# How do you solve the equation and identify any extraneous solutions for 6/(m+5) = 1 - 1/(m-5)?

Jun 18, 2015

I found:
${m}_{1} = 0$
${m}_{2} = 7$

#### Explanation:

Consider the common denominator $\left(m + 5\right) \left(m - 5\right)$ and get:

$\frac{6 \left(m - 5\right)}{\left(m + 5\right) \left(m - 5\right)} = \frac{1 \cdot \left(m + 5\right) \left(m - 5\right) - 1 \cdot \left(m + 5\right)}{\left(m + 5\right) \left(m - 5\right)}$

$\frac{6 \left(m - 5\right)}{\cancel{\left(m + 5\right) \left(m - 5\right)}} = \frac{1 \cdot \left(m + 5\right) \left(m - 5\right) - 1 \cdot \left(m + 5\right)}{\cancel{\left(m + 5\right) \left(m - 5\right)}}$

$6 m - 30 = {m}^{2} - 25 - m - 5$

${m}^{2} - 7 m = 0$

$m \left(m - 7\right) = 0$

${m}_{1} = 0$
${m}_{2} = 7$