# How do you solve 8/(x-4) -3=1/(x-10)?

Oct 21, 2015

${x}_{1 , 2} = \frac{49 \pm 7}{6}$

#### Explanation:

The first thing to notice here is that you have two values of $x$ for which the denominators are equal to zero.

This means that any possible solution set will not include thes values. In other words, you need

$x - 4 \ne 0 \implies x \ne 4 \text{ }$ and $\text{ } x - 10 \ne 0 \implies x \ne 10$

The next thing to do is use the common denominator of the two fractions, which is equal to $\left(x - 4\right) \left(x - 10\right)$, to rewrite the equation without denominators.

To do that, multiply the first fraction by $1 = \frac{x - 10}{x - 10}$, $3$ by $1 = \frac{\left(x - 4\right) \left(x - 10\right)}{\left(x - 4\right) \left(x - 10\right)}$, and the second fraction by $1 = \frac{x - 4}{x - 4}$.

This will get you

$\frac{8}{x - 4} \cdot \frac{x - 10}{x - 10} - 3 \cdot \frac{\left(x - 4\right) \left(x - 10\right)}{\left(x - 4\right) \left(x - 10\right)} = \frac{1}{x - 10} \cdot \frac{x - 4}{x - 4}$

$\frac{8 \left(x - 10\right)}{\left(x - 4\right) \left(x - 10\right)} - \frac{3 \left(x - 4\right) \left(x - 10\right)}{\left(x - 4\right) \left(x - 10\right)} = \frac{x - 4}{\left(x - 4\right) \left(x - 10\right)}$

This is of course equivalent to

$8 x - 80 - 3 \left({x}^{2} - 14 x + 40\right) = x - 4$

$7 x - 76 - 3 {x}^{2} + 42 x - 120 = 0$

$3 {x}^{2} - 49 x + 196 = 0$

Use the quadratic formula to find the two roots of this quadratic equation

${x}_{1 , 2} = \frac{- \left(- 49\right) \pm \sqrt{{\left(- 49\right)}^{2} - 4 \cdot 3 \cdot 196}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{49 \pm \sqrt{49}}{6} = \frac{49 \pm 7}{6}$

Therefore, you have

${x}_{1} = \frac{49 - 7}{6} = 7 \text{ }$ and ${x}_{2} = \frac{49 + 7}{6} = \frac{28}{3}$

Since both solutions satisfy the condtions $x \ne 4$ and $x \ne 10$, both will be valid solutions to the original equation.