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?