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

Apr 24, 2018

$m = 0$ is an extraneous solution and $m = 5$ is a solution.

#### Explanation:

$\frac{1}{{m}^{2} - m} + \frac{1}{m} = \frac{5}{{m}^{2} - m}$. Multiplying by $m \left({m}^{2} - m\right)$

on both sides we get, $m + \left({m}^{2} - m\right) = 5 m$ or

$\cancel{m} + {m}^{2} - \cancel{m} - 5 m = 0$ or

$m \left(m - 5\right) = 0 \therefore m = 0 , m = 5$

On check , $m \ne 0$ , since function is undefined at $m = 0$

So $m = 0$ is an extraneous solution and $m = 5$ is a solution. [Ans]

May 3, 2018

In a problem like this we can assume the denominators are not zero, so we never really run into extraneous solutions $m = 0$ or $m = 1$ and get right to

$m = 5$

#### Explanation:

Sometimes extraneous solutions are unavoidable, as often happens when we have to square both sides of an equation.

That's not the case in a problem like this. We can at the outset assume the denominators are not zero, i.e assume

${m}^{2} - m \ne 0$

$m \left(m - 1\right) \ne 0$

$m \ne 0$ and $m \ne 1$.

The $\frac{1}{m}$ tells us $m \ne 0$ which we already know.

Usually we must always factor, never cancel. But when we know factors are non-zero, we can cancel them.

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

$\frac{1 + 1 \left(m - 1\right)}{m \left(m - 1\right)} = \frac{5}{m \left(m - 1\right)}$

$1 + 1 \left(m - 1\right) = 5$

$m = 5$

Check:

$\frac{1}{{5}^{2} - 5} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} = \frac{1}{4}$

5/{5^2-5}=5/20 quad sqrt