# How do you solve 1/(2x-1) - 1/(2x+1) = 1/40 ?

Feb 8, 2016

$x = - \frac{9}{2} , x = \frac{9}{2}$

#### Explanation:

Get a common denominator on the left hand side of the equation.

$\frac{2 x + 1}{\left(2 x - 1\right) \left(2 x + 1\right)} - \frac{2 x - 1}{\left(2 x - 1\right) \left(2 x + 1\right)} = \frac{1}{40}$

Note that that the fractions can be combined and that the denominator is $\left(2 x - 1\right) \left(2 x + 1\right) = 4 {x}^{2} - 1$.

$\frac{2 x + 1 - \left(2 x - 1\right)}{4 {x}^{2} - 1} = \frac{1}{40}$

$\frac{2 x + 1 - 2 x + 1}{4 {x}^{2} - 1} = \frac{1}{40}$

$\frac{2}{4 {x}^{2} - 1} = \frac{1}{40}$

Cross-multiply.

$80 = 4 {x}^{2} - 1$

$0 = 4 {x}^{2} - 81$

Factor this as a difference of squares.

$0 = \left(2 x + 9\right) \left(2 x - 9\right)$

Now, set these both equal to $0$ to find the two values of $x$ that satisfy this equation.

$2 x + 9 = 0 \text{ "=>" } x = - \frac{9}{2}$

$2 x - 9 = 0 \text{ "=>" } x = \frac{9}{2}$