Question

How do you solve $ \dfrac{n+5}{n+8}=1+\dfrac{6}{n+1} $ and check for extraneous solutions?

Answer 1

Hint: We will first understand the concept of extraneous solution. We will then simplify the above equation by transforming the equation by taking LCM on the RHS side and cross multiplying the equation obtained and then we will find all the possible values of ‘n’ and then we will check that the obtained value of ‘n’ satisfies the above equation or not. If it does not satisfy the equation then that value of ‘n’ will be the extraneous solution.

Complete step by step answer:
We know that the extraneous solution is the roots or we can say the solution of the transformed equation does not satisfy the original equation because it is excluded from the domain of the original equation.
We have the equation: $ \dfrac{n+5}{n+8}=1+\dfrac{6}{n+1} $ .
Since we know that denominator can not be equal to 0. So, (n+8) and (n+1) can not be equal to zero.
 $ \Rightarrow n+8\not{=}0\text{ and }n+1\not{=}0 $
 $ \Rightarrow n\not{=}-8\text{ and }n\not{=}-1 $
So, -8, and -1 are not in the domain of the above-given equation.
Now, we will take (n+1) as LCM on the RHS, then we will get:
 $ \Rightarrow \dfrac{n+5}{n+8}=\dfrac{\left( n+1 \right)+6}{n+1} $
 $ \Rightarrow \dfrac{n+5}{n+8}=\dfrac{n+7}{n+1} $
Now, after cross multiplying both the side we will get:
 $ \Rightarrow \left( n+5 \right)\left( n+1 \right)=\left( n+7 \right)\left( n+8 \right) $
Now, we will open the bracket and multiply the first term with each of second term inside the bracket:
 $ \Rightarrow n\left( n+1 \right)+5\left( n+1 \right)=n\left( n+8 \right)+7\left( n+8 \right) $
 $ \Rightarrow {{n}^{2}}+n+5n+5={{n}^{2}}+8n+7n+56 $
 $ \Rightarrow {{n}^{2}}+6n+5={{n}^{2}}+15n+56 $
Now, we will take all the terms on the RHS to the LHS then we will get:
 $ \Rightarrow {{n}^{2}}-{{n}^{2}}+6n-15n+5-56=0 $
 $ \Rightarrow -9n-51=0 $
 $ \Rightarrow 9n+51=0 $
 $ \begin{align}
  & \Rightarrow n=-\dfrac{51}{9} \\
 & \therefore n=-\dfrac{17}{3} \\
\end{align} $
Since, $ n=-\dfrac{17}{3} $ is not equal to -1 or -8. So, $ -\dfrac{17}{3} $ is the solution of the above equation.
Hence, $ -\dfrac{17}{3} $ is not the extraneous solution of the above equation.

Note:
 Students are required to note that when we are given an equation that can not be solved without transformation such as doing square roots, cross multiplying, etc. then there are chances of getting extraneous solutions so we need to verify all the roots obtained from the transformed equation by putting them into the original equation.