How do you solve the system #x+2y=13# and #3x-5y=6# using substitution?

1 Answer
Mar 17, 2018

Answer:

#x=7#
#y=3#

Explanation:

So you can pick either one of the 2 equations listed to get 1 term alone. In explanation I will be using the second one #(3x-5y=6)#

So we have #3x-5y=6#

First we add #5y# to both sides giving us
#3x=5y+6#

Next we divide #3# on both sides to get #x# alone, which gives us
#x=5/3y+6/3#

#x=5/3y+2#

Now that have #x# we can plug this into the first equation, which is #(x+2y=13)#

#5/3y+2+2y=13#

Now we subtract the #2# on both sides giving us
#5/3y+2y=11#

Now we add the #y# values by finding common denominators
#5/3y+"(3)2"/"(3)1"y=11#

Then add the #y# values
#5/3y+6/3y=11#

#(5y+6y)/3=11#

#(11y)/3=11#

Now we multiply by #3# on both sides
#11y=33#

Next divide by #11#
#y=3#

So now that we have the value for #y# we can find #x# easily by plugging this into any equation that has #y#. For this part I'm going to use the first equation which was, #x+2y=13#

#y=3#
#x+2y=13#

#x+2(3)=13#

#x+6=13#

Subtracting 6 from both sides gives us
#x=7#

There fore
#x=7#
#y=3#