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How do you solve the system #x = 5 - y# and #3y = 3x + 1# by substitution?

1 Answer

May 28, 2015

We can substitute the #x# of the first equation into the second:

#->3y=3*(5-y)+1#
#->3y=15-3y+1#
#->6y=16->y=16/6=8/3#

Now plug that into the first equation:
#x=5-8/3 =15/3-8/3=7/3#

So #x=7/3 and y=8/3#

Check your answer with the second equation:

#3* 8/3 = 3* 7/3 +1 ?-> 8=7+1# Check!

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