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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
chetan2u Bunuel

i have one doubt , can you help ?

Can we write |5-x| as -|x-5| in this equation ? or is it only applicable on multiplication problems ?
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
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Harshjha001 wrote:
chetan2u Bunuel

i have one doubt , can you help ?

Can we write |5-x| as -|x-5| in this equation ? or is it only applicable on multiplication problems ?


No |5-x| is not equal to -|x-5|
|5-x|\(\geq 0\), while \(-|x-5|\leq 0\)

However, if x>5, making (5-x) as negative, then |5-x|=-(5-x)
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
Hello,
Is there a quicker way to solve this than to check the values of x in each internal?

Is there someway to arrive at the answer by looking at the check points?

Posted from my mobile device
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
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Hi rahulms94,

This is a great 'concept' question, meaning that if you recognize the concepts involved, you don't have to do much math to get to the correct answer. By extension, the fastest way to answer it is NOT to look at the 'check points.'

To start, notice that we're adding two absolute values that must sum to a total that is greater than a third individual absolute value. Consider what happens when X becomes really large (for example, X = 1,000). The differences in the values of the individual absolute values becomes negligible, meaning that we're essentially ending up with 1000 + 1000 > 1000. This will occur for every integer value of X at higher and higher values - thus, there's really no reason to try to count them all up - there's an infinite set of solutions.

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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
chetan2u, sir can you explain this using number line?
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
To be honest (and I could be wrong), I didn't think that 2 minutes would be enough to solve this question.

I guess I decided to... guess.

I simply decided on 2 values of x and tested what happened. In both cases, the left side was greater than the right side...it was soon clear to me that no matter what - LEFT > RIGHT.

*I am not advocating my method here, but all I am saying is that sometimes a prudent guess can turned out to be a good one.
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Re: For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
As long as you choose any number/integer less than 2: be it 1 , 0 , -1, -5, -10, -100, -1000 , -1000000, -Trillion, the left side will always be greater than the right side, meaning Infinite!

Choice E
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For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|? [#permalink]
Can this problem be solved by first solving the inequality with a + sign in front of every absolute value and then a negative sign in front of every absolute value? Please let me know if this procedure is wrong: since the answers are infinite on the number line, then the answer is E?
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Re: For how many integer values of x, is |x 8| + |5 x| > |x + 7|? [#permalink]
can we do this problem using the wavy line method with breakpoints at 8,5 and -7.
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For how many integer values of x, is |x 8| + |5 x| > |x + 7|? [#permalink]
I have a question on the last range (where x>8) , how does x fall in the range. Doesn't it mean X>20 but not from 8 to 19?


Bunuel wrote:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite

STEP-BY-STEP DETAILED SOLUTION:


There are three transition points at -7, 5 and 8 (transition points are points for which expression in modulus is equal to 0). So, four ranges to check:


1. \(x < -7\)

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is negative, so |x + 7| = -(x + 7).

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > -(x + 7), which in turn is x < 20. Since, we consider x < -7 range, then finally for it we get x < -7.


2. \(-7 \leq x \leq 5\)

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > x + 7, which in turn is x < 2. Since, we consider \(-7 \leq x \leq 5\) range, then finally for it we get \(-7 \leq x < 2\).


3. \(5 < x < 8\)

For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) -(5 - x) > x + 7, which in turn is x < -4. Since, we consider \(5 < x < 8\) range and x < -4 is out of it, then for this range the inequality does not have a solution (no x from this range satisfied the inequality).


3. \(x \geq 8\)

For this range:
x - 8 is positive, so |x - 8| = x - 8.
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.

Thus, |x – 8| + |5 – x| > |x + 7| becomes x - 8 -(5 - x) > x + 7, which in turn is [highlight]x > 20 (already entirely in the range we consider)


So, we got that |x – 8| + |5 – x| > |x + 7| is true for x < 2 (combined range from cases 1 and 2) and x > 20. Thus, infinitely many integer values of x satisfy |x – 8| + |5 – x| > |x + 7|.

Answer: E.

Hope it's clear.
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Re: For how many integer values of x, is |x 8| + |5 x| > |x + 7|? [#permalink]
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ayushimani wrote:
I have a question on the last range (where x>8) , how does x fall in the range. Doesn't it mean X>20 but not from 8 to 19?



8 is a critical point for the expression as x>8 will give some values in the modulus as positive and some as negative. Accordingly, you change the sign while opening the modulus.

Now, when you do that you realise x>20, which is in the range x>8, so our assumption on the start is correct. Only thing is that we get more specific range that matches with our initial assumption.

A simpler example would be
|x|=2
Take x>0, then x=2.
So, the value of x is 2 although we took range as x>0.
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Re: For how many integer values of x, is |x 8| + |5 x| > |x + 7|? [#permalink]
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