# How do you solve 1/ (2x+ 3) + 1/ (4x-3) = 5?

Sep 28, 2017

$x = \frac{- 3 \pm 3 \sqrt{6}}{10}$

#### Explanation:

$\frac{1}{2 x + 3} + \frac{1}{4 x - 3} = 5$
Take LCM of denominators $\left(2 x + 3\right) \left(4 x - 3\right)$
Multiply both sides by $\left(2 x + 3\right) \left(4 x - 3\right)$
$\left(2 x + 3\right) \left(4 x - 3\right) \left(\frac{1}{2 x + 3} + \frac{1}{4 x - 3}\right) = 5 \left(2 x + 3\right) \left(4 x - 3\right)$
$\frac{\left(2 x + 3\right) \left(4 x - 3\right)}{2 x + 3} + \frac{\left(2 x + 3\right) \left(4 x - 3\right)}{4 x - 3} = 5 \left(2 x + 3\right) \left(4 x - 3\right)$

Cancel Equal like terms
$\left(4 x - 3\right) + \left(2 x + 3\right) = 5 \left(2 x + 3\right) \left(4 x - 3\right)$
$6 x = 5 \left(8 {x}^{2} - 6 x + 12 x - 9\right)$
$6 x = 5 \left(8 {x}^{2} + 6 x - 9\right)$
$6 x = 40 {x}^{2} + 30 x - 45$
Subtract $6 x$ from both side
$6 x - 6 x = 40 {x}^{2} + 30 x - 45 - 6 x$
$0 = 40 {x}^{2} + 24 x - 45$

$40 {x}^{2} + 24 x - 45 = 0$
comparing with $a {x}^{2} + b x + c = 0$
$a = 40 , b = 24 , c = - 45$
Solution $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 24 \pm \sqrt{{\left(24\right)}^{2} - 4 \left(40\right) \left(- 45\right)}}{2 \left(40\right)}$
$x = \frac{- 24 \pm \sqrt{576 + 7200}}{80}$
$x = \frac{- 24 \pm \sqrt{7776}}{80}$
$x = \frac{- 24 \pm \sqrt{1296 \times 6}}{80}$
$x = \frac{- 24 \pm 36 \sqrt{6}}{80}$
$x = \frac{- 3 \pm 3 \sqrt{6}}{10}$