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

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Math Expert
Joined: 02 Sep 2009
Posts: 51261
If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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14 Aug 2018, 01:35
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45% (medium)

Question Stats:

62% (01:31) correct 38% (01:31) wrong based on 209 sessions

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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. $$\frac{-19}{5} \leq x \leq 5$$

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Joined: 02 Aug 2009
Posts: 7107
If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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14 Aug 2018, 01:44
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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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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

### Show Tags

14 Aug 2018, 01:44
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:

$$-3x - 7 \geq2x + 12$$

-$$3x - 2x \geq19$$

$$x\leq-19/5$$

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Joined: 21 Aug 2017
Posts: 2
If |3x + 7| ≥ 2x + 12, then which of the following is true?  [#permalink]

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14 Aug 2018, 08:38
|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

If |3x + 7| ≥ 2x + 12, then which of the following is true? &nbs [#permalink] 14 Aug 2018, 08:38
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