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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 10 Jan 2021, 12:37
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If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x


M36-88

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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 12 Nov 2021, 02:36
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Bunuel
If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x


M36-88
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Official Solution:


If \(x\) is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of \(x\) ?

The critical points (aka key points or transition points) are -8, -4 and 3 (the values of \(x\) for which the expressions in the absolute values become 0).

So, we should consider the following ranges:

If \(x < - 8\), then \(x - 3 < 0\), \(x + 4 < 0\) and \(x + 8 < 0\), so \(|x - 3| = -(x-3)\), \(|x + 4| = -(x + 4)\) and \(|x + 8| = -(x + 8)\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(-(x - 3) - (x + 4) \leq -(x + 8)\). This gives \(x \geq 7\). Since we consider \(x < - 8\) range, then for this range given inequality does not have a solution

If \(-8 \leq x \leq - 4\), then \(x - 3 < 0\), \(x + 4 \leq 0\) and \(x + 8 \geq 0\), so \(|x - 3| = -(x-3)\), \(|x + 4| = -(x + 4)\) and \(|x + 8| = x + 8\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(-(x - 3) - (x + 4) \leq x + 8\). This gives \(x \geq -3\). Since we consider \(-8 \leq x \leq - 4\) range, then for this range given inequality does not have a solution

If \(-4 < x \leq 3\), then \(x - 3 \leq 0\), \(x + 4 > 0\) and \(x + 8 > 0\), so \(|x - 3| = x-3\), \(|x + 4| = -(x + 4)\) and \(|x + 8| = x + 8\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(-(x - 3) + (x + 4) \leq x + 8\). This gives \(x \geq -1\). Since we consider \(-4 < x \leq 3\) range, then for this range given inequality gives the following integer solutions: -1, 0, 1, 2, and 3.

If \(x > 3\), then \(x - 3 > 0\), \(x + 4 > 0\) and \(x + 8 > 0\), so \(|x - 3| = x-3\), \(|x + 4| = x + 4\) and \(|x + 8| = x + 8\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(x - 3 + x + 4 \leq x + 8\). This gives \(x \leq 7\). Since we consider \(x > 3\) range, then for this range given inequality gives the following integer solutions: 4, 5, 6, and 7.



So, \(|x - 3| + |x + 4| \leq |x + 8|\) holds true for the following integer values of \(x\): -1, 0, 1, 2, 3, 4, 5, 6, and 7.

(1) \(x < 1\)

\(x\) can be 0 or -1. Not sufficient.

(2) \(|x| > x\)

The above implies that \(x\) is negative, therefore \(x=-1\). Sufficient.


Answer: B
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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 10 Jan 2021, 21:51
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Bunuel
If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x
Show more

S1: if x is less then 1, when we drop the modulus, the signs of the expression inside the first modulus will change to the opposite, the rest will stay the same (put 0, which is less than 1, inside every modulus)
\(-x+3+x+4 \leq x+8\)
\(x\geq{-1}\)
there are two possibilities, -1 and 0
not sufficient

S2: |x| > x, means x<0
in that case again we have following equation
\(-x+3+x+4 \leq x+8\)
\(x \geq {-1}\)
as x is strictly less than 0, the only integer satisfying the inequality is -1.
Sufficient.

Answer B
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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 11 Jan 2021, 04:58
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Given, x is an integer and
|x-3| + |x+4| <= |x+8|

I have solved this by plugging in values.

S1: x<1

If x = 0, LHS < RHS
If x = -1, LHS = RHS

Here, we have two values of x.

Hence, Not sufficient

S2: |x|>x
This means, x<0

If x= -1, LHS = RHS

Any other integer less than -1 will make LHS > RHS.

Therefore, we have only one value of x.
Hence, Sufficient.

The answer will be B.

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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 05 Aug 2022, 10:45
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Bunuel
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If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x


M36-88

Official Solution:


If \(x\) is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of \(x\) ?

