If ||x+3| - 12|

Question

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

A. (-4, 22)
B. (-28, -2)
C. (-28, 22)
D. (-28, 2) U (2, 22)
E. (-28, -2) U (4, 22)

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

Previous Question

To read the article:      Different methods to solve absolute value equations and inequalities

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-Algebra-e1533292163516.png

EgmatQuantExpert · Accepted Answer

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

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-Algebra-e1533292089856.jpg

T1101 · Answer

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!!!

SB2004 · Answer

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

fskilnik · Answer

?\,\,\,:\,\,\,{	ext{solution}}\,\,{	ext{set}}

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

\left| {z - w} ight| = {	ext{dist}}\left( {z,w} ight)

\left| {k + m} ight| = \left| {k - \left( { - m} ight)} ight| = {	ext{dist}}\left( {k, - m} ight)

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

a = \left| {x + 3} ight|

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

See image attached.

Right answer: (C)

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

Regards,
Fabio.

KARISHMA315 · Answer

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

rkoshlyak · Answer

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

elincopi · Answer

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

trangnguyen1116 · Answer

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

dave13 · Answer

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

pushpitkc · Answer

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!

dave13 · Answer

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

mbafinhopeful · Answer

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 .

rever08 · Answer

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 ------------------------------------------- Please hit Kudos if it helps!

generis · Answer

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. I used a shortcut to get from -(u-12)<25 to u-12>-25 . The shortcut is in the footnote*) 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

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.

generis · Answer

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: -- type the '@' sign -- put the person's username right next to it, NO space, but -- leave ONE space afterwards. (In other words, if you want to call on KARISHMA315 and put a comma after the tag, you have to leave a space after the last letter of the tag, then put the comma.)

venkivety · Answer

As  generis  explained the absolute value is the measure of the distance from 0. 
that is why the absolute value can never be negative.
Eg.1 |x|=3 which can also be written as |x-0|=3 
means the distance from zero is 3 units. i.e. x can be 3 or -3

Eg.2 |x-2|=5
means the distance from 2 is 5 units. i.e. 7 or -3

Eg.3 |x+5|=13 can be written as |x-(-5)|=13
means the distance from -5 is 13 units i.e. -18 or 8

Eg.4 |x+3|<13
we generally solve the equation as follows

-13<x+3<13
-16< x<10

but when it comes to this question it says the absolute value lies between -1 < |x + 3| < 25

Since, the absolute value couldn't be less than zero, we write the inequality as follows
0 ≤ |x + 3| < 25

Hence solving for |x + 3| < 25 gives the answer as (-28,22)

ShankSouljaBoi · Answer

I did get to the correct answer. Just want to know the difference between option C and D ?

Kinshook · Answer

Asked: If ||x+3| - 12| < 13, what is the range of x?
||x+3| - 12| < 13
-13 <|x+3| -12 < 13
-1 <0 < |x+3| < 25
-25 < x+3 < 25
-28 < x < 22

Range of x = (-28,22)

IMO C

chayma · Answer

Since, the value of |x + 3| is always greater than or equal to 0

can i know how we assume that ?

Posted from my mobile device

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards