For how many integer values of x, is |3x-3|+|2x+8|

Question

For how many integer values of x, is |3x-3|+|2x+8|<15 ? A. 2 B. 3 C. 4 D. 5 E. 6 Untitled.jpg

tanmay056 · Accepted Answer

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.

KarishmaB · Answer

|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).

Answer (D)

HKD1710 · Answer

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 Answer D.

Buttercup3 · Answer

Are there any more questions apart from the mentioned above on the same topic?

Bunuel · Answer

9. Inequalities 
     Theory   
 Inequalities Made Easy! 
 Inequalities: Tips and hints 
 Arithmetic with Inequalities 
 Wavy Line Method Application - Complex Algebraic Inequalities 
 Solving Quadratic Inequalities: Graphic Approach 
 Inequalities - Quadratic Inequalities 
 Graphic approach to problems with inequalities 
 Inequalities trick 
 INEQUATIONS(Inequalities) Part 1 
 INEQUATIONS(Inequalities) Part 2 
     Questions   
 Inequality and absolute value questions from my collection 
 DS Questions 
 PS Questions

10. Absolute Value 
     Theory   
 Math: Absolute value (Modulus) 
 Absolute Value: Tips and hints 
 Properties of Absolute Values on the GMAT 
 Absolute modulus : A better understanding 
 GMAT Question Patterns: Abolute Value 
 The Reason Behind Absolute Value Questions on the GMAT 
     Questions   
 Inequality and absolute value questions from my collection 
 The E-GMAT Question Series on ABSOLUTE VALUE  
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

exc4libur · Answer

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:-4-4:invalid=x<-4 x= =5 Ans (D)

GMAT695 · Answer

-15 < 3x-3+2x+8 < 15
-15 < 5x + 5 < 15
-20 < 5x < 10
-4 < x < 2
X can be -3, -2, -1, 0 , 1, 2 . ( D )

asad9242 · Answer

The value hold true for a -3,-2,-1,0 and 1

All these can be solved very quickly. Total of 5, therefore answer is  D

Jason13 · Answer

One can always break the modulus and then proceed with the usual process as mentioned is other solutions but I find solving such problems by plotting the graph of the inequality. The visual is far better to interpret and avoids missing any values even if there are multiple such modulus.
Below is a descriptive solution for each step.

Posted from my mobile device

Vivek1707 · Answer

Faster way of solving such questions would be to test values

Since the quesiton mentions that we need 'integer' values, the scope gets narrowed down

x = 0 --> |3|+|8 | < 15 yes
x = 1 --> |0|+|10| < 15 yes
x = 2 --> |3|+|12| < 15 no

So positive nos beyond 2 wont help

Lets try negetive numbers

x = -1 ---> |6| + |6| < 15 yes
x = -2 ---> |9| + |4| < 15 yes
x = -3 --->|12| + |2| < 15 yes
x = -4 --->|15| + |0| < 15 no

So numbers less than-4 wont help
Hence we have 5 numbers : Answer is D

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

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