# How do you simplify (k^2-26)/(k-5)-1/(5-k)?

May 4, 2017

See the solution process below:

#### Explanation:

To subtract or add and simplify these two fractions they must be over a common denominator. Multiply the fraction on the right by the appropriate form of $1$ which is ($\frac{- 1}{-} 1$):

$\frac{{k}^{2} - 26}{k - 5} - \left(\frac{- 1}{-} 1 \times \frac{1}{5 - k}\right) \implies$

$\frac{{k}^{2} - 26}{k - 5} - \left(- \frac{1}{- 1 \left(5 - k\right)}\right) \implies$

$\frac{{k}^{2} - 26}{k - 5} - \left(- \frac{1}{\left(- 1 \times 5\right) - \left(- 1 \times k\right)}\right) \implies$

$\frac{{k}^{2} - 26}{k - 5} - \left(- \frac{1}{- 5 - \left(- 1 k\right)}\right) \implies$

(k^2 - 26)/(k - 5) - (-1/(-5 + 1k))) =>

(k^2 - 26)/(k - 5) - (-1/(-5 + k))) =>

$\frac{{k}^{2} - 26}{k - 5} - \left(- \frac{1}{k - 5}\right) \implies$

$\frac{{k}^{2} - 26}{k - 5} + \frac{1}{k - 5}$

We can now add the numerator over the common denominator:

$\frac{{k}^{2} - 26 + 1}{k - 5}$

$\frac{{k}^{2} - 25}{k - 5}$

The numerator is a special for of the quadratic:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

We can factor the remaining fraction as:

$\frac{\left(k + 5\right) \left(k - 5\right)}{k - 5}$

We can now cancel common terms in the numerator and denominator:

$\frac{\left(k + 5\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(k - 5\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{k - 5}}}} \implies$

$k + 5$ Where, from the original expression $k \ne 5$