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Senior SC Moderator V
Joined: 14 Nov 2016
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For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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18 00:00

Difficulty:   65% (hard)

Question Stats: 60% (02:18) correct 40% (02:27) wrong based on 231 sessions

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For how many integer values of x, is $$|3x-3|+|2x+8|<15$$?

A. 2
B. 3
C. 4
D. 5
E. 6

Attachment: Untitled.jpg [ 61.01 KiB | Viewed 3026 times ]

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Originally posted by hazelnut on 13 Feb 2017, 06:42.
Last edited by hazelnut on 13 Feb 2017, 07:20, edited 1 time in total.
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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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3
4
we can solve this by following ways -

1. First take the positive parts of mod values -

|3x-3| = 3x-3
|2x+8| = 2x+8

So, it boils down to -

3x-3+2x+8 < 15
x<2

2. Now consider the negative parts of them -

|3x-3| = -3x+3
|2x+8| = -2x-8

-3x+3-2x-8<15
-x<4
x>-4

So , we can conclude that: -4<x<2
Thus, we will have - 5 integers, which will satisfy this equation.
##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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1
3
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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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ziyuenlau wrote:
How to resolve this question?

For how many integer values of x, is $$|3x-3|+|2x+8|<15$$?

ziyuenlau

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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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1
$$3x-3 = 0$$ then $$x = 1$$
$$2x+8 = 0$$ then $$x = -4$$

there can be 3 range
$$x\leq{-4}$$ | $$-4 < x < 1$$ | $$x\geq{1}$$

1. When $$x\leq{-4}$$, then both $$|3x-3|$$ and $$|2x+8|$$ will be negative.

$$-3x+3 -2x-8 < 15$$

$$-5x-5 < 15$$

$$x > -4$$ (this is opposite to $$x\leq{-4}$$. )

Hence $$x\leq{-4}$$ is not a possibility.

2. When $$-4 < x < 1$$, then $$|3x-3|$$ would still be negative but $$|2x+8|$$ will be positive. hence

$$-3x+3 +2x+8 < 15$$

-x+11 < 15

$$-x < 4$$ i.e. $$x > -4$$ (this lies within $$-4 < x < 1$$) so correct range.

3. When $$x\geq{1}$$, then both $$|3x-3|$$ and $$|2x+8|$$ will be positive.

$$3x-3+2x+8 < 15$$

$$5x+5 < 15$$

$$x+1 < 3$$

$$x < 2$$ this range is also fine because $$x\geq{1}$$. so x must be 1.

So our range is $$-4 <$$$$x\geq{1}$$, so howmany integers within this range?

-3, -2, -1, 0 and 1

Count is 5

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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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2
2
ziyuenlau wrote:
For how many integer values of x, is $$|3x-3|+|2x+8|<15$$?

A. 2
B. 3
C. 4
D. 5
E. 6

$$|3x-3|+|2x+8|<15$$

$$3*|x-1|+2*|x+4|<15$$

We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from -4" is less than 15.

Let's try to find the point where this distance is equal to 15.

........................ (-4) ...................................... (0) ........... (1) ..........................

The distance between -4 and 1 is 5. Thrice this distance is 15. So at the point x = -4, the sum will be 15. As we move to the right of -4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10.
What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now.
At x = 2, the sum becomes 3*1 + 2*6 = 15.
To the right of 2, the sum will keep increasing.

So the sum will be less than 15 between -4 and 2. This gives us 5 integer values (-3, -2, -1, 0, 1).

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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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Bunuel wrote:

Are there any more questions apart from the mentioned above on the same topic?
Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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Status: It's near - I can see.
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For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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VeritasPrepKarishma wrote:
ziyuenlau wrote:
For how many integer values of x, is $$|3x-3|+|2x+8|<15$$?

A. 2
B. 3
C. 4
D. 5
E. 6

$$|3x-3|+|2x+8|<15$$

$$3*|x-1|+2*|x+4|<15$$

We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from -4" is less than 15.

Let's try to find the point where this distance is equal to 15.

........................ (-4) ...................................... (0) ........... (1) ..........................

The distance between -4 and 1 is 5. Thrice this distance is 15. So at the point x = -4, the sum will be 15. As we move to the right of -4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10.
What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now.
At x = 2, the sum becomes 3*1 + 2*6 = 15.
To the right of 2, the sum will keep increasing.

