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

Question

Nathan Axel · Accepted Answer

#x=5# (no extraneous solutions)

Multiply both sides by #x#. #x/x+(2x)/(3x)=x/3# #1+2/3=x/3# #5/3 = x/3# Multiply by three now. #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# It is not unnecessary since the solution functions.

Genesis Allen · Answer

undefined To solve the equation 1/x + 2/(3x) = 1/3 and find any extraneous solutions, we can follow these steps:

1. Find a common denominator for the fractions on the left side of the equation. In this case, the common denominator is 3x.

2. Multiply each term by the common denominator to eliminate the fractions. This gives us: 3x * (1/x) + 3x * (2/(3x)) = 3x * (1/3).

3. Simplify the equation by canceling out common factors. This results in: 3 + 2 = x.

4. Combine like terms on the left side of the equation: 5 = x.

5. Therefore, the solution to the equation is x = 5.

6. To check for extraneous solutions, substitute the found value of x back into the original equation: 1/5 + 2/(3*5) = 1/3.

7. Simplify the equation: 1/5 + 2/15 = 1/3.

8. Find a common denominator for the fractions on the left side: 3/15 + 2/15 = 1/3.

9. Combine the fractions: 5/15 = 1/3.

10. Simplify the left side: 1/3 = 1/3.

11. Since the equation is true, the solution x = 5 is valid and there are no extraneous solutions.