# How do you solve (-1)/(x-3)=(x-4)/(x^2-27)?

Jul 30, 2016

$x = - \frac{3}{2}$ or $x = 5$

#### Explanation:

In $\frac{- 1}{x - 3} = \frac{x - 4}{{x}^{2} - 27}$, let us multiply each side by $\left(x - 3\right) \left({x}^{2} - 27\right)$ to simplify we get

$\frac{- 1}{x - 3} \times \left(x - 3\right) \left({x}^{2} - 27\right) = \frac{x - 4}{{x}^{2} - 27} \times \left(x - 3\right) \left({x}^{2} - 27\right)$ or

$\frac{- 1}{\cancel{\left(x - 3\right)}} \times \cancel{x - 3} \left({x}^{2} - 27\right) = \frac{x - 4}{\cancel{{x}^{2} - 27}} \times \left(x - 3\right) \cancel{\left({x}^{2} - 27\right)}$ or

$- 1 \times \left({x}^{2} - 27\right) = \left(x - 4\right) \left(x - 3\right)$ or

$- {x}^{2} + 27 = x \left(x - 3\right) - 4 \left(x - 3\right)$ or

$- {x}^{2} + 27 = {x}^{2} - 3 x - 4 x + 12$ or

$- {x}^{2} + 27 - {x}^{2} + 7 x - 12 = 0$ or

$- 2 {x}^{2} + 7 x + 15 = 0$

$2 {x}^{2} - 7 x - 15 = 0$

$2 {x}^{2} - 10 x + 3 x - 15 = 0$ or

$2 x \left(x - 5\right) + 3 \left(x - 5\right) = 0$ or

$\left(2 x + 3\right) \left(x - 5\right) = 0$ and hence

ether $2 x + 3 = 0$ i.e. $x = - \frac{3}{2}$

or $x - 5 = 0$ i.e. $x = 5$

Jul 30, 2016

$x = 5$
$x = - \frac{3}{2}$

#### Explanation:

-1/(x-3)=(x-4)/(x^2-27
or
$- {x}^{2} + 27 = \left(x - 3\right) \left(x - 4\right)$
or
$- {x}^{2} + 27 = {x}^{2} - 3 x - 4 x + 12$
or
$- {x}^{2} + 27 = {x}^{2} - 7 x + 12$
or
$2 {x}^{2} - 7 x - 15 = 0$
or
$2 {x}^{2} - 10 x + 3 x - 15 = 0$
or
$2 x \left(x - 5\right) + 3 \left(x - 5\right) = 0$
or
$\left(x - 5\right) \left(2 x + 3\right) = 0$
or
$x - 5 = 0$
or
$x = 5$=======Ans $1$
or
$2 x + 3 = 0$
or
$2 x = - 3$
or
$x = - \frac{3}{2}$=======Ans $2$