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# Absolute fundamentals

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Director
Joined: 07 Jun 2004
Posts: 610

Kudos [?]: 948 [0], given: 22

Location: PA

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31 Jul 2010, 04:30
Can some one explain in steps how to solve the absolute equations algebraically and when do we flip the inequality sign

EG : | x + 3 | > -10

how do we come up with the solutions for this inequality
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Kudos [?]: 948 [0], given: 22

Manager
Joined: 16 Jun 2010
Posts: 181

Kudos [?]: 103 [1], given: 5

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31 Jul 2010, 08:38
1
KUDOS
Try to remember these fundamentals, they will help you open the modulus:
#1
if you have |quantity| < 4
--> -4 < quantity < 4.
#2
if you have quantity^2 < 16
--> -4 < quantity < 4.

#3
if you have |quantity| > 4
--> quantity > 4 OR quantity < -4.
#4
if you have quantity^2 > 16
--> quantity > 4 OR quantity < -4.

So according to these, your question becomes type 3:
i.e: x+3> -10 or x+3< 10 i.e x>-13 or x<7

You should be checking the Math book post on Absolute values to get a better understnading.
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Please give me kudos, if you like the above post.
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Kudos [?]: 103 [1], given: 5

Director
Joined: 07 Jun 2004
Posts: 610

Kudos [?]: 948 [0], given: 22

Location: PA

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31 Jul 2010, 09:42
Thanks how do we handle this with multiple

| x + 3 | > | x - 14 | + 1

how many solutions does the above inequality have
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Kudos [?]: 948 [0], given: 22

Manager
Joined: 16 Jun 2010
Posts: 181

Kudos [?]: 103 [0], given: 5

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31 Jul 2010, 14:31
I think, your question above is incorrect.
math-absolute-value-modulus-86462.html

If the above post helped then would appreciate Kudos
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Re: Absolute fundamentals   [#permalink] 31 Jul 2010, 14:31
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