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If ||x+3| - 12| < 13, what is the range of x?

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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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Different methods to solve absolute value equations and inequalities- Exercise Question #5

If ||x+3| - 12| < 13, what is the range of x?

Options

    a) (-4, 22)
    b) (-28, -2)
    c) (-28, 22)
    d) (-28, 2) U (2, 22)
    e) (-28, -2) U (4, 22)

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Originally posted by EgmatQuantExpert on 13 Sep 2018, 03:07.
Last edited by EgmatQuantExpert on 13 Sep 2018, 08:20, edited 2 times in total.
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post Updated on: 17 Sep 2018, 04:04
1
2

Solution


Given:
    We are given an absolute value inequality, ||x+3| - 12| < 13
    Note that we do not have any constraint on x. So, x can be an integer or a decimal

To find:
    We need to find the range of values of x, that satisfies the given inequality

Approach and Working:
First, let us substitute the inner modulus function, |x + 3|, with t, which gives, |t - 12| < 13
    We know that, the range of x, for |x| < a, is a < x < -a
    So, the range of x, for which |t - 12| < 13, is
      -13 < t – 12 < 13,
      Adding 12 on all the sides, we get, -1 < t < 25
      Now, substituting, back the value of t as |x + 3|, we get, -1 < |x + 3| < 25
        Since, the value of |x + 3| is always greater than or equal to 0
      Hence, 0 ≤ |x + 3| < 25
      Now, solving the inequality, |x + 3| < 25, we get the range of x as (-28, 22)

Therefore, the range of x, that satisfies the inequality, ||x+3| - 12| < 13, is (-28, 22)

Hence, the correct answer is option C.

Answer: C

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Originally posted by EgmatQuantExpert on 13 Sep 2018, 03:16.
Last edited by EgmatQuantExpert on 17 Sep 2018, 04:04, edited 1 time in total.
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post Updated on: 09 Oct 2018, 06:36
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My Approach:

|x+3| will always be positive. Therefore |x+3|<25 in order for ||x+3|- 12|< 13 to be true!

|x+3|<25 --> -28<x<22

Answer: C

Please correct me if I went wrong somewhere!!!
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Originally posted by T1101 on 13 Sep 2018, 08:35.
Last edited by T1101 on 09 Oct 2018, 06:36, edited 1 time in total.
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 13 Sep 2018, 14:47
1
Hi Payal: Is my process okay for this problem?
I solved the problem this way:
Since the equation is pretty messy(by my standards), I decided to use substitution.
||x+3| - 12| < 13

1.) I let |x+3| = a and then rephrased the eqn
|a-12| < 13
a - 12 < 13 and -(a-25) < 13 --> Since |x+3| > 0 I didn't bother to solve the 2nd eqn
a < 25

2.) I plugged in the original value of a to find the range
|x+3| < 25
x+3 < 25 and -(x+3) < 25
x < 22 x> -28

Answer C
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Sep 2018, 13:15
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EgmatQuantExpert wrote:

If ||x+3| - 12| < 13, what is the range of x?

Options

    a) (-4, 22)
    b) (-28, -2)
    c) (-28, 22)
    d) (-28, 2) U (2, 22)
    e) (-28, -2) U (4, 22)


\(?\,\,\,:\,\,\,{\text{solution}}\,\,{\text{set}}\)

The geometric interpretation of (some) absolute values is essential to solve this sort of question quickly and safely:

\(\left| {z - w} \right| = {\text{dist}}\left( {z,w} \right)\)

\(\left| {k + m} \right| = \left| {k - \left( { - m} \right)} \right| = {\text{dist}}\left( {k, - m} \right)\)

The application of these ideas is used in the particular case:

\(a = \left| {x + 3} \right|\)

\(?\,\,\,:\,\,\,\,\left| {a - 12} \right| < 13\,\,\,\,\,\,\,\left( {{\text{variable}}\,\,x} \right)\,\,\,\,\,\)

See image attached.

