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

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Answer 1 · 2017-11-02T09:21:35

$x = 1.436$
and
$x = 3.064$

$1/(2(x-3))+3/(2-x)=5 x$

Make the denominators equal:

$1/(2(x-3)) xx(2-x)/(2-x) +3/(2-x) xx (2(x-3))/(2(x-3))=5 x$

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

Now we can add the numerators:

$=>( (2-x) + 6(x-3))/(2(x-3)(2-x)) =5$

$(2-x +6x-18)/(2(x-3)(2-x)) =5$

Transposition :

$=> (2-x +6x-18) = 5xx2(x-3)(2-x)$

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

$=> 5x -16 = 10(2x-x^2 -6+ 3x)$

$=> 5x -16 = 10(5x-x^2 -6)$

$=> 5x -16 = 50x-10x^2 -60$

$=> 50x-5x-10x^2 -60+16 = 0$

$=> -10x^2 +45x -44 =0$

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

$x=( -b +-sqrt(b^2 -4ac))/(2a)$

Here $a= 10, b= -45 and c= 44$

$b^2 -4ac = 2025-1760 = 265$

We get :
$x = 1.436$ or $x = 3.064$

How do you solve 1/(2(x-3))+3/(2-x)=5 x1/(2(x-3))+3/(2-x)=5 x?