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Absolute Value Problem-Taken From Manhattan Gmat!

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by pkw209 » Wed Feb 17, 2010 4:03 pm
What is the value of y?

(1) 3|x^2 - 4| = y - 2

(2) |3 - y| = 11

Answer is C

My question is why can't you have two separate equations for statement 1----> -3x^2 + 4 = y-2 and 3x^2 - 12 = y-2? This would lead to x=2 or -2 and then y=2 and statement 1 would be sufficient.

Experts please help!

Thanks.
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by Stuart@KaplanGMAT » Wed Feb 17, 2010 6:52 pm
pkw209 wrote:What is the value of y?

(1) 3|x^2 - 4| = y - 2

(2) |3 - y| = 11

Answer is C

My question is why can't you have two separate equations for statement 1----> -3x^2 + 4 = y-2 and 3x^2 - 12 = y-2? This would lead to x=2 or -2 and then y=2 and statement 1 would be sufficient.

Experts please help!

Thanks.
The two equations you've set up (small math errors aside) are ALTERNATIVE solutions, not MULTIPLE solutions.

In other words, when we have:

|x| = 2, we solve as:

+x = 2 OR -x = 2

whereas you've solved it as:

+x = 2 AND -x = 2.
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by pkw209 » Wed Feb 17, 2010 7:03 pm
Thanks, Stuart. However, I'm still a little confused. I looked at both x = 2 or x = -2 and the answer is y=2 for both so how come statement 1 isn't sufficient?

Sorry for not getting it.
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by Stuart@KaplanGMAT » Wed Feb 17, 2010 7:10 pm
pkw209 wrote:Thanks, Stuart. However, I'm still a little confused. I looked at both x = 2 or x = -2 and the answer is y=2 for both so how come statement 1 isn't sufficient?

Sorry for not getting it.
Because you haven't proven that x=2 or -2.

First, fixing the math errors:

3|x^2 - 4| = y - 2

|x^2 - 4| = (y - 2)/3

(to be safe, you should isolate the inequality before solving, although it doesn't matter here)

+(x^2 - 4) = (y-2)/3 OR -(x^2 - 4) = (y-2)/3

x^2 = (y-2)/3 + 4 OR -(x^2) = (y-2)/3 - 4

Since we have two variables in each equation, there's no possible way to solve either one.

You combined the two equations to solve, which as I explained in my original post, you're not allowed to do; they're alternative, not joint, equations.
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by pkw209 » Wed Feb 17, 2010 7:38 pm
got it. thanks
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by vertigo05 » Wed Mar 17, 2010 9:50 am
i'm getting the answer as "E". How come its "C". Can anybody explain?
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by boazkhan » Wed Mar 17, 2010 10:07 am
This is how I solved it....I solved eq 2 first since its easier. We get two values of Y, -8 & 14. This alone is insufficent.
Eq 1 alone is also insufficent as we have two variables and one equation. Taking both equations together since Mod x^2-4 = y-2/3 will always be positive only the positive value of Y (y=14) will satisfy this equation. Therefore C is the answer.
Anyone..is this the correct approach?
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by gmatmachoman » Wed Mar 17, 2010 11:00 am
vertigo05 wrote:i'm getting the answer as "E". How come its "C". Can anybody explain?
yeah. u r partially true.But going a step further u can solve for Y without knowing value of X.

Now we have 2 values of Y (-8 & 14)

In st 1, the LHS will be positive for any value of X. So ,RHS also needs to be Postive .So we are left with Y =14.

Going bit more ahead....solve for X:(Not necessary here..just for ur happiness) hehehe

X can be 2 Sqrt 2 or -2sqrt 2 ,0
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by Testluv » Wed Mar 17, 2010 11:45 am
I gave an explanation for this question here: https://www.beatthegmat.com/value-of-y-t ... tml#201008
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by vertigo05 » Wed Mar 17, 2010 4:37 pm
Thanx Testluv. Gr8 explanation.
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