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

Nov 2, 2017

$x = 1.436$
and
$x = 3.064$

#### Explanation:

$\frac{1}{2 \left(x - 3\right)} + \frac{3}{2 - x} = 5 x$

Make the denominators equal:

$\frac{1}{2 \left(x - 3\right)} \times \frac{2 - x}{2 - x} + \frac{3}{2 - x} \times \frac{2 \left(x - 3\right)}{2 \left(x - 3\right)} = 5 x$

(2-x)/(2(x-3)(2-x)) + (3xx2(x-3))/(2(x-3)(2-x) =5

Now we can add the numerators:

$\implies \frac{\left(2 - x\right) + 6 \left(x - 3\right)}{2 \left(x - 3\right) \left(2 - x\right)} = 5$

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

Transposition :

$\implies \left(2 - x + 6 x - 18\right) = 5 \times 2 \left(x - 3\right) \left(2 - x\right)$

=> 5x -16 = 10(x(2-x) -3(2-x)

$\implies 5 x - 16 = 10 \left(2 x - {x}^{2} - 6 + 3 x\right)$

$\implies 5 x - 16 = 10 \left(5 x - {x}^{2} - 6\right)$

$\implies 5 x - 16 = 50 x - 10 {x}^{2} - 60$

$\implies 50 x - 5 x - 10 {x}^{2} - 60 + 16 = 0$

$\implies - 10 {x}^{2} + 45 x - 44 = 0$

$= 10 {x}^{2} - 45 x + 44 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Here$a = 10 , b = - 45 \mathmr{and} c = 44$
${b}^{2} - 4 a c = 2025 - 1760 = 265$
$x = 1.436$ or $x = 3.064$