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Re: For How many integer values of x, is |2x-5|+|x+1|+|x|<10? [#permalink]
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Hi ziyuenlau,

We're told that X must be an INTEGER, so since we're adding the sums of 3 absolute values - AND we're looking for sums that are LESS than 10 - there cannot be an infinite number of possibilities. Thus, based on the answer choices, there can't be more than 5 possible integer values for N. As such, we can use a bit of 'brute force' to find all of the possible values without too much trouble.

We're given the inequality: |2X - 5| + |X + 1| + |X| < 10.

Let's start with...
X = 0.... |-5| + |1| + |0| = 6 so X could be 0

Now let's work our way "up"...
X = 1.... |-3| + |2| + |1| = 6 so X could be 1
X = 2.... |-1| + |3| + |2| = 6 so X could be 2
X = 3.... |1| + |4| + |3| = 8 so X could be 3
X = 4.... |3| + |5| + |3| = 11 so X CANNOT be 4

Now let's work our way "down"....
X = -1.... |-7| + |0| + |-1| = 8 so X could be -1

At this point, we have 5 possible values of X; since that it the largest of the possible values, we can stop working.

Final Answer:

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Re: For How many integer values of x, is |2x-5|+|x+1|+|x|<10? [#permalink]
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BillyZ wrote:
For how many integer values of x, is \(|2x-5|+|x+1|+|x|<10\)?

A. 1
B. 2
C. 4
D. 5
E. Infinite


here I first simply opened the modulus:

=> 2x-5+x+1+x<10
=> 4x-4<10
=> 4x<14
=> x<7/2
=> x<3.5

then I open modulus and multiply left side by -1

=> -2x+5-x-1-x<10
=> -4x+4<10
=> -4x<6
=> x>-6/4
=> x>-1.5

thus -1.5 <x<3.5

therefore the integer values are -1,0,1,2,3 ==> 5 integer values

Answer is D.

But honestly I don't know the logic of multiplying left side by -1. This is what I had read somewhere and applied. Although the answer is correct however if the methodology is correct or not I am not sure, also even if this method is correct, why it is correct, I don't have an answer.
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For How many integer values of x, is |2x-5|+|x+1|+|x|<10? [#permalink]
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Given that |2x−5|+|x+1|+|x|<10 and we need to find how many integer values can x take

When we have multiple Absolute values in a problem then we need to reduce the number of cases. Let's see how to do that.

Lets find out the points on the number line where the values inside each absolute value will change sign.
We can do that by putting each term inside the absolute value individually equal to zero.

For |2x-5| -> 2x-5 = 0 => x = \(\frac{5}{2}\) = 2.5
For |x+1| -> x+1 = 0 => x = -1
For |x| -> x = 0

Attachment:
problem.JPG
problem.JPG [ 20.43 KiB | Viewed 5927 times ]


Case 1: x < -1
Let's take any value of x < -1, lets say -2 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*-2)-5 = -9 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> -2+1 = -1 => -ve => |x+1| = -(x+1)
For |x| -> -2 => -ve => |x| = -x

|2x−5|+|x+1|+|x|<10 => -(2x-5) - (x+1) -x < 10
=> -2x + 5 -x - 1 -x < 10
=> -4x < 10 -4
=> -4x < 6 => x > \(\frac{-6}{4}\) => x > -1.5
And conditions was x < -1
So, intersection will be -1.5 < x < -1 => NO Integer solution in this range

Case 2: -1 ≤ x < 0
Let's take any value of x which satisfies -1 ≤ x < 0, lets say -0.5 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*-0.5)-5 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> -0.5+1 => +ve => |x+1| = x+1
For |x| -> -0.5 => -ve => |x| = -x

|2x−5|+|x+1|+|x|<10 => -(2x-5) + (x+1) -x < 10
=> -2x + 5 + x + 1 -x < 10
=> -2x < 10 -6
=> -2x < 4 => x > \(\frac{-4}{2}\) => x > -2
And conditions was -1 ≤ x < 0
So, intersection will be -1 ≤ x < 0 => x=-1 is an integer solution in this range

