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

A. 15
B. 17
C. 19
D. 21
E. 23

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

Cross multiply to get: (2)(x² - 3x) = (7)(4x)
Expand to get: 2x² - 6x = 28x
Subtract 28x from both sides to get: 2x² - 34x = 0
Factor to get: 2x(x - 17) = 0
So, EITHER x = 0 OR x = 17

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

Answer: B

Cheers,
Brent

Dear GMATPrepNow Brent

This problem stimulates a question. Suppose the question does not say x >0. Can x=0 be a valid solution? I do not think so as it will turn the original fraction into 0/0 which is undefined. My take away that I need to refer to orginal question after finding solution.

Is my thinking right. What do you think?

Thanks in advance
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Dear GMATPrepNow Brent

This problem stimulates a question. Suppose the question does not say x >0. Can x=0 be a valid solution? I do not think so as it will turn the original fraction into 0/0 which is undefined. My take away that I need to refer to orginal question after finding solution.

Is my thinking right. What do you think?

Thanks in advance

You're absolutely right; x = 0 cannot be a valid solution.
So, regardless of the proviso that x>0, the ONLY solution is x = 17

Cheers,
Brent
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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 B
Hope it helps!
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