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If |3x + 7| ≥ 2x + 12, then which of the following is true?

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If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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New post 14 Aug 2018, 02:35
00:00
A
B
C
D
E

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  45% (medium)

Question Stats:

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If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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New post 14 Aug 2018, 02:44
Bunuel wrote:
If \(|3x + 7| ≥ 2x + 12\), then which of the following is true?


A. \(x \leq \frac{-19}{5}\)

B. \(x \geq \frac{-19}{5}\)

C. \(x \geq 5\)

D. \(x \leq \frac{-19}{5}\) or \(x \geq 5\)

E. \(\frac{-19}{5} \leq x \leq 5\)


Square both sides..
\(9x^2+42x+49\geq{4x^2+48x+144}........5x^2-6x-95\geq{0}..........5x^2-25x+19x-95\geq{0}...........(5x+19)(x-5)\geq{0}.....\)
So either \(x\leq{-19/5}\) or \(x\geq{5}\)

D

Or open the mod
I.. \(3x+7\geq{2x+12}........x\geq{5}\)
II. \(-3x-7\geq{2x+12}............5x\leq{-19}\)

D


Or the choices should help us..
A,B,C are subset of D and E so if any of D and E is correct, one of A,B and C will also be correct..
So our answer has to be D or E..

Now to choose between D and E, take X as 0
So equation becomes 7>12...NO
So our ans should not contain 0 as a value...
Eliminate E

D
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If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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New post 14 Aug 2018, 02:44
Bunuel wrote:
If \(|3x + 7| ≥ 2x + 12\), then which of the following is true?


A. \(x \leq \frac{-19}{5}\)

B. \(x \geq \frac{-19}{5}\)

C. \(x \geq 5\)

D. \(x \leq \frac{-19}{5}\) or \(x \geq 5\)

E. \(\frac{-19}{5} \leq x \leq 5\)



As we are dealing with Absolute value , we have assess both negative and positive possibility .

Case 1: When 3x + 7 is positive.

\(3x + 7 \geq2x + 12\)

\(x \geq5\)

Case 2: when 3x + 7 is negative:

\(-3x - 7 \geq2x + 12\)

-\(3x - 2x \geq19\)

\(x\leq-19/5\)

The best answer is D.
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If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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New post 14 Aug 2018, 09:38
1
|3x+7| \(\geq{2x+12}\)

The critical points of the above equation are -6 and \(\frac{-7}{3}\)
To find the critical Points equate the either side of the equations to 0
i.e. 3x + 7 = 0 ==> x =\(\frac{-7}{3}\)
and 2x + 12 = 0 ==> x = -6

So 3 ranges to consider :
Range 1 : x<-6
Take any value in this area say for Ex : -7

Look for the Expr. inside mod 3x +7 will become 3(-7)+7
So this expression will evaluate to a -ve value

Therefore this equation can be written as :

(-1)(3x+7) \(\geq{2x+12}\)
-3x-7 \(\geq{2x+12}\)
-5x \(\geq{19}\)
x \(\leq{-19/5}\)

Now this range is x<-6 , so based on the solution all values of x in this range should satisfy the equation

Range 2 : -6<x<\(\frac{-7}{3}\)
Take any value in this area say for Ex : -5

Look for the Expr. inside mod 3x +7 will become 3(-5)+7
So this expression will evaluate to a -ve value

Therefore this equation can be written as :

(-1)(3x+7) \(\geq{2x+12}\)
-3x-7 \(\geq{2x+12}\)
-5x \(\geq{19}\)
x \(\leq{-19/5}\)

Now this range is -6<x<\(\frac{-7}{3}\) , so only values of x \(\leq{-19/5}\) will satisfy

Range 3 : x>\(\frac{-7}{3}\)
Take any value in this area say for Ex : -1

Look for the Expr. inside mod 3x +7 will become 3(-1)+7
So this expression will evaluate to a +ve value

Therefore this equation can be written as :

(3x+7) \(\geq{2x+12}\)
3x+7 \(\geq{2x+12}\)
3x-2x \(\geq{5}\)
x \(\geq{5}\)

Now this range is x>\(\frac{-7}{3}\), so only values of x\(\geq{5}\) will satisfy

So Combining ,x \(\leq{-19/5}\) OR x \(\geq{5}\) will hold true

Hence Answer : D
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Re: If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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Re: If |3x + 7| ≥ 2x + 12, then which of the following is true?   [#permalink] 08 Sep 2019, 21:08
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