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If ab>0 and (a^2)*(b^2) + 2ab – 3 = 0

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by Brent@GMATPrepNow » Wed Jul 15, 2015 6:14 am
TheGmatTutor wrote:If ab>0 and (a^2)*(b^2) + 2ab - 3 = 0, which of the following could be a value of a?

I. -1
II. 1
III. 3

(A) I only
(B) II only
(C) I and II only
(D) II and III
(E) I, II and III
Given: (a²)(b²) + 2ab - 3 = 0
Rewrite as: (ab)² + 2ab - 3 = 0

Let's use a technique called u-substitution.
Let u = ab
So, take (ab)² + 2ab - 3 = 0 and replace ab with u to get: u² + 2u - 3 = 0
Factor: (u + 3)(u - 1) = 0
Solve: u = -3 or u = 1
Since u = ab, we can conclude that ab = -3 or ab = 1
Since we're told that ab > 0, we can eliminate the case that ab = -3, which means ab = 1

Now check the options:
I. a = -1. If ab = 1, and a = -1, then b = -1 (perfect - this adheres to the given info)
If we plug those values of a and b into the original equation, everything works

II. a = 1. If ab = 1, and a = 1, then b = 1 (perfect - this adheres to the given info)
If we plug those values of a and b into the original equation, everything works

III. a = 3. If ab = 1, and a = 3, then b = 1/3 (perfect - this adheres to the given info)
If we plug those values of a and b into the original equation, everything works

So, a can equal any of those values.
Answer: E

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Brent
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by vineet.nitd » Wed Jul 15, 2015 8:19 pm
Solve the quadratic in ab to get the values of ab as -3 and 1. However, as per problem ab>0, hence ab=-3 can be discarded.

So, ab = 1.

Now, since there is no restriction on a or b belonging only to set of integers, we can assign any value to 'a' (except 0), such that we always have its multiplicative inverse in b, such that ab=1 is always satisfied.

Hence, option E is the right choice.
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by GMATGuruNY » Thu Jul 16, 2015 2:16 am
TheGmatTutor wrote:If ab>0 and (a^2)*(b^2) + 2ab - 3 = 0, which of the following could be a value of a?

I. -1
II. 1
III. 3

(A) I only
(B) II only
(C) I and II only
(D) II and III
(E) I, II and III
An alternate approach is to test I, II and III in the original quadratic.

I: a=-1
Plugging a=-1 into a²b² + 2ab - 3 = 0, we get:
(-1)²b² + 2(-1)b - 3 = 0
b² - 2b - 3 = 0
(b-3)(b+1) = 0.
Here, it's possible that b+1=0, implying that b=-1 and that ab = (-1)(-1) = 1.

II: a=1
Plugging a=1 into a²b² + 2ab - 3 = 0, we get:
1²b² + 2(1)b - 3 = 0
b² + 2b - 3 = 0
(b+3)(b-1) = 0.
Here, it's possible that b-1=0, implying that b=1 and that ab = 1*1 = 1.

III: a=3
Plugging a=3 into a²b² + 2ab - 3 = 0, we get:
3²b² + 2(3)b - 3 = 0
3b² + 2b - 1 = 0
(3b-1)(b+1) = 0.
Here, it's possible that 3b-1=0, implying that b=1/3 that ab = (3)(1/3) = 1.

Since I, II and III are all possible, the correct answer is E.
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by [email protected] » Thu Jul 16, 2015 12:27 pm
Hi All,

The above prompt is remarkably similar to the following GMAC question:

If xy ≠ 0 and x^2*y^2 - xy = 6, which of the following could be y in terms of x?

I. 1/(2x)
II. - 2/x
III. 3/x

A. I only
B. II only
C. I and II
D. I and III
E. II and III

Final Answer: E

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by GMATGuruNY » Thu Jul 16, 2015 1:34 pm
If xy ≠ 0 and x²y² - xy = 6, which of the following could be y in terms of x?

A. I only
B. II only
C. I and II
D. I and III
E. II and III
Here, we can plug in a value for x and solve for y.
Let x=1.
Plugging x=1 into x²y² - xy = 6, we get:
1²y² - 1y = 6
y² - y - 6 = 0
(y+2)(y-3) = 0
y=-2 or y=3.
Since the question stem asks for possible values of y, our targets for y are -2 and 3.

Now plug x=1 into I, II and III to see whether any are equal to -2 or 3.
I. 1/(2x) = 1/(2*1) = 1/2.
II. - 2/x = -2/1 = -2.
III. 3/x = 3/1 = 3.

