If \(\frac{(ab)^2+3ab-18}{(a-1)(a-2)}=0\) where \(a\) and

[VJesus12]

If \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\) where \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)

I. 1
II. 2
III. 3

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

[Jay@ManhattanReview]

If \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\) where \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)

I. 1
II. 2
III. 3

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

Given that \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\), we have (ab)^2 + 3ab - 18 = 0

=> (ab)^2 + 6ab - 3ab - 18 = 0

ab(ab + 6) - 3(ab + 6) = 0

(ab + 6)*(ab - 3) = 0

=> ab = -6 or 3

Let's take each option one by one.

I. 1 : If b = 1, then with ab = -6, we have a = -6 (valid value for a); similarly, with ab = 3, we have a = 3 (valid value for a); so b can be 1.

II. 2: If b = 2, then with ab = -6, we have a = -3 (valid value for a); however, with ab = 3, we cannot have b = 2, else a would not be an integer; so b can be 2.

III. 3: If b = 3, then with ab = -6, we have a = -2 (invalid value for a since at a = -2, the denominator of the expression would turn 0, and the value of the expression would be indeterminable. Similarly, with ab = 3, we have a = 1 (invalid value for a since at a = 1, again, the denominator of the expression would turn 0, and the value of the expression would be indeterminable. Thus, b can't be 3.

The correct answer: C

Hope this helps!

-Jay
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[GMATGuruNY]

If \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\) where \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)

I. 1
II. 2
III. 3

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only

The equation is valid only if (ab)²+3ab-18 = 0.

I: b=1
Plugging b=1 into (ab)²+3ab-18 = 0, we get:
a²+3a-18 = 0
(a+6)(a-3) = 0
a=-6 or a=3
This works.
Since the correct answer must include I, eliminate B.

II: b=2
Plugging b=2 into (ab)²+3ab-18 = 0, we get:
4a²+6a-18 = 0
2a²+3a-9 = 0
(2a-3)(a+3) = 0
Since a must be an integer, a=-3.
This works.
Since the correct answer must include II, eliminate A and D.

III: b=3
Plugging b=3 into (ab)²+3ab-18 = 0, we get:
9a²+9a-18 = 0
a²+a-2 = 0
(a+2)(a-1) = 0
The expression in red constitutes the denominator of the equation in the prompt.
Since division by 0 is not allowed, this expression cannot be equal to 0.
Thus, b=3 is not viable.
Since the correct answer cannot include III, eliminate E.

The correct answer is C.

[Scott@TargetTestPrep]

If \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\) where \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)

I. 1
II. 2
III. 3

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

If the value of a fraction is 0, the numerator must be 0 (providing that the denominator is not 0). Therefore, we have:

(ab)^2 + 3ab - 18 = 0

(ab + 6)(ab - 3) = 0

ab = -6 or ab = 3

Notice that a can't be 1 or -2 (otherwise the denominator will be 0), so for ab = -6, we have:

if a = 2, b = -3; if a = 3, b = -2; if a = 6, b = -1; if a = -1, b = 6; if a = -3, b = 2; and if a = -6, b = 1.

Similarly, for ab = 3; we have:

if a = 3, b = 1; if a = -1, b = -3; and if a = -3, b = -1.

We see that b can be either 1 or 2, but it can't be 3.

Answer: C