So the sum will be less than 15 between -4 and 2. This gives us 5 integer values (-3, -2, -1, 0, 1).

Mam, my question in when x < = -4, then for if x = -5 , |2x+8| is negative. but we are also using x = -4, then |2x+8| becomes "0". then why we are not considering |2x+8| = 0 for x = -4. I don't know whether I made myself clear to you or not
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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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HKD1710 wrote:
$$3x-3 = 0$$ then $$x = 1$$
$$2x+8 = 0$$ then $$x = -4$$

there can be 3 range
$$x\leq{-4}$$ | $$-4 < x < 1$$ | $$x\geq{1}$$

1. When $$x\leq{-4}$$, then both $$|3x-3|$$ and $$|2x+8|$$ will be negative.

$$-3x+3 -2x-8 < 15$$

$$-5x-5 < 15$$

$$x > -4$$ (this is opposite to $$x\leq{-4}$$. )

Hence $$x\leq{-4}$$ is not a possibility.

2. When $$-4 < x < 1$$, then $$|3x-3|$$ would still be negative but $$|2x+8|$$ will be positive. hence

$$-3x+3 +2x+8 < 15$$

-x+11 < 15

$$-x < 4$$ i.e. $$x > -4$$ (this lies within $$-4 < x < 1$$) so correct range.

3. When $$x\geq{1}$$, then both $$|3x-3|$$ and $$|2x+8|$$ will be positive.

$$3x-3+2x+8 < 15$$

$$5x+5 < 15$$

$$x+1 < 3$$

$$x < 2$$ this range is also fine because $$x\geq{1}$$. so x must be 1.

So our range is $$-4 <$$$$x\geq{1}$$, so howmany integers within this range?

-3, -2, -1, 0 and 1

Count is 5

but took 3.5 mins, i'd expect quicker approach
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Location: United States
Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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1
hazelnut wrote:
For how many integer values of x, is $$|3x-3|+|2x+8|<15$$?

A. 2
B. 3
C. 4
D. 5
E. 6

$$positive:|3x-3|≥0…3x≥3…x≥1…negative:x<1$$
$$positive:|2x+8|≥0…2x≥-8…x≥-4…negative:x<-4$$
$$range:--(neg)--(-4)---(pos,neg)--(1)--(pos)---$$

$$x≥1:|3x-3|+|2x+8|<15…3x-3+2x+8<15…5x<10…x<2:1≤x<2=$$
$$4≤x<1:…-3x+3+2x+8<15…-x<4…x>-4:-4<x<1=[-3,-2,-1,0]$$
$$x<-4:…-3x+3-2x-8<15…-5x<20…x>-4:invalid=x<-4$$

$$x=[-3,-2,-1,0,1]=5$$

Ans (D)
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Re: For how many integer values of x, is |3x-3|+|2x+8|<15?  [#permalink]

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faltan

faltan wrote:
HKD1710 wrote:
$$3x-3 = 0$$ then $$x = 1$$
$$2x+8 = 0$$ then $$x = -4$$

there can be 3 range
$$x\leq{-4}$$ | $$-4 < x < 1$$ | $$x\geq{1}$$

1. When $$x\leq{-4}$$, then both $$|3x-3|$$ and $$|2x+8|$$ will be negative.

$$-3x+3 -2x-8 < 15$$

$$-5x-5 < 15$$

$$x > -4$$ (this is opposite to $$x\leq{-4}$$. )

Hence $$x\leq{-4}$$ is not a possibility.

2. When $$-4 < x < 1$$, then $$|3x-3|$$ would still be negative but $$|2x+8|$$ will be positive. hence

$$-3x+3 +2x+8 < 15$$

-x+11 < 15

$$-x < 4$$ i.e. $$x > -4$$ (this lies within $$-4 < x < 1$$) so correct range.

3. When $$x\geq{1}$$, then both $$|3x-3|$$ and $$|2x+8|$$ will be positive.

$$3x-3+2x+8 < 15$$

$$5x+5 < 15$$

$$x+1 < 3$$

$$x < 2$$ this range is also fine because $$x\geq{1}$$. so x must be 1.

So our range is $$-4 <$$$$x\geq{1}$$, so howmany integers within this range?

-3, -2, -1, 0 and 1

Count is 5

but took 3.5 mins, i'd expect quicker approach

Did you came up with one? Re: For how many integer values of x, is |3x-3|+|2x+8|<15?   [#permalink] 12 Dec 2019, 09:16
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