Question

How do you solve \[\dfrac{1}{x}+\dfrac{2}{3x}=\dfrac{1}{3}\] and find any extraneous solutions?

Answer 1

Hint: For the question \[\dfrac{1}{x}+\dfrac{2}{3x}=\dfrac{1}{3}\] it is asked to solve for the value of x and find if there are any extraneous solutions. For this question by multiplying the both sides with the variable x and doing some arithmetical calculations and finding the solution and to find whether it has extraneous solutions or not we will do the process which is discussed below according to the given question.

Complete step by step solution:
So, firstly for the equation we will multiply with x on both sides of the equation. The equation will be reduced as follows.
\[\Rightarrow \dfrac{1x}{x}+\dfrac{2x}{3x}=\dfrac{1x}{3}\]
For this equation we will cancel the numerator and denominator which are same then the equation will be reduced as follows.
\[\Rightarrow 1+\dfrac{2}{3}=\dfrac{x}{3}\]
Here we will do the basic arithmetic property of maths that is addition on the left hand side then after doing it the equation will be reduced as follows.
\[\Rightarrow \dfrac{2+3}{3}=\dfrac{x}{3}\]
Here for this equation we will add the numerator on the left hand side of the equation then it will be reduced as follows.
\[\Rightarrow \dfrac{5}{3}=\dfrac{x}{3}\]
Here we will multiply the whole equation with 3 that is on both sides of equation then the equation will be simplified as follows.
\[\Rightarrow x=5\]
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 it back that is 5 for x into the original equation and check to make sure we get the right answer.
\[\Rightarrow \dfrac{1}{x}+\dfrac{2}{3x}=\dfrac{1}{3}\]
Here in this we will substitute 5 which we got as solution in the question then the equation will be reduced as follows.
\[\Rightarrow \dfrac{1}{5}+\dfrac{2}{3\times 5}=\dfrac{1}{3}\]
\[\Rightarrow \dfrac{3}{5\times 3}+\dfrac{2}{3\times 5}=\dfrac{1}{3}\]
\[\Rightarrow \dfrac{3}{15}+\dfrac{2}{15}=\dfrac{1}{3}\]
\[\Rightarrow \dfrac{5}{15}=\dfrac{1}{3}\]
\[\Rightarrow \dfrac{1}{3}=\dfrac{1}{3}\]
Here the solution works, so it is not extraneous.

Note: Students must be very careful in doing the calculations. To find whether the given question has an extraneous solution or not we have to substitute the value which we got in the same equation if it satisfies the given question is not extraneous if it doesn’t satisfy then it will be extraneous solutions.