Case 3: 0 ≤ x < 2.5
Let's take any value of x which satisfies 0 ≤ x < 2.5, lets say 1 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*1)-5 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> 1+1 => +ve => |x+1| = x+1
For |x| -> 1 => +ve => |x| = x

|2x−5|+|x+1|+|x|<10 => -(2x-5) + (x+1) + x < 10
=> -2x + 5 + x + 1 + x < 10
=> 6 < 10
Which is always true
So, entire range 0 ≤ x < 2.5 is possible
x=0, 1, 2 are integer solution in this range

Case 4: x ≥ 2.5
Let's take any value of x which satisfies x ≥ 2.5, lets say 10 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*10)-5 => +ve =>|2x-5| = 2x-5
For |x+1| -> 10+1 => +ve => |x+1| = x+1
For |x| -> 10 => +ve => |x| = x

|2x−5|+|x+1|+|x|<10 => 2x-5 + (x+1) + x < 10
=> 2x - 5 + x + 1 + x < 10
=> 4x - 4 < 10
=> 4x < 14
=> x < \(\frac{14}{4}\)
=> x < 3.5
And conditions was x ≥ 2.5
So, intersection will be 2.5 ≤ x < 3.5=> x=3 is an integer solution in this range

So, integer values which are possible are -1 , 0, 1, 2, 3
=> 5 values

So, Answer will be D
Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems

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Re: For How many integer values of x, is |2x-5|+|x+1|+|x|<10? [#permalink]
Expert Reply
BillyZ wrote:
For how many integer values of x, is \(|2x-5|+|x+1|+|x|<10\)?

A. 1
B. 2
C. 4
D. 5
E. Infinite



\(2|x-\frac{5}{2}|+|x+1|+|x|<10\)

'Twice the distance of x from 5/2' + 'Distance of x from -1' + 'Distance of x from 0' < 10

Draw the number line:

------------(-3) -- (-2) -- (-1) -- 0 -- 1 -- 2 -- 3 -- 4 -- 5 ------------

Consider the point -1 first, x = -1

'Twice the distance of x from 2.5' + 'Distance of x from -1' + 'Distance of x from 0' = 7 + 0 + 1 = 8 (which is less than 10)

If we move further to the left of -1 i.e. x = -2, then 'Twice the distance of x from 2.5' will increase by 2 and already the sum is 10 so this is not possible.

Now consider the point to the right of 2.5 i.e. x = 3 (looking for integer values at both extremes)

'Twice the distance of x from 2.5' + 'Distance of x from -1' + 'Distance of x from 0' = 1 + 4 + 3 = 8

If we move further to the right of 3 i.e. x = 4, then 'Twice the distance of x from 2.5' will increase by 2 and already the sum is 10 so this is not possible.

Hence, values of x will range from -1 to 3 i.e. 5 integer values.

Answer (D)
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Re: For How many integer values of x, is |2x-5|+|x+1|+|x|<10? [#permalink]
I did get the answer but using such long drawn methods in the real exam seems to be a choice not many people would want to make. Is there any shortcut method for these modulus questions?
BrushMyQuant wrote:
Given that |2x−5|+|x+1|+|x|<10 and we need to find how many integer values can x take

When we have multiple Absolute values in a problem then we need to reduce the number of cases. Let's see how to do that.

Lets find out the points on the number line where the values inside each absolute value will change sign.
We can do that by putting each term inside the absolute value individually equal to zero.

