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# What is the value of y? (1) 3|(x^2) 4| = y 2 (2) |3 y| = 11

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Director
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What is the value of y? (1) 3|(x^2) 4| = y 2 (2) |3 y| = 11 [#permalink]

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24 Jul 2008, 15:47
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What is the value of y?

(1) 3|(x^2) – 4| = y – 2

(2) |3 – y| = 11

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Director
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24 Jul 2008, 17:01
jimmyjamesdonkey wrote:
What is the value of y?

(1) 3|(x^2) – 4| = y – 2

(2) |3 – y| = 11

1: clearly insufficient and by the way y >=2
2: insufficient two solutions (one for y < 3, second for y >= 3)

1&2: sufficient because y>= 2 reduces the number of solutions in 2 from 2 to 1. Too lazy to calculate.

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Director
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24 Jul 2008, 17:16
The answer is C, but I have been having a really hard time grasping the concepts....

Let me walk through...please show me my error(s)....

3|(x^2) – 4| = y – 2

agree y >= 2

But still insuff...

Statement B.

Case 1: If 3-y > 0 (or 3 > y) then y = -8
Case 2: If -(3-y) > 0 (or y > 3), then y = 14

Together we know y>= 2....So I try to pick what case we in...but y>= 2 then both our case statements are sufficient. y>=2 is true for both 3>y and y>3.

I get stuck here. Can anyone elaborate on this?

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24 Jul 2008, 17:29
jimmyjamesdonkey wrote:
What is the value of y?

(1) 3|(x^2) – 4| = y – 2

(2) |3 – y| = 11

3*|abs| =+ positive number, thus y-2>0 i.e y>2

|3-y|=11 i.e y=-8 or y=14

1) 2) together we know that y has to be positive thus y=14 C it is..

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Director
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24 Jul 2008, 17:32
I guess I'm confused with the case system...

We essentially have the case If Y < 3 then negative If Y > 3 then positive answer. How do we know what case we are in using Statement 1? Statement 1 tells us y>= 2 which fits both cases.

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24 Jul 2008, 18:03
jimmyjamesdonkey wrote:
Case 1: If 3-y > 0 (or 3 > y) then y = -8
Case 2: If -(3-y) > 0 (or y > 3), then y = 14

Together we know y>= 2....So I try to pick what case we in...but y>= 2 then both our case statements are sufficient. y>=2 is true for both 3>y and y>3.

I get stuck here. Can anyone elaborate on this?

condition 2 gives y = -8 or y = 14
condition 1 gives y >= 2

then y = -8 doesn't satisfy both conditions -> y = 14 is the only solution. Does that help?

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24 Jul 2008, 18:27
could someone please walk me through statement 1 to get y >=2?
thx

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Director
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24 Jul 2008, 18:31
maratikus, I absolutely see what your saying...but I always learned this abs value equations as IF THEN statements. So I guess I disagree what you are saying statement 2 is saying. I think it is a IF, THEN statement.

If Y < 3, THEN Y = -8
If Y > 3, THEN Y = 14

So before I solve these...I need to figure out what case I am in. Basically, is Y > 3 or Y < 3, and this will tell me my answer to Y. So my question is how can we tell if Y > 3 or Y < 3.

From Statement 1 we only know Y > = 2. That doesn't help me answer if Y > 3 or Y < 3.

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Director
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24 Jul 2008, 18:42
jimmyjamesdonkey wrote:
maratikus, I absolutely see what your saying...but I always learned this abs value equations as IF THEN statements. So I guess I disagree what you are saying statement 2 is saying. I think it is a IF, THEN statement.

If Y < 3, THEN Y = -8
If Y > 3, THEN Y = 14

So before I solve these...I need to figure out what case I am in. Basically, is Y > 3 or Y < 3, and this will tell me my answer to Y. So my question is how can we tell if Y > 3 or Y < 3.

