SOLUTION: Solve for x | 2(x+1) + 4 | < 10

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Question 1202784: Solve for x
| 2(x+1) + 4 | < 10

Found 3 solutions by mananth, ikleyn, josgarithmetic:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
| 2(x+1) + 4 | < 10
Apply absolute rule
-10< 2(x+1) + 4 < 10
1)
2(x+1) + 4 < 10
subtract 4 from both sides
2(x+1)<6
divide by 2
x+1<3
x<2
II)
2(x+1) + 4>-10
subtract 4 both sides
2(x+1)>-14
divide by2
x+1>-7
x>-8
combining both
-8 < x < 2



Answer by ikleyn(52887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x
| 2(x+1) + 4 | < 10
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


Below I present the solution in very compact form.


Your starting inequality is

    | 2(x+1) + 4 | < 10.


It is equivalent to this compound inequality (which represents, actually, two inequalities at the same time)

    -10 < 2(x+1) + 4 < 10.


Subtract 4 from all three terms (left term, middle term and right term).  You will get

    -10 - 4 < 2(x+1) < 10 - 4

or, equivalently,

    -14 < 2(x+1) < 6.


Now divide all three terms by 2.  You will get

    -7 < x+1 < 3.


Last step is to subtract 1 from all three terms.  You will get

    -8 < x < 2,     which is your  ANSWER

Solved.

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The major lesson to learn from my post is that
you can make these equivalent transformations
SIMULTANEOUSLY with all three terms of a compound inequality.

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To see many other similar and different SOLVED problems on absolute value inequalities,  look into the lesson
    - Solving absolute value inequalities
in this site.



Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
abs%282x%2B2%2B4%29-10%3C0
abs%282x%2B6%29-10%3C0
abs%28x%2B3%29-5%3C0

If x+3 is POS. or ZERO
x%2B3-5%3C0
x%3C-3%2B5
x%3C2

If x+3 is NEG.
-%28x%2B3%29-5%3C0
-x-3-5%3C0
-x%3C8
x%3E-8

ANSWER: highlight%28-8%3Cx%3C2%29