# 7/(z+1) - z - 5/(z^2-1) = 6/z ? P.S. I have no idea on how to solve this problem.

Mar 16, 2017

Given:$\text{ } \frac{7}{z + 1} - z - \frac{5}{{z}^{2} - 1} = \frac{6}{z}$

We need to find a common denominator

Observe that ${z}^{2} - 1$ is the same as ${z}^{2} - {1}^{2} = \left(z - 1\right) \left(z + 1\right)$.
So I choose the common denominator to be $\left(z - 1\right) \left(z + 1\right)$

We need to 'force' each fraction to have this as its denominator.

$\frac{7}{z + 1} - z - \frac{5}{\left(z - 1\right) \left(z + 1\right)} = \frac{6}{z}$

Now we start to change things. I am not changing the RHS side as I can 'get rid' of the z denominator by cross multiplying.

$\textcolor{g r e e n}{\left[\frac{7}{z + 1} \textcolor{red}{\times 1}\right] - \left[z \textcolor{red}{\times 1}\right] - \left[\frac{5}{\left(z - 1\right) \left(z + 1\right)}\right] = \left[\frac{6}{z}\right]}$

$\textcolor{w h i t e}{.}$
The equation gets folded over 2 lines due to its length

color(green)([7/(z+1)color(red)(xx(z-1)/(z-1))] -[zcolor(red)(xx((z-1)(z+1))/((z-1)(z+1)))]-[5/((z-1)(z+1))] =[6/z]

$\textcolor{g r e e n}{\frac{7 \left(z - 1\right) - z \left(z - 1\right) \left(z + 1\right) - 5}{\left(z - 1\right) \left(z + 1\right)} = \frac{6}{z}}$

Multiply both sides by $z$

$\textcolor{g r e e n}{\frac{7 z \left(z - 1\right) - {z}^{2} \left(z - 1\right) \left(z + 1\right) - 5 z}{\left(z - 1\right) \left(z + 1\right)} = 6}$

Multiply both sides by $\left(z - 1\right) \left(z + 1\right)$

color(green)(7z(z-1)-z^2(z-1)(z+1)-5z=6(z-1)(z+1)

$7 {z}^{2} - 7 z - {z}^{4} + {z}^{2} - 5 z = 6 {z}^{2} - 6$

Hence

${z}^{4} - 2 {z}^{2} + 12 z - 6 = 0$

I do not know how to take it on from this point!
I will ask another contributor I know to take a look!

Mar 16, 2017

$z = 3$ for corrected question.

#### Explanation:

I think the equation in the question should be:

$\frac{7}{z + 1} - \frac{z - 5}{{z}^{2} - 1} = \frac{6}{z}$

Note that ${z}^{2} - 1 = \left(z - 1\right) \left(z + 1\right)$

So in order to change this rational equation into a polynomial one, we can multiply by $z \left(z - 1\right) \left(z + 1\right)$ to get:

$7 z \left(z - 1\right) - z \left(z - 5\right) = 6 \left({z}^{2} - 1\right)$

which multiplies out to give:

$\textcolor{red}{\cancel{\textcolor{b l a c k}{7 {z}^{2}}}} - 7 z - \textcolor{red}{\cancel{\textcolor{b l a c k}{{z}^{2}}}} + 5 z = \textcolor{red}{\cancel{\textcolor{b l a c k}{6 {z}^{2}}}} - 6$

The terms in ${z}^{2}$ cancel out and we can combine the remaining terms to get:

$- 2 z = - 6$

Divide both sides by $- 2$ to get:

$z = 3$

Finally we need to check that this is a valid solution by making sure that none of the denominators in the original equation are $0$ when we substitute this value of $z$. No problem.