From Statement 1 we only know Y > = 2. That doesn't help me answer if Y > 3 or Y < 3.

you learned it wrong. If statements only help find solutions. the best way to check a solution is put it back in the equation:

|3-(-8)| = |3+8| = 11 -> y = -8 satisfies |3-y| = 11
|3-14| = |-11| = 11 -> y = 14 satisfies |3-y| = 11

Now you know that -8 and 14 are solutions of the second equation. Combine that with y >= 2 -> y = 14 is the only solution

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24 Jul 2008, 18:43
young_gun wrote:
could someone please walk me through statement 1 to get y >=2?
thx

3|(x^2) – 4| = y – 2
since the left side as absolute value is greater or equal than zero, then the right side is also non-negative -> y - 2 >= 0

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24 Jul 2008, 18:45

~~~~~~~~~~~~~~
(1) INSUFFICIENT: Since this equation contains two variables, we cannot determine the value of y. We can, however, note that the absolute value expression |x2 – 4| must be greater than or equal to 0. Therefore, 3|x2 – 4| must be greater than or equal to 0, which in turn means that y – 2 must be greater than or equal to 0. If y – 2 > 0, then y > 2.

(2) INSUFFICIENT: To solve this equation for y, we must consider both the positive and negative values of the absolute value expression:

If 3 – y > 0, then 3 – y = 11
y = -8

If 3 – y < 0, then 3 – y = -11
y = 14

Since there are two possible values for y, this statement is insufficient.

(1) AND (2) SUFFICIENT: Statement (1) tells us that y is greater than or equal to 2, and statement (2) tells us that y = -8 or 14. Of the two possible values, only 14 is greater than or equal to 2. Therefore, the two statements together tell us that y must equal 14.

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Director
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24 Jul 2008, 18:46
I guess in this question is doesn't even matter if Y > 3 or Y < 3....We know we have a positive and negative solution either way...and statement 1 tells us our OA must be positive.

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24 Jul 2008, 23:47
jimmyjamesdonkey wrote:
What is the value of y?

(1) 3|(x^2) – 4| = y – 2

(2) |3 – y| = 11

I try to elaborate this problem step by step.

1. Statement 1. the left side of the statement is always positive and can be in a range [0..+inf). Therefore, y can be in a range [2,+inf), or y>=2. The statement insufficient.

2. Statement 2. Apply If/then approach:

3. if (3-y)>=0 then 3-y=11 --> y=-8

4. Our answer must satisfy our condition: (3-(-8))=11>=0 - it satisfies. the condition (3-y)>=0 need only for checking our answer.

5. if (3-y)<0 then -3+y=11 --> y=14

6. Our answer must satisfy our condition: (3-14)=-11<0 - it satisfies. The condition (3-y)<0 need only for checking our answer.

7. Now, you should forget about the equation in the second statement and remember only that the second statement says: there is two possibilities: y=-8 and y=14. Your conditions were necessary only for opening modulus and verification of solutions.

8. Therefore, the second statement is insufficient to answer the question.

9. 1&2 statements. only y=14 satisfies y>=2 condition. Therefore, the answer is C.

Jimmyjamesdonkey, to clarify this question, you should ask yourself: For what purpose do we use conditions y>=3 and y<3. The purpose is to open modulus and check solution. No more.
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Director
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25 Jul 2008, 08:32
Thanks all for the help understanding. This is a semi-related question, but when dealing with abs values and solving for a variable. Do you plug back into the original equation with the abs values OR do you plug into the modified question (equations generated from positive case/negative case)?

Actually, it probably doesn't matter right?

Also, does checking against the conditions take the place of plugging the solution back into the equation. For example, can you have a condition that x > 2, x = 1, and x = 1 satisfies the equation but not the condition? Prob no, right?

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25 Jul 2008, 09:58
look remember this.. |abs| is always positive..

so if the Right Hand Side 3*|abs| is positive then the Left hand side has to be positive..which means y has to be greater than 2..

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25 Jul 2008, 10:15
jimmyjamesdonkey wrote:
What is the value of y?

(1) 3|(x^2) – 4| = y – 2

(2) |3 – y| = 11

I took a simplified approach. Please let me know if my thought process is constipated:

1- Insuff, anytime there are two unidentified variables, it cannot be solved.

C- We know value of (y)= -8, 14. y cannot = -8 based on 3|(x^2) – 4| = y – 2
y=14 is positive and possible.

Too simple?

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Re: manhattan abs value   [#permalink] 25 Jul 2008, 10:15
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