For |2x-5| -> 2x-5 = 0 => x = \(\frac{5}{2}\) = 2.5
For |x+1| -> x+1 = 0 => x = -1
For |x| -> x = 0

Attachment:
problem.JPG


Case 1: x < -1
Let's take any value of x < -1, lets say -2 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*-2)-5 = -9 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> -2+1 = -1 => -ve => |x+1| = -(x+1)
For |x| -> -2 => -ve => |x| = -x

|2x−5|+|x+1|+|x|<10 => -(2x-5) - (x+1) -x < 10
=> -2x + 5 -x - 1 -x < 10
=> -4x < 10 -4
=> -4x < 6 => x > \(\frac{-6}{4}\) => x > -1.5
And conditions was x < -1
So, intersection will be -1.5 < x < -1 => NO Integer solution in this range

Case 2: -1 ≤ x < 0
Let's take any value of x which satisfies -1 ≤ x < 0, lets say -0.5 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*-0.5)-5 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> -0.5+1 => +ve => |x+1| = x+1
For |x| -> -0.5 => -ve => |x| = -x

|2x−5|+|x+1|+|x|<10 => -(2x-5) + (x+1) -x < 10
=> -2x + 5 + x + 1 -x < 10
=> -2x < 10 -6
=> -2x < 4 => x > \(\frac{-4}{2}\) => x > -2
And conditions was -1 ≤ x < 0
So, intersection will be -1 ≤ x < 0 => x=-1 is an integer solution in this range

Case 3: 0 ≤ x < 2.5
Let's take any value of x which satisfies 0 ≤ x < 2.5, lets say 1 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*1)-5 => -ve =>|2x-5| = -(2x-5)
For |x+1| -> 1+1 => +ve => |x+1| = x+1
For |x| -> 1 => +ve => |x| = x

|2x−5|+|x+1|+|x|<10 => -(2x-5) + (x+1) + x < 10
=> -2x + 5 + x + 1 + x < 10
=> 6 < 10
Which is always true
So, entire range 0 ≤ x < 2.5 is possible
x=0, 1, 2 are integer solution in this range

Case 4: x ≥ 2.5
Let's take any value of x which satisfies x ≥ 2.5, lets say 10 and substitute and see whether the values inside the absolute value is +ve or -ve or 0

For |2x-5| -> (2*10)-5 => +ve =>|2x-5| = 2x-5
For |x+1| -> 10+1 => +ve => |x+1| = x+1
For |x| -> 10 => +ve => |x| = x

|2x−5|+|x+1|+|x|<10 => 2x-5 + (x+1) + x < 10
=> 2x - 5 + x + 1 + x < 10
=> 4x - 4 < 10
=> 4x < 14
=> x < \(\frac{14}{4}\)
=> x < 3.5
And conditions was x ≥ 2.5
So, intersection will be 2.5 ≤ x < 3.5=> x=3 is an integer solution in this range

So, integer values which are possible are -1 , 0, 1, 2, 3
=> 5 values

So, Answer will be D
Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems


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For How many integer values of x, is |2x-5|+|x+1|+|x|<10? [#permalink]
KarishmaB wrote:
BillyZ wrote:
For how many integer values of x, is \(|2x-5|+|x+1|+|x|<10\)?

A. 1
B. 2
C. 4
D. 5
E. Infinite

\(2|x-\frac{5}{2}|+|x+1|+|x|<10\)

'Twice the distance of x from 5/2' + 'Distance of x from -1' + 'Distance of x from 0' < 10

Draw the number line:

------------(-3) -- (-2) -- (-1) -- 0 -- 1 -- 2 -- 3 -- 4 -- 5 ------------

Consider the point -1 first, x = -1

'Twice the distance of x from 2.5' + 'Distance of x from -1' + 'Distance of x from 0' = 7 + 0 + 1 = 8 (which is less than 10)

If we move further to the left of -1 i.e. x = -2, then 'Twice the distance of x from 2.5' will increase by 2 and already the sum is 10 so this is not possible.

Now consider the point to the right of 2.5 i.e. x = 3 (looking for integer values at both extremes)

'Twice the distance of x from 2.5' + 'Distance of x from -1' + 'Distance of x from 0' = 1 + 4 + 3 = 8

If we move further to the right of 3 i.e. x = 4, then 'Twice the distance of x from 2.5' will increase by 2 and already the sum is 10 so this is not possible.

Hence, values of x will range from -1 to 3 i.e. 5 integer values.

Answer (D)
 

­How are you getting your values for distances from ref point?
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