# How do you solve (p+5)/(p^2+p)=1/(p^2+p)-(p-6)/(p+1) and check for extraneous solutions?

Mar 23, 2017

$p = 1 , 4$

#### Explanation:

$\frac{p + 5}{p \left(p + 1\right)} = \frac{1}{p \left(p + 1\right)} - \frac{p - 6}{p + 1}$
$\frac{p + 5}{p \left(p + 1\right)} = \frac{1 - p \left(p - 6\right)}{p \left(p + 1\right)}$ => $p \ne 0 , - 1$
$p + 5 = 1 - {p}^{2} + 6 p$
${p}^{2} - 5 p + 4 = 0$
$\left(p - 1\right) \left(p - 4\right) = 0$
$p = 1 , 4$ => check both solutions in the original equation to verify they both work and not extraneous solutions.
$p = 1$:
$\frac{1 + 5}{1 + 1} = \frac{1}{1 + 1} - \frac{1 - 6}{1 + 1}$
$\frac{6}{2} = \frac{1}{2} - \frac{- 5}{2}$
$3 = \frac{1 + 5}{2}$
$3 = 3$ => verified p=1 is a valid solution, do the same for 4.

Apr 10, 2017

color(blue)(p=1 or color(blue)(p=4

#### Explanation:

$\frac{p + 5}{{p}^{2} + p} = \frac{1}{{p}^{2} + p} - \frac{p - 6}{p + 1}$

$\therefore \frac{p + 5}{p \left(p + 1\right)} = \frac{1}{p \left(p + 1\right)} - \frac{p - 6}{p + 1}$

$\therefore \frac{p + 5 = 1 - p \left(p - 6\right)}{p \left(p + 1\right)}$

$\therefore \frac{p + 5}{p \left(p + 1\right)} = \frac{1 - p \left(p - 6\right)}{p \left(p + 1\right)}$

multiply L.H.S. and R.H.S. by color(blue)( (p(p+1))/1

$\therefore {\cancel{\textcolor{b l u e}{p \left(p + 1\right)}}}^{\textcolor{red}{1}} / 1 \times \frac{p + 5}{\cancel{p \left(p + 1\right)}} ^ \textcolor{red}{1} = \frac{1 - p \left(p - 6\right)}{\cancel{p \left(p + 1\right)}} ^ \textcolor{red}{1} \times {\cancel{\textcolor{b l u e}{p \left(p + 1\right)}}}^{\textcolor{red}{1}} / 1$

$\therefore p + 5 = 1 - p \left(p - 6\right)$

$\therefore p + 5 = 1 - {p}^{2} + 6 p$

$\therefore {p}^{2} - 5 p + 4 = 0$

$\therefore \left(p - 1\right) \left(p - 4\right) = 0$

:.color(blue)(p=1 or p=4

substitute color(blue)(p=1

$\therefore \frac{\textcolor{b l u e}{1} + 5}{{\textcolor{b l u e}{1}}^{2} + \textcolor{b l u e}{1}} = \frac{1}{{\textcolor{b l u e}{1}}^{2} + \textcolor{b l u e}{1}} - \frac{\textcolor{b l u e}{1} - 6}{\textcolor{b l u e}{1} + 1}$

$\therefore \frac{6}{2} = \frac{1}{2} - \left(\frac{- 5}{2}\right)$

$\therefore \frac{6}{2} = \frac{1}{2} + \frac{5}{2}$

$\therefore \frac{6}{2} = \frac{6}{2}$

substitute color(blue)(p=4

$\therefore \frac{\textcolor{b l u e}{4} + 5}{{\textcolor{b l u e}{4}}^{2} + \textcolor{b l u e}{4}} = \frac{1}{{\textcolor{b l u e}{4}}^{2} + \textcolor{b l u e}{4}} - \frac{\textcolor{b l u e}{4} - 6}{\textcolor{b l u e}{4} + 1}$

$\therefore \frac{9}{20} = \frac{1}{20} - \left(\frac{- 2}{5}\right)$

$\therefore \frac{9}{20} = \frac{1}{20} + \frac{2}{5}$ $\left(\frac{2}{5} \times \frac{4}{4} = \frac{8}{20}\right)$

$\therefore \frac{9}{20} = \frac{1}{20} + \frac{8}{20}$

$\therefore \frac{9}{20} = \frac{9}{20}$