Right answer: (C)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post Updated on: 23 Sep 2018, 04:17
EgmatQuantExpert wrote:

Solution


Given:
    We are given an absolute value inequality, ||x+3| - 12| < 13
    Note that we do not have any constraint on x. So, x can be an integer or a decimal

To find:
    We need to find the range of values of x, that satisfies the given inequality

Approach and Working:
First, let us substitute the inner modulus function, |x + 3|, with t, which gives, |t - 12| < 13
    We know that, the range of x, for |x| < a, is a < x < -a
    So, the range of x, for which |t - 12| < 13, is
      -13 < t – 12 < 13,
      Adding 12 on all the sides, we get, -1 < t < 25
      Now, substituting, back the value of t as |x + 3|, we get, -1 < |x + 3| < 25
        Since, the value of |x + 3| is always greater than or equal to 0
      Hence, 0 ≤ |x + 3| < 25
      Now, solving the inequality, |x + 3| < 25, we get the range of x as (-28, 22)

Therefore, the range of x, that satisfies the inequality, ||x+3| - 12| < 13, is (-28, 22)

Hence, the correct answer is option C.

Answer: C

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Hi Payal & EgmatQuantExpert , thanks for explanation but I have a doubt which may be stupid . Normally for |x| < a we do have answer as -a < x < a , then why for -1 < |x + 3| < 25 we made it 0 ≤ |x + 3| < 25. Again in our final answer we had negative range inclusive -> (-28, 22) . Please clear this doubt .

Thanks

Originally posted by KARISHMA315 on 23 Sep 2018, 03:56.
Last edited by KARISHMA315 on 23 Sep 2018, 04:17, edited 1 time in total.
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 23 Sep 2018, 04:17
2
In this problem, the easiest way would to mind answers for information.
If ||x+3| - 12| < 13, what is the range of x?
Our bounderis should satisfy ||x+3|-12| = 13 equation.
a) (-4, 22)
Let's put -4 in our equation:
||-4+3|-12| = 11 => A out
b) (-28, -2)
||-28+3|-12| = 13
||-2+3|-12| = 11 => B out
c) (-28, 22)
-28 - good
||22+3|-12| = 13
good, but we need to check others
d) (-28, 2) U (2, 22)
-28 - good
||2+3|-12| = 7 => D out
e) (-28, -2) U (4, 22)
-28 - good
-2 - bad => E out

So, we know that only C
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 26 Sep 2018, 13:45
1
KARISHMA315, sorry that I can't quote but here is the answer:

Hi! That is because the absolute value can't be negative, so zero is the lesser value that |x+3| can have. Then, you can rewrite the equation:

\(0 ≤ |x + 3| < 25\)
\(|x + 3| < 25\)

The second equation is analog because an inequation with absolute value is non-negative by definition. Thus, the solution is straightforward:

\(-25 < x + 3 < 25\)
\(-28 < x < 22\)

Hope it will be useful :)
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 09 Oct 2018, 06:33
You can do this by substitution:
E+D+B --> Try x = 2 --> the equation is still correct --> only A, C left
A,C --> Try x = -5 --> the equation is still correct --> C left
--> C is the correct answer
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Oct 2018, 03:44
EgmatQuantExpert wrote:

Solution


Given:
    We are given an absolute value inequality, ||x+3| - 12| < 13
    Note that we do not have any constraint on x. So, x can be an integer or a decimal

To find:
    We need to find the range of values of x, that satisfies the given inequality

Approach and Working:
First, let us substitute the inner modulus function, |x + 3|, with t, which gives, |t - 12| < 13
    We know that, the range of x, for |x| < a, is a < x < -a
    So, the range of x, for which |t - 12| < 13, is
      -13 < t – 12 < 13,
      Adding 12 on all the sides, we get, -1 < t < 25
      Now, substituting, back the value of t as |x + 3|, we get, -1 < |x + 3| < 25
        Since, the value of |x + 3| is always greater than or equal to 0
      Hence, 0 ≤ |x + 3| < 25
      Now, solving the inequality, |x + 3| < 25, we get the range of x as (-28, 22)

Therefore, the range of x, that satisfies the inequality, ||x+3| - 12| < 13, is (-28, 22)

Hence, the correct answer is option C.

