# How do you solve  [18/(x^2-3x)]-[6/(x-3)]=5/x and find any extraneous solutions?

Aug 1, 2016

Therefore, the eqn. has No Soln. , but has $x = 0 , \mathmr{and} , x = 3$ Extraneous Solns.

#### Explanation:

We rewriting the given eqn. as, $\frac{18}{{x}^{2} - 3 x} = \frac{6}{x - 3} + \frac{5}{x}$

$\Rightarrow \frac{18}{{x}^{2} - 3 x} = \frac{6 x + 5 \left(x - 3\right)}{x \left(x - 3\right)} = \frac{11 x - 15}{{x}^{2} - 3 x}$

$\Rightarrow 18 \left({x}^{2} - 3 x\right) = \left(11 x - 15\right) \left({x}^{2} - 3 x\right)$

$\Rightarrow 18 \left({x}^{2} - 3 x\right) - \left(11 x - 15\right) \left({x}^{2} - 3 x\right) = 0$

$\Rightarrow \left({x}^{2} - 3 x\right) \left\{18 - \left(11 x - 15\right)\right\} = 0$

$\Rightarrow x \left(x - 3\right) \left(33 - 11 x\right) = 0$

$\Rightarrow 11 x \left(x - 3\right) \left(3 - x\right) = 0$

$\Rightarrow x = 0 , \mathmr{and} , x = 3$

But, let us observe that these roots make all the fractions of the eqn. Meaningless.

Therefore, the eqn. has No Soln. , but has $x = 0 , \mathmr{and} , x = 3$ Extraneous Solns.