Which of the following inequalities is equivalent to 2 < : PS Archive
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# Which of the following inequalities is equivalent to 2 <

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Which of the following inequalities is equivalent to 2 < [#permalink]

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03 Jul 2004, 07:56
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15. Which of the following inequalities is equivalent to â€“2 < x < 4 ?
(A) | x â€“ 2 | < 4
(B) | x â€“ 1 | < 3
(C) | x + 1 | < 3
(D) | x + 2 | < 4
(E) None of the above
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03 Jul 2004, 08:29
D looks tempting but I will choose E.

-2<x<4

Add 2 to each side of the inequality

0<x+2<6

Therefore non of the available answers will fit with this inequality.

Correct me if I am mistaken.
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03 Jul 2004, 08:33
Quote:
Add 2 to each side of the inequality

0<x+2<6

I don't think you can do this
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Paul

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03 Jul 2004, 08:41
Paul wrote:
Quote:
Add 2 to each side of the inequality

0<x+2<6

I don't think you can do this

Why not! in inequality you can add/ subtract/ multiply/divide any number to each part of an inequality.
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03 Jul 2004, 08:51
dr_sabr wrote:
D looks tempting but I will choose E.

-2<x<4

Add 2 to each side of the inequality

0<x+2<6

Therefore non of the available answers will fit with this inequality.

Correct me if I am mistaken.

adding numbers doesn't change the inequality. the inequality remains the same. you said that x+2<6 ==> x<4 ----so we're back to square one.

you have to do it this way.

To check for B.
if x>1 , |x-1| = x-1 then inequality is x-1<3 or x<4
x>1 and x<4 are consistent

if x<1, |x-1| = -(x-1) then inequality is -(x-1)<3 or x>-2
x>-2 and x<1 is again consistent.

Therefore B is correct.
If you try with other answers, you will land up with inconsistent inequalities.

I don't know if there is a shorter method.

Does anyone know a shorter(and quicker) method ?

- ash
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03 Jul 2004, 09:02
ashkg wrote:
adding numbers doesn't change the inequality. the inequality remains the same. you said that x+2<6 ==> x<4 ----so we're back to square one.

you have to do it this way.

To check for B.
if x>1 , |x-1| = x-1 then inequality is x-1<3 or x<4
x>1 and x<4 are consistent

if x<1, |x-1| = -(x-1) then inequality is -(x-1)<3 or x>-2
x>-2 and x<1 is again consistent.

Therefore B is correct.
If you try with other answers, you will land up with inconsistent inequalities.

I don't know if there is a shorter method.

Does anyone know a shorter(and quicker) method ?

- ash

Very impressive explaination ash, Thank you
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03 Jul 2004, 10:14

It is good to remember

|x| < 4 ==> -4 < x < 4
|x| <= 4 ==> -4 <= x <= 4

|x| > 4 ==> x > 4 or x < -4

Apply the same concepts here

a. |x - 2| < 4 ==> -4 < x-2 < 4 ==> -2 < x < 6 (wrong)
b. |x - 1| < 3 ==> -3 < x-1 < 3 ==> -2 < x < 4 (correct)

Ans B
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03 Jul 2004, 11:16
sathya76 wrote:

It is good to remember

|x| < 4 ==> -4 < x < 4
|x| <= 4 ==> -4 <= x <= 4

|x| > 4 ==> x > 4 or x < -4

Apply the same concepts here

a. |x - 2| < 4 ==> -4 < x-2 < 4 ==> -2 < x < 6 (wrong)
b. |x - 1| < 3 ==> -3 < x-1 < 3 ==> -2 < x < 4 (correct)

Ans B

yep......if you remember this, its the best ( i dont know how i forgot this,its silly of me)

anyway.........here's and easy way to remember it(visual)
case 1 . if |x| > 4

draw the number line..........mark 4 and -4 on it.
all numbers beyond 4 and beyond -4 are satisfy our inequality.

so x > 4 or x < -4

case 2 . if |x| < 4

draw the number line..........mark 4 and -4 on it.
all numbers betweeen 4 and -4 satisfy our inequality.

so -4 < x < 4

THanks Sathya.
- ash
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ash
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03 Jul 2004, 11:16
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