Solve the following equation for x. 3x/x-2 - 1/2 = 6/x-2 .?

3x/(x - 2) - 1/2 = 6/(x - 2)

=> 3x/(x - 2) - 6/(x - 2) = 1/2

=> (3x - 6)/(x - 2) = 1/2

=> 3(x - 2)/(x- 2) = 1/2

here (x - 2) cancel each other assuming x # 2. So solutions are not possible. But if we don't cancel them then

=>6(x - 2) = (x - 2)

=> 6x - 12 = x - 2

=> 5x = 10

=> x = 2

now x= 2

So, either our assumption is wrong or the solutions is not possible.

if we put x = 2 in the given equation, it does not satisfy the given equation. It becomes

(3 x 2)/(2 - 2) - 6/( 2 - 2)

= 6 /0 - 6/0 ==>Math error

So, we can say that here solutions are not possible.

3x/x-2 - 1/2 = 6/x-2

switch -1/2 and 6/x-2 and you get:
3x/x-2 - 6/x-2 = 1/2
same denominator (x-2) so you just subtract 6 from 3x:
3x-6/x-2 = 1/2
cross multiply:
2(3x-6) = x-2
solve for x:
x = 2

hope it helps :)

3x/(x-2)-1/2=6/(x-2)
3x/(x-2)-6/(x-2)=1/2
(3x-6)/(x-2)=1/2
2(3x-6)=x-2
6x-12=x-2
5x=10
x=10/5
x=2

x=2

EDIT: What do you mean you you made a mistake?! lmao anyways you would still get x=2 as your solution.

Question MUST have brackets :-

3x / (x - 2) - 1 / 2 = 6 / (x - 2)
This is NOT the question you have typed.

6x - (x - 2) = 12
5x + 2 = 12
5x = 10
x = 2

REPEAT
Brackets are a MUST !

3x/(x - 2) - 1/2 = 6/(x - 2)
3x/(x - 2) - 6/(x - 2) = 1/2
2[x - 2][(3x - 6)/(x - 2)] = 2[x - 2][1/2]
2(3x - 6) = x - 2
2*3x - 2*6 = x - 2
6x - 12 = x - 2
6x - x = 12 - 2
5x = 10
x = 10/5
x = 2

3x/x-2 - 1/2 = 6\x-2
3x/x-2 - 6x/x-2 =1/2
3x-6\x-2=1/2

we cross multiply
2(3x-6)=x-2
6x-12=x-2
5x=10 x = 2