Answer: C

Image



Hi EgmatQuantExpert

based on which formula/rule do you assume that the value of |x + 3| is always greater than or equal to 0
:? we always have two cases positive and negative.... isn't it so ? :?
can you please explain :)

hey :) Are you there pushpitkc :) maybe you know answer to my beautiful question :)
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Oct 2018, 07:11
1
Hey dave13

The cardinal rule for an absolute value expression is that it can never be negative.
For example |-2| = |+2| = 2 (whatever comes outside the is ALWAYS positive)

In the problem we have the expression ||x+3| - 12| < 13. The minimum value for
the part |x+3| is 0, when x = -3. For anything else, the value will be greater than
0. For instance, if x = -4, |x+3| = |-1| = 1(which is again positive)

Hope that helps!
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post Updated on: 14 Oct 2018, 13:37
pushpitkc wrote:
Hey dave13

The cardinal rule for an absolute value expression is that it can never be negative.
For example |-2| = |+2| = 2 (whatever comes outside the is ALWAYS positive)

In the problem we have the expression ||x+3| - 12| < 13. The minimum value for
the part |x+3| is 0, when x = -3. For anything else, the value will be greater than
0. For instance, if x = -4, |x+3| = |-1| = 1(which is again positive)

Hope that helps!




Hi pushpitkc, thanks for nice explanation, if so then why in the example below when we get 2 < |x - 3| < 14 consider two cases negative and positive ? :? ... whereas in this question when we get, -1 < |x + 3| < 25 we don't consider TWO cases ? Any idea generis ? who knows, may be you know :)
    Since, the value of |x + 3| is always greater than or equal to 0
Hence, 0 ≤ |x + 3| < 25
Now, solving the inequality, |x + 3| < 25, we get the range of x as (-28, 22)





Example 5

If ||x - 3| - 8| < 6, what is the range of x?

Solution

We solved a question of double modulus in absolute value equations.
So, we will follow the similar approach.

First, substitute |x - 3| as y, and write the inequality as,
    • |y - 8| < 6
    • Now, this is in the form of |x| < a
    • Therefore, the range of x is -a < x < a.
      o Similarly, we can find the range of y if we substitute x by y-8 and a by 6.
         Thus, -6 < y-8< 6
         Adding 8 on all the sides of the inequality, we get: 2< y< 14.

Now, we can substitute back the value of y as |x - 3| and get:
    2 < |x - 3| < 14
    • Here, we need to consider two cases, when x – 3 is positive and when x - 3 is negative
      o Considering first case, x – 3 is positive, we have:
         2 < x - 3 < 14
         Simplifying this inequality, we get
          • 5 < x < 17 ………….…. (1)
      o Now, considering second case, x – 3 is negative, we have:
         2 < - (x – 3) < 14
         So, let us now multiply by -1 on all the sides, we get
          • -2 > x – 3 > -14 or -14 < x – 3 < -2

Note: Did you notice that, the sign of the inequality is changed when you multiply the whole expression by -1?


So, always keep in mind to change the signs when you multiply an inequality by a negative number.

      o Simplifying this inequality, we get
         -11 < x < 1 ……………. (2)

Therefore, the range of x is (1) ⋃ (2), which is equal to (-11, 1) ⋃ (5, 17).

https://gmatclub.com/forum/different-me ... l#p2130767

Originally posted by dave13 on 14 Oct 2018, 07:22.
Last edited by dave13 on 14 Oct 2018, 13:37, edited 1 time in total.
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Oct 2018, 07:40
1
EgmatQuantExpert wrote:
Different methods to solve absolute value equations and inequalities- Exercise Question #5

If ||x+3| - 12| < 13, what is the range of x?

