=12-3 -> x=9#