How do you solve #1/x+2/(3x)=1/3# and find any extraneous solutions?

1 Answer
May 22, 2017

Answer:

#x=5# (no extraneous solutions)

Explanation:

Multiply both sides by #x#.

#x/x+(2x)/(3x)=x/3#

#1+2/3=x/3#

#5/3 = x/3#

Now multiply by 3.

#5=x#

This is the only solution. All that is left to do is check whether or not it is extraneous (that is, it doesn't actually work when we plug it back in due causing division by 0 or something of that nature). To do this, we have to just plug in #5# for #x# into the original equation and check to make sure we get the right answer.

#1/x+2/(3x)=^?1/3#

#1/5+2/(3*5)=^?1/3#

#3/15+2/15=^?1/3#

#5/15 =^? 1/3#

#1/3=1/3#

The solution works, so it is not extraneous.