Question

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

Answer 1

x=3/2 , -3

Answer 2

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

Answer 3

The denominator of the LHS of the equation factorises into
(x+3)(x-3).
Multiply the whole equation by (x+3)(x-3) to get this:

4x^2 -2x (x+3) = 3(x-3)
4x^2 -2x^2 -6x =3x -9
2x^2 -6x = 3x -9
Collecting all the terms on the LHS of the equation, and thus reducing the RHS to zero.
2x^2 -6x -3x +9 =0

2x^2-9x +9 = 0


So, factorising 2x^2 -9x +9, ie, the LHS gives y ou

(2x -3)(x - 3) for the LHS and zero for the RHS.
So, (2x-3)(x-3) =0

Therefore, 2x-3 =0, and x-3 =0.

So, x has two values, ie, 3/2 and 3.

But lolking at the denominators, you'll find that if we put x=3 in
x^2-9 gives us zero.

Putting x=3 in the denominator (x-3) also equals zero.

ie, putting x=3 in the equation gives us this:

4 (9)/(9-9) -2(3)/(3+3) = 3/(3-3)

So here we have two denominators which become zero, when x=3.

In other words, we are dividing by zero. Which is not acceptable.

So x=3 is not acceptable as a solution.

The only acceptable solution is therefore 2x=3, ie x=3/2.



Put x=3/2 in your equation, and the LHS gives you:

(9)(4)/(9-36) - (3)(2)/(3+6)

ie, 36/-27 - 6/9
ie -4/3 -2/3, which equals -2.

So for the LHS of the equation, x=3/2 gives us -2.

The RHS of the equation (the orignal in the question) is:

3/(x-3)

Wheln x=3/2, this becomes 6/(-3) which equals -2.

But for x=3/2, the LHS is also -2.

Hence although simplifying the equation gave us two answers (ie, 3 and 3/2), only x=3 is acceptable

Answer 4

Sorry, the last two lines should read:
"Hence although simplifying the equation gave us two answers (ie, 3 and 3/2), only x=3/2 is acceptable".