Options

    a) (-4, 22)
    b) (-28, -2)
    c) (-28, 22)
    d) (-28, 2) U (2, 22)
    e) (-28, -2) U (4, 22)


There are 2 possible cases:
(i) \((x+3)-12 < 13\)
(ii) \(-(x+3)-12 < 13\)

Case (i):
\(x+3-12 < 13\)
\(x-9 < 13\)
\(x < 22\)

Case (ii):
\(-x-3-12 < 13\)
\(-x-15 < 13\)
\(-x < 28\)
\(x > -28\)

Range of \(x: -28 < x < 22\).
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Re: If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Oct 2018, 08:01
1
dave13 wrote:
pushpitkc wrote:
Hey dave13

The cardinal rule for an absolute value expression is that it can never be negative.
For example |-2| = |+2| = 2 (whatever comes outside the is ALWAYS positive)

In the problem we have the expression ||x+3| - 12| < 13. The minimum value for
the part |x+3| is 0, when x = -3. For anything else, the value will be greater than
0. For instance, if x = -4, |x+3| = |-1| = 1(which is again positive)

Hope that helps!




Hi pushpitkc, thanks for nice explanation, if so then why in the example below when we get 2 < |x - 3| < 14 consider two cases negative and positive ? :? ... whereas in this question when we get, -1 < |x + 3| < 25 we don't consider TWO cases ?
    Since, the value of |x + 3| is always greater than or equal to 0
Hence, 0 ≤ |x + 3| < 25
Now, solving the inequality, |x + 3| < 25, we get the range of x as (-28, 22)



Example 5

If ||x - 3| - 8| < 6, what is the range of x?

Solution

We solved a question of double modulus in absolute value equations.
So, we will follow the similar approach.

First, substitute |x - 3| as y, and write the inequality as,
    • |y - 8| < 6
    • Now, this is in the form of |x| < a
    • Therefore, the range of x is -a < x < a.
      o Similarly, we can find the range of y if we substitute x by y-8 and a by 6.
         Thus, -6 < y-8< 6
         Adding 8 on all the sides of the inequality, we get: 2< y< 14.

Now, we can substitute back the value of y as |x - 3| and get:
    2 < |x - 3| < 14
    • Here, we need to consider two cases, when x – 3 is positive and when x - 3 is negative
      o Considering first case, x – 3 is positive, we have:
         2 < x - 3 < 14
         Simplifying this inequality, we get
          • 5 < x < 17 ………….…. (1)
      o Now, considering second case, x – 3 is negative, we have:
         2 < - (x – 3) < 14
         So, let us now multiply by -1 on all the sides, we get
          • -2 > x – 3 > -14 or -14 < x – 3 < -2

Note: Did you notice that, the sign of the inequality is changed when you multiply the whole expression by -1?


So, always keep in mind to change the signs when you multiply an inequality by a negative number.

      o Simplifying this inequality, we get
         -11 < x < 1 ……………. (2)

Therefore, the range of x is (1) ⋃ (2), which is equal to (-11, 1) ⋃ (5, 17).

https://gmatclub.com/forum/different-me ... l#p2130767


Why |x| is written as -x, when absolute value represents the positive value of the number.

• However, if you pay attention to the constraint that “if x is negative”, then |x| is equal to -x.
• And, we all know that negative of a negative number is always positive.
o Hence, |x| is always positive.

Hope this helps.

For more on inequality please refer: https://gmatclub.com/forum/different-me ... l#p2130767

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Please hit Kudos if it helps! :)
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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Oct 2018, 22:55
2
EgmatQuantExpert wrote:
Different methods to solve absolute value equations and inequalities- Exercise Question #5

If ||x+3| - 12| < 13, what is the range of x?

Options

    a) (-4, 22)
    b) (-28, -2)
    c) (-28, 22)
    d) (-28, 2) U (2, 22)
    e) (-28, -2) U (4, 22)

dave13 - whenever I see "nested" absolute value inequalities,
I break the problem into pieces or stages.
That process often includes considering negative and positive cases.
I think that this approach is easier and more basic than the theoretical references I see.

Essentially, we simplify one part of the LHS for part of the time that we solve.

We have to find the range of solutions for a "nested" absolute value inequality.
That situation is complicated. We want to make it simpler.

1) Make the LHS simpler.

Change the inside absolute value expression with a placeholder for a little while.
I will let \(u\) stand for \(|x+3|\)

\(|x+3|\) becomes \(u\). That is, \(u=|x+3|\)

I will "put back" \(|x+3|\) later.

