# How do you solve 1/(r-2)+1/(r^2-7r+10)=6/(r-2) and check for extraneous solutions?

Oct 29, 2016

$r = \frac{26}{5}$

#### Explanation:

Note that ${r}^{2} - 7 r + 10 = \left(r - 2\right) \left(r - 5\right)$

So the given equation can be written:

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

Multiply through by $\left(r - 2\right)$ to get:

$1 + \frac{1}{r - 5} = 6$

(noting that $r = 2$ would be a spurious solution, if it occurs)

Subtract $1$ from both sides to get:

$\frac{1}{r - 5} = 5$

Take the reciprocal of both sides to get:

$r - 5 = \frac{1}{5}$

Add $5$ to both sides to get:

$r = \frac{26}{5}$

Since $\frac{26}{5} \ne 2$ this is not a spurious solution.