Given: $\frac{4x}{(x^2 -3x)} - \frac{2}{7} = 0$

Add $\frac{2}{7}$ to both sides to get: $\frac{4x}{(x^2 -3x)} = \frac{2}{7}$

Cross multiply to get: $(7)(4x) = (2)(x^2 -3x)$

Expand both sides: $28x = 2x^2-6x$

Subtract $28x$ from both sides: $0 = 2x^2-34x$

Divide both sides by 2: $0 = x^2-17x$

Factor: $0 = x(x-17)$

So, EITHER x = 0 OR x = 17

Since we're told that x > 0, we can conclude that x = 17

Answer: 17

If x>0 and $\frac{4x}{x^2-3x}-\frac{2}{7}=0$, what is the value of x?

Given that x>0 and $\frac{4x}{x^2-3x}-\frac{2}{7}=0$ and we need to find the value of x

$\frac{4x}{x^2-3x}-\frac{2}{7}=0$
=> $\frac{4x}{x^2-3x}=\frac{2}{7}$

Cross multiplying by 7 and $x^2-3x$ we get

7 *4x = 2 * ($x^2-3x$)

Dividing both sides by 2 we get
7*2x = $x^2-3x$
=> $x^2 - 3x$ - 14x = 0
=> $x^2 - 17x$ = 0
=> x*(x-17) = 0
=> x = 0 or x = 17

Given that x > 0
=> x = 17

So, Answer will be 17
Hope it helps!

this is how I solved it... multiply both sides by 7*(x^2 - 3x) to get 7(4x) - 2 (x^2 - 3x) = 0
which gives us 28x - 2x^2 + 6x = 0. which is also 34x - 2x^2 = 0.
then we add 2x^2 to both sides to get 34x = 2x^2. divide both sides by 2 to get 17x = x^2. since x > 0, we divide x by both sides to get x = 17. Does this work Brent?