Change the original: \(||x+3| - 12| < 13\)

Changed: NOW we have: \(|u - 12| < 13\)

I use \(u\) without brackets to solve more easily
BUT I do keep in mind that the brackets are "in the background"
\(u\) stands for an absolute value expression, |x + 3|
\(u\) is "really" an absolute value expression

One way to think about absolute value: it measures distance from 0.
By definition, distances cannot be negative.
Absolute value must be positive or 0.
Absolute value can never be negative.

3) Using \(|u - 12| < 13\) proceed as usual

Case 1: the value inside \(|u-12|\) is positive. (LHS is positive)
Simply remove the absolute value bars
\(u-12<13\)
\(u<25\)


That result is okay. RHS is positive.

Case 2: \(|u-12|\) is negative* (LHS is negative)
\(u-12>-25\)
\(u>-13\)

No. Cannot use this case.
RHS is negative.
\(u\) is actually an absolute value expression (is really |x+3|).
Absolute values cannot be negative.

Ignore that case. Toss it in the trash.
Case 1 works. Case 2 does not work. Next stage.

5) Use \(u<25\) from Case 1. "Put back" \(|x+3|\). Proceed as usual

We have yet another inequality: \(|x+3|<25\)

Case 1: LHS is positive
\(x+3<25\)
\(x<22\)


Case 2: LHS is negative
\(x+3>-25\)
\(x>-28\)
, which is the same as
\(-28<x\)

Those two values give us our range of solutions

\(-28<x<22\)
End values that specify this range are
\((-28,22)\)

I hope that helps. :)

Answer C

*Finding the negative case in an absolute value inequality. I took a shortcut.
Use |x + 3| < 25, from #5, above. The case in which the value inside the brackets is negative is written
x + 3 > -25
That inequality does not appear to be "LHS is negative." It is. I used a shortcut / mechanical rule.

Mechanical Rule - In an absolute value inequality, to find Case 2 in which LHS is negative:
Remove absolute value brackets, add a negative sign to RHS, and flip the sign.
Start: |x + 3| < 25 , negative case
EITHER
- (x + 3) < 25
(-1) (x + 3) < 25
Divide by negative 1. Dividing an inequality by a negative number flips the sign.

\((\frac{-1)(x+3)}{(-1)}> \frac{25}{-1}\)
x + 3 > -25
Then simplify to x > -28
OR
- (x + 3) < 25
-x - 3 < 25
-x < 28
(-1) * (x) < 28

\(\frac{(-1)(x)}{(-1)}>\frac{28}{-1}\)
x > -28

**Regarding the theoretical references (I am not sure they're helpful, but . . .)
In the statement "if x is less than 0, then |x| = -x "
The right hand side is NOT a negative number. It is the negative of a negative number: RHS is positive.
Try ANY negative number:
|-3| = 3
3 = -(-3)
The absolute value of -3 is 3, and that absolute value, 3, is the opposite of negative 3.
The negative sign in variable x "hides."

You may want to take a look at Bunuel 's post starting HERE. My post follows: Approaches: If x is negative, then |x| = -x . Finally, just below that post Bunuel here: finishes the discussion.

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If ||x+3| - 12| < 13, what is the range of x?  [#permalink]

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New post 14 Oct 2018, 23:08
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elincopi wrote:
KARISHMA315, sorry that I can't quote but here is the answer:

Hi! That is because the absolute value can't be negative, so zero is the lesser value that |x+3| can have. Then, you can rewrite the equation:

\(0 ≤ |x + 3| < 25\)
\(|x + 3| < 25\)

The second equation is analog because an inequation with absolute value is non-negative by definition. Thus, the solution is straightforward:

\(-25 < x + 3 < 25\)
\(-28 < x < 22\)

Hope it will be useful :)

elincopi your explanation was good, and more importantly, you went out of your way to help quite awhile ago.
I think it's time you were thanked. +1 for both the explanation (which will indeed be useful) and the graciousness
:)
P. S. In order to make sure that a person sees a post, make sure that the person is tagged. To tag:
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If ||x+3| - 12| < 13, what is the range of x? &nbs [#permalink] 14 Oct 2018, 23:08
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