Which of the following inequalities specifies the shaded reg

Which of the following inequalities specifies the shaded region to the left?

From the diagram we know that \(-1<x<4\).

A. \(\sqrt{x^2+ 1}< 3\) --> square the inequality (we can do this since both sides are non-negative): \(x^2+1<9\) --> \(x^2<8\) --> \(-2\sqrt{2}<x<2\sqrt{2}\).

B. |2x - 3| < 5 --> \(-5<2x - 3<5\) --> add 3 to each part: \(-2<2x<8\) --> reduce by 2: \(-1<x<4\). BINGO!

C. |x + 1| > -1. No need to test this one. The absolute value is always more than or equal to zero, thus this inequality holds for any value of x.

D. x - 2 < 2 --> \(x<4\).

E. |x - 1| < 4 --> \(-4<x - 1<4\) --> add 1 to each part: \(-3<x<5\).

Answer: B.

Does this make sense?
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No need to plug here , from the shaded region we know that -1<x<4 so all you have to do is to check the answer choices whether they match with the given inequality

l 2x-3 l< 5 means that -5 < 2x - 3 < 5
--> -2 < 2x < 8
--> -1 < x < 4

Answer : B

Hi happyface101,

Your calculation includes values that are GREATER than 4 and LESS than -1, and those values fall outside the range of the drawing (which is -1 to +4).

Notice how you wrote (5 - 3/2)....

GMAT assassins aren't born, they're made,
Rich

Not sure what to add...

|x + 1| is an absolute value --> absolute value is always more than or equal to zero \(|some \ expresseion|\geq{0}\), thus \(|x + 1|\geq{0}\) no matter the value of x --> \(|x + 1|\geq{0}>-1\).

Hope it's clear.
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Which of the following inequalities specifies the shaded region to the left?

A. √(x^2+ 1) < 3
B. |2x - 3| < 5
C. |x + 1| > -1
D. x - 2 < 2
E. |x - 1| < 4

Quickly rule out D as it is simply x<4 (i.e. x is between 4 and negative infinity)

Quickly going through the list, B.) looks to be a promising choice, so let's try that:

|2x - 3| < 5
2x - 3 < 5
2x < 8
x<4
OR
-(2x - 3) < 5
-2x + 3 < 5
-2x < 2
x > -1

So: -1 < x < 4

(B)

Dear Team,

Shouldn't the answer choice B be -1<=x<=4 instead of -1<x<4?

Exatly, since it doesn't hold true for the endpoints, how can (B) be correct? Even though the endpoints in the picture make it seems as if those integer values are included in the range, (B) omits them. This is not intuitive, if (B) tells us that its inequality does NOTnecessarily hold true for the endpoints, that doesn't implicity mean that (B) asumes the endpoints in fact are "= 5", right?

What Im trying to say is that I interpret the shaded region as saying \(-1<=x<=4\), since both -1 and 4 are "covered" by the shaded area (at least it seems so to me). Thus, x could be -1 or it could be 4. But \(-1<x<4\) means that x cannot be either -1 or 4, and this is what confuses me. Either I am misinterpreting the shaded area, or I lack a fundamental understanding of inequalities.

Using midpoint and distance, write an absolute value inequality in the form

|x - (midpoint)| < distance


1) Find the midpoint of the region, exactly halfway between -1 and 4, which is \(\frac{3}{2}\)

2) Find the distances of the endpoints from the midpoint

-1 is a distance of \(\frac{5}{2}\) from \(\frac{3}{2}\), and

4 is a distance of \(\frac{5}{2}\) from \(\frac{3}{2}\)

The distance cannot equal \(\frac{5}{2}\) because the endpoints aren't included. The distance can be anything up to, but less than, \(\frac{5}{2}\)

3. Set up the inequality

From above, the distance of x from the midpoint,* namely |x - \(\frac{3}{2}\)|, is < \(\frac{5}{2}\)

4. Write the inequality.

|x - (midpoint)| < distance

|x - \(\frac{3}{2}\)| < \(\frac{5}{2}\)

5. Find the answer that matches that inequality.

Eliminate C, which is always true (absolute value is always \(\geq0\), hence also always \(>-1\)), and D, which, without absolute value bars, does not cover two directions.

From the remaining choices: The answer must account for \(\frac{3}{2}\) somehow. There is only one answer with a 3 on LHS:

B) |2x - 3| < 5

That fits; just multiply all terms of the inequality derived from midpoint and distance by 2:

|x - \(\frac{3}{2}\)| < \(\frac{5}{2}\) (each term * 2)

|2x - 3| < 5

ANSWER B

*|x - (some number)| is the distance of x from (some number)
|x - 4| is the distance of x from 4
|x + 4| is the distance of x from -4

Blackbox, I'm not sure whether or not how I solved the problem is what you seek. I think it might be.
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You can read the above thread for clarity. Also, it is not correct to discard all the answer choices as at least one answer choice has to be the answer. Extreme points are clearly marked, if they are inclusive in range.
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Hi Bunuel,

This does make sense. A lil doubt here
I would like to understand the red colored bit above. How did you reach to the conclusion? I am sorry if the question sounds silly, but I am trying to learn the correct ways to handle such problems and I take it from the forum that you are the best goto guy for this.
Thanks in advance,
Vaibhav.
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We have strict inequality in |2x - 3| < 5, so solution should be as written: -1 < x < 4. Substitute the endpoints of the range (-1 and 4) to see that for them |2x - 3| < 5 does not hold true.
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Not sure I understand what you mean. The shaded region gives \(-1<x<4\) and option B also gives \(-1<x<4\). Thus B is correct. No other option gives this range, no matter whether you include endpoints on the diagram in the inequality or not.
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Can someone help me understand why my approach is not getting me to the right answer?

First, I found the length from -1 to 4, which is 5. Then I found the center, which is (4-1)/2, so 3/2.

Then I tried to write the equation: |x - 3/2| < (5-3/2) ---> |x - 3/2| < 7/2 ---> |(2x - 3)/2| < 7/2 ---> |2x-3| < 7

Why is |2x-3| < 7 wrong and the actual answer is |2x-3| < 5? What did I do wrong? Thanks!!

Ahh!! A silly mistake ... it always gets me ... truly terrible.

Thank you very much for your help!! +1 Kudo!

Case 1
0 < 5t - 8 < 1
8/5 < t < 9/5
Case 2
0< -5t + 8 < 1
8/5 > t > 7/5
D

EMPOWERgmatRichC - Hi, Could you be a bit more specific on what mistake happyface101 did in his equation? It'd help a great deal if you actually showed me how to answer this question along the lines of what happyface101 did.

Can some body tell me what is the meaning of the highlighted part?
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QZ , I got thrown, too, for a bit.

I suspect that the question was cut and pasted, and that on the original page, the number line literally was placed to the left of the question.

I think the poster just forgot to change the wording to "the highlighted part [in the diagram] below."

Hope that helps.
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