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Bunuel
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Nidzo
1) Looking at the top line alone: 9ab = 18 tells us that either one of a or b is 2 and the other is 1.
Looking at the bottom line: 3+a = 5 tells us that a = 2, and therefore b = 1

SUFFICIENT

2) Only gives us the value of b, and we cannot deduce a from this.

INSUFFICIENT

Answer A

Your answer is correct, but your approach is incorrect.
Why are you assuming that a and b are integers?
If you simplify the equation using the (a^2-b^2) = (a-b)(a+b) formula in the denominator, after applying the LCM you will get the exact expression mentioned in statement (1).
Hence, option A is sufficient alone to answer the target question!
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Hi chetan2u Bunuel

Why are we not considering the fact that "a" cannot be equal to +-3 as (9-a^2) is in denominator ? If we use b=1, it ensures that the a =2, and hence removes the possibility of "a" being +-3. Why isn't the correct answer C ? Using the equation in statement 1, we just get x = 18(3-a)/5(3-a) which would be equal to 18/5 IF it was mentioned a is not equal to 3.

What am I missing ? Please clarify.
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Bunuel
If \(x = \frac{9b - 3ab}{\frac{3}{a} - \frac{a}{3}}\), what is x?


(1) \(\frac{9ab}{3+a} = \frac{18}{5}\)

(2) \(b = 1\)


Statement I should tell you that the initial equation requires to be simplified to check statement I.

\(x = \frac{9b - 3ab}{\frac{3}{a} - \frac{a}{3}}\)

\(x = \frac{9b - 3ab}{\frac{9-a^2}{3a}}\)

\(x=\frac{3a(9b-3ab)}{9-a^2}\)

\(x=\frac{3a*3b(3-a)}{(3-a)(3+a)}=\frac{9ab}{3+a}\)

Statement I gives this value as 18/5.
Sufficient

Statement II gives b=1
So, \(x=\frac{9*a*1}{3+a}=\frac{9a}{3+a}\)
x depends on a.
If a is 1, x is 9/4
If a is 2, x is 18/5
If a is 3, x is 27/6 or 9/2 and so on
Insufficient

A
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