Question

How do you solve $\dfrac{{p + 5}}{{{p^2} + p}} = \dfrac{1}{{{p^2} + p}} - \dfrac{{p - 6}}{{p + 1}}$ and check for extraneous solutions?

Answer 1

Hint: To solve such questions start by making the denominators of the left-hand side and right-hand side equal then cancel them out. Simplify further the given equation to get the desired solution. After finding the solution substitute the obtained values in the given equation to check whether the solution obtained is extraneous or not.

Complete step by step answer:
Given the equation $\dfrac{{p + 5}}{{{p^2} + p}} = \dfrac{1}{{{p^2} + p}} - \dfrac{{p - 6}}{{p + 1}}$.
Here it is asked to solve the given equation and check if the solution is extraneous or not.
First, start by simplifying the denominator term of the given equation.
That is,
$\dfrac{{p + 5}}{{p\left( {p + 1} \right)}} = \dfrac{1}{{p\left( {p + 1} \right)}} - \dfrac{{p - 6}}{{p + 1}}$
$\dfrac{{p + 5}}{{p\left( {p + 1} \right)}} = \dfrac{{1 - p\left( {p - 6} \right)}}{{p\left( {p + 1} \right)}}$
Here in this case $P \ne 0, - 1$as these will make the equation undefined.
Next, we have to cancel the denominators, that is
$p + 5 = 1 - {p^2} + 6p$
Taking all the terms to the left-hand side and simplifying further we get,
${p^2} - 5p + 4 = 0$
$\left( {p - 1} \right)\left( {p - 4} \right) = 0$
Therefore equating $p - 1 = 0$ and $p - 4 = 0$, we get
$p = 1,4$
Next, we have to check whether the solution is extraneous or not.
First substitute$p = 1$ in the equation $\dfrac{{p + 5}}{{{p^2} + p}} = \dfrac{1}{{{p^2} + p}} - \dfrac{{p - 6}}{{p + 1}}$.
That is,
$\dfrac{{1 + 5}}{{1 + 1}} = \dfrac{1}{{1 + 1}} - \dfrac{{1 - 6}}{{1 + 1}}$
$\dfrac{6}{2} = \dfrac{1}{2} - \dfrac{{ - 5}}{2}$
Further simplifying, we get,
$3 = \dfrac{{1 + 5}}{2}$
 $3 = 3$
Therefore, $p = 1$ is a correct solution.
Next check for $p = 4$ in the equation $\dfrac{{p + 5}}{{{p^2} + p}} = \dfrac{1}{{{p^2} + p}} - \dfrac{{p - 6}}{{p + 1}}$.
Substituting the value,
$\dfrac{{4 + 5}}{{{4^2} + 4}} = \dfrac{1}{{{4^2} + 4}} - \dfrac{{4 - 6}}{{4 + 1}}$
$\dfrac{9}{{20}} = \dfrac{1}{{20}} - \dfrac{{ - 2}}{5}$
Next, make the denominator of RHS same, that is,
$\dfrac{9}{{20}} = \dfrac{1}{{20}} + \dfrac{{2 \times 4}}{{5 \times 4}}$
$\dfrac{9}{{20}} = \dfrac{{1 + 8}}{{20}}$
Simplifying further we get,
$\dfrac{9}{{20}} = \dfrac{9}{{20}}$

Therefore, the solution for the equation $\dfrac{{p + 5}}{{{p^2} + p}} = \dfrac{1}{{{p^2} + p}} - \dfrac{{p - 6}}{{p + 1}}$ is $p = 1,4$.
Also verified that the obtained solution $p = 1$ and $p = 4$ is not an extraneous solution.

Note: Always remember that extraneous solutions are those solutions that are not correct even if we have got them through correct calculation. To find whether the solution is extraneous or not, we have to substitute the obtained solutions in the given equation and check whether they are written. After verification, if the obtained solution does not satisfy the given solution then that solution is considered to be an extraneous solution.