Solve the equation for the variable x?

2/x-2 + 3/x+2 = -4x/x^2-4 --- > (1)

Take the LHS

LHS = (2(x + 2) + 3(x - 2)) /(x^2 - 4)

= (2x + 4 + 3x - 6) / (x^2 - 4)

= (5x - 2) / (x^2 - 4)

Now (1) becomes,

(5x - 2) / (x^2 - 4) = -4x / (x^2 - 4)

5x - 2 = -4x

9x = 2

x = 2/9

SHOULD be written as :-
2 / (x - 2) + 3/(x + 2) = (- 4x) / (x² - 4)
2 / (x - 2) + 3 / (x + 2) = (- 4x) / [ (x - 2)(x + 2) ]
2(x + 2) + 3(x - 2) = - 4x
5x - 2 = - 4x
9x = 2
x = 2/9

NOTE
2/x - 2 and 2/(x - 2) are NOT THE SAME

2/x-2 + 3/x+2 = -4x/x^2-4
2(x+2)/(x-2)(x+2) + 3(x-2)/(x+2)(x-2) = -4x/x^2-4
2x+4+3x-6/x^2-4 = -4x/x^2-4
5x - 2 = -4x
9x = 2
x = 2/9

2/(x - 2) + 3/(x + 2) = -4x/(x^2 - 4)
(x + 2)(x - 2)[2/(x - 2) + 3/(x + 2)] = (x + 2)(x - 2)[-4x/(x + 2)(x - 2)
2(x + 2) + 3(x - 2) = -4x
2x + 4 + 3x - 6 = -4x
2x + 3x + 4x = -4 + 6
9x = 2
x = 2/9

2/x-2 + 3/x+2 +4x/x^2-4 = 0
(2(x+2)+3(x-2)+4x)/x^2-4 = 0
(2x+4+3x-6+4x)/x^2-4 = 0
9x-2/x^2-4 = 0
x=2/9

2/x-2 +3/x+2 = -4x/x^2 -4
(cancel out -2 and +2 left hand side i.e they become zero)
2/x+3/x = -4x/x^2-4
5/x= -4x/x^2-4
(cancel out x from -4x with x^2 it remains -4/x)
5/x=-4/x-4
( now take all common at one side to solve rest of problem i.e i mean x and when - 4/x goes to left side it become + 4/x)
5/x+4/x = -4
9/x=-4
so x=-9/4

[2(x+2)+3(x-2)]/[(x-2)(x+2)] = -4x/(x^2-4)

the last term is x^2-4...use difference of 2 squares and factor.

[2(x+2)+3(x-2)]/[(x-2)(x+2)] = -4x/[(x+2)(x-2)]

the denominators are the same now so cross them out.

2(x+2)+3(x-2) = -4x

expand.

2x+4+3x-6 = -4x

group like terms.

5x-2 = -4x
9x = 2

x = 2/9

i think.

2/(x-2) + 3/(x+2)= -4x/(x^2 -4)
= - 4x/{(x-2)(x+2)}
take lcm on lhs;

{2(x+2) + 3(x-2)} / (x+2)(x-2) = -4x/{(x-2)(x+2)}

cancel {(x-2)(x+2)} from the denominator of both sides...

2(x+2) + 3(x-2)} =-4x
=>2x+4+3x-6=-4x
=>9x=2
=>x= 2/9

2/x-2 + 3/x+2 = -4x/x^2-4

equal to

2(x+2) + 3(x-2) = -4x

2x +4 + 3x -6 = -4x

9x - 2

therefore...

x = 2/9

2/x-2 + 3/x+2 = -4x/x^2-4

x^2-4=(x-2)(x+2):


[2(x+2)+ 3(x-2)+4x]/x^2-4=0

2x+4+3x-6+4x=0
9x=2
x=2/9.