The critical points (aka key points or transition points) are -8, -4 and 3 (the values of \(x\) for which the expressions in the absolute values become 0).

So, we should consider the following ranges:

If \(x < - 8\), then \(x - 3 < 0\), \(x + 4 < 0\) and \(x + 8 < 0\), so \(|x - 3| = -(x-3)\), \(|x + 4| = -(x + 4)\) and \(|x + 8| = -(x + 8)\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(-(x - 3) - (x + 4) \leq -(x + 8)\). This gives \(x \geq 7\). Since we consider \(x < - 8\) range, then for this range given inequality does not have a solution

If \(-8 \leq x \leq - 4\), then \(x - 3 < 0\), \(x + 4 \leq 0\) and \(x + 8 \geq 0\), so \(|x - 3| = -(x-3)\), \(|x + 4| = -(x + 4)\) and \(|x + 8| = x + 8\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(-(x - 3) - (x + 4) \leq x + 8\). This gives \(x \geq -3\). Since we consider \(-8 \leq x \leq - 4\) range, then for this range given inequality does not have a solution

If \(-4 < x \leq 3\), then \(x - 3 \leq 0\), \(x + 4 > 0\) and \(x + 8 > 0\), so \(|x - 3| = x-3\), \(|x + 4| = -(x + 4)\) and \(|x + 8| = x + 8\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(-(x - 3) + (x + 4) \leq x + 8\). This gives \(x \geq -1\). Since we consider \(-4 < x \leq 3\) range, then for this range given inequality gives the following integer solutions: -1, 0, 1, 2, and 3.

If \(x > 3\), then \(x - 3 > 0\), \(x + 4 > 0\) and \(x + 8 > 0\), so \(|x - 3| = x-3\), \(|x + 4| = x + 4\) and \(|x + 8| = x + 8\). Thus in this range \(|x - 3| + |x + 4| \leq |x + 8|\) becomes \(x - 3 + x + 4 \leq x + 8\). This gives \(x \leq 7\). Since we consider \(x > 3\) range, then for this range given inequality gives the following integer solutions: 4, 5, 6, and 7.



So, \(|x - 3| + |x + 4| \leq |x + 8|\) holds true for the following integer values of \(x\): -1, 0, 1, 2, 3, 4, 5, 6, and 7.

(1) \(x < 1\)

\(x\) can be 0 or -1. Not sufficient.

(2) \(|x| > x\)

The above implies that \(x\) is negative, therefore \(x=-1\). Sufficient.


Answer: B
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Can we see if the ranges are valid by picking a number from the range and substituting it in the inequality?
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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 15 May 2023, 07:01
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SherzodAzamov
Bunuel
If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x

S1: if x is less then 1, when we drop the modulus, the signs of the expression inside the first modulus will change to the opposite, the rest will stay the same (put 0, which is less than 1, inside every modulus)
\(-x+3+x+4 \leq x+8\)
\(x\geq{-1}\)
there are two possibilities, -1 and 0
not sufficient

S2: |x| > x, means x<0
in that case again we have following equation
\(-x+3+x+4 \leq x+8\)
\(x \geq {-1}\)
as x is strictly less than 0, the only integer satisfying the inequality is -1.
Sufficient.

Answer B
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When you are removing your modulus signs for S2, why only check for x=0? What about x=-5, or -9? Then you have to manipulate the expressions inside modulus sign differently. With this method how do you know to stop there?

For the first expression I understand since we have already shown it is insufficient.
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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 30 May 2024, 06:53
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Bunuel
If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x


M36-88

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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 30 May 2024, 06:56
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Bunuel
If x is an integer and \(|x - 3| + |x + 4| \leq |x + 8|\), what is the value of x ?


(1) x < 1

(2) |x| > x


M36-88

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­Bunuel – What does "M36-88" refer to?
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If x is an integer and |x - 3| + |x + 4| <= |x + 8|, what is the value 05 Aug 2025, 04:23
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