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What is the value of │x + 7│?

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Manager
Joined: 30 Jan 2006
Posts: 64
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What is the value of │x + 7│? [#permalink]  12 Aug 2007, 11:53
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Q: What is the value of │x + 7│?

(1) │x + 3│= 14
(2) (x + 2)2 = 169
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 98 [0], given: 0

(C) for me

│x + 7│ = ?

from 1
│x + 3│= 14
<=> x+3 = 14 or -(x+3) = 14
<=> x = 11 or x = -17

Then, we should check both values in |x+7|
o |11+7| = 18
o |-17+7| = 10

INSUFF.

from 2
(x + 2)^2 = 169 = 13^2 = (-13)^2
<=> x+2 = 13 or x+2 = -13
<=> x = 11 or x = -15

Then, we should check both values in |x+7|
o |11+7| = 18
o |-15+7| = 8

INSUFF.

Both (1) & (2)
We have to be with x = 11.

SUFF.

Last edited by Fig on 12 Aug 2007, 13:49, edited 1 time in total.
Director
Joined: 11 Jun 2007
Posts: 932
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Re: Mod-DS [#permalink]  12 Aug 2007, 13:46
smily_buddy wrote:
Q: What is the value of │x + 7│?

(1) │x + 3│= 14
(2) (x + 2)2 = 169

i get C for the answer.

(1) │x + 3│= 14
x + 3 = 14 -> x = 11
x + 3 = -14 -> x = -17
not sufficient

(2) (x + 2)2 = 169

x = 11, -15
not sufficient

putting (1) and (2) together, we know that x should be 11
so we can solve for │x + 7│
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 70 [0], given: 0

Fig wrote:
(C) for me

│x + 7│ = ?

from 1
│x + 3│= 14
<=> x+3 = 14 or -(x+3) = 14
<=> x = 11 or x = -17

Then, we should check both values in |x+7|
o |11+7| = 18
o |-17+7| = 10

INSUFF.

from 2
(x + 2)^2 = 169 = 13^2 = (-13)^2
<=> x+2 = 13 or x+2 = -13
<=> x = 11 or x = -15

Then, we should check both values in |x+7|
o |11+7| = 18
o |-15+7| = 8

INSUFF.

Both (1) & (2)
We have to be with x = 11.

SUFF.

whoa we both posted around the same time! i can kinda rusty at checking at the values but the only time values may be undefined is when we have zero for the denominator? just to make sure, so in this case, would -7 work? or do we just worry about it if we had zero for denominator as i had mentioned? thanks!
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 98 [0], given: 0

beckee529 wrote:
Fig wrote:
(C) for me

│x + 7│ = ?

from 1
│x + 3│= 14
<=> x+3 = 14 or -(x+3) = 14
<=> x = 11 or x = -17

Then, we should check both values in |x+7|
o |11+7| = 18
o |-17+7| = 10

INSUFF.

from 2
(x + 2)^2 = 169 = 13^2 = (-13)^2
<=> x+2 = 13 or x+2 = -13
<=> x = 11 or x = -15

Then, we should check both values in |x+7|
o |11+7| = 18
o |-15+7| = 8

INSUFF.

Both (1) & (2)
We have to be with x = 11.

SUFF.

whoa we both posted around the same time! i can kinda rusty at checking at the values but the only time values may be undefined is when we have zero for the denominator? just to make sure, so in this case, would -7 work? or do we just worry about it if we had zero for denominator as i had mentioned? thanks!

Actually, we should check the values because we could obtain with 2 different values of x a similar value of |x+7|

For instance, we could be with x=8 and x=-22. We would be with |x+7| = 15 and so be able to determine a unique value, answering to the question
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