Since II and III yield our target values for y, the correct answer is E.
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by [email protected] » Thu Jul 16, 2015 7:24 pm
Hi All,

This question is quirky in that it tests you on math rules and patterns that you probably know, but in ways that you're not used to thinking about...

We're told that neither X nor Y are equal to 0. We're also told that (X^2)(Y^2) - XY = 6. We're asked which of the following COULD be the value of Y in terms of X...

The first interesting thing about this question is the use of the word COULD....that word implies that there's MORE THAN ONE possible solution.
The second interesting thing is that the 'term' (XY) can be factored out of the 'left side' of the equation. Normally, you look to factor our a single variable or number, but here, it's the product of two variables that you can factor out. Doing so gives us...

XY(XY - 1) = 6

While this looks complicated, there's an easy pattern here:

(number)(number - 1) = 6

Can you think of 2 numbers, that differ by 1, that you can multiply to get 6?

You should be thinking 2 and 3... because (3)(3-1) = 6

So XY = 3 is a possible solution. In this case, Y = 3/X. The wording of the prompt makes me think that there should be another solution though, so is there ANOTHER pair of numbers, that differ by 1, that you can multiply together to get 6? Hint: the numbers do NOT have to be positive....

How about -2 and -3....

(-2)(-2-1) = 6

So XY = -2 is another possible solution. In this case, Y = -2/X

There's only one answer that includes both of those solutions...

Final Answer: E

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by mallika hunsur » Fri Jul 17, 2015 4:33 am
[email protected] wrote:Hi All,

This question is quirky in that it tests you on math rules and patterns that you probably know, but in ways that you're not used to thinking about...

We're told that neither X nor Y are equal to 0. We're also told that (X^2)(Y^2) - XY = 6. We're asked which of the following COULD be the value of Y in terms of X...

The first interesting thing about this question is the use of the word COULD....that word implies that there's MORE THAN ONE possible solution.
The second interesting thing is that the 'term' (XY) can be factored out of the 'left side' of the equation. Normally, you look to factor our a single variable or number, but here, it's the product of two variables that you can factor out. Doing so gives us...

XY(XY - 1) = 6

While this looks complicated, there's an easy pattern here:

(number)(number - 1) = 6

Can you think of 2 numbers, that differ by 1, that you can multiply to get 6?

You should be thinking 2 and 3... because (3)(3-1) = 6

So XY = 3 is a possible solution. In this case, Y = 3/X. The wording of the prompt makes me think that there should be another solution though, so is there ANOTHER pair of numbers, that differ by 1, that you can multiply together to get 6? Hint: the numbers do NOT have to be positive....

How about -2 and -3....

(-2)(-2-1) = 6

So XY = -2 is another possible solution. In this case, Y = -2/X

There's only one answer that includes both of those solutions...

Final Answer: E

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Rich
Hi Rich, Mitch,

I am thinking-
xy(xy-1)=6 can be
1) xy=6
2) xy-1=6 or xy=7

so y=6/x or 7/x

Why is this line of reasoning not working..?

Thanks,
Mallika
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by GMATGuruNY » Fri Jul 17, 2015 5:06 am
mallika hunsur wrote:I am thinking-
xy(xy-1)=6 can be
1) xy=6
2) xy-1=6 or xy=7

Why is this line of reasoning not working..?
If we substitute xy=6 into xy(xy-1)=6, we get:
6(6-1) = 6
30 = 6.
Doesn't work.

If we substitute xy=7 into xy(xy-1)=6, we get:
7(7-1) = 6
42 = 6.
Doesn't work.

One way to avoid incorrect solutions is to solve as I did in my post above -- by plugging in an easy value for x and solving for y.
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by [email protected] » Fri Jul 17, 2015 7:50 am
Hi Mallika,

Mitch has properly explained why the (X)(Y) cannot be 6 or 7. I'd like to know more about the logic you used to get to that point though. What did you 'see' in that original equation that made you think that (X)(Y) might equal 6?

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by mallika hunsur » Sat Jul 18, 2015 9:07 am
[email protected] wrote:Hi Mallika,

Mitch has properly explained why the (X)(Y) cannot be 6 or 7. I'd like to know more about the logic you used to get to that point though. What did you 'see' in that original equation that made you think that (X)(Y) might equal 6?

GMAT assassins aren't born, they're made,
Rich
Hi Rich,

I have no clue now why i did that. I should've just assigned xy=a and solved a quadratic equation.
Need to get back into the groove :|

Thanks,
Mallika
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by nikhilgmat31 » Wed Jul 29, 2015 5:48 am
good question to select values of b to fit in both equations.
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