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If x is a real number, which of the following cannot be equa 08 Mar 2026, 00:54
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If x is a real number, which of the following cannot be equal to

|x - 4| + |2x + 3| + |x| ?

A. 1 - 4x
B. 7
C. 7 + 2x
D. 4x - 1
E. 2x - 1
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E
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Dereno
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If x is a real number, which of the following cannot be equa 08 Mar 2026, 03:34
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kevincan
If x is a real number, which of the following cannot be equal to

|x - 4| + |2x + 3| + |x| ?

A. 1 - 4x
B. 7
C. 7 + 2x
D. 4x - 1
E. 2x - 1
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The Critical points are x= 4, -(3/2) and 0.

So, if we plot the points on the number line, we get 4 regions. Visualise it in a number line.

Region 1: (-3/2) to -ve infinity.

Region 2: Between -(3/2) and 0

Region 3: Between 0 and 4

Region 4: Between (4 and +ve infinity)

Let’s assume the value |x - 4| as A, |2x + 3| as B, and |x| as C.

For Region 1: All A,B,C are negative.

So, the modulus opens in -ve.

-(x-4) -(2x+3) -x

solving we get Option A (1-4x)

For Region 2: A is -ve and B is +ve (as it’s to the right of the critical Point (-3/2)). And C as -ve.

-(x-4)+(2x+3)-x

Solving , we get option B :7

For region 3: A is negative and B,C are both positive.

-(x-4) +2x+3+x

solving, we get Option C: 2x+7

For region 4, All are positive.

(x-4)+(2x+3)+x

Solving, we get Option D: 4x -1

So, the answer is Option E
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If x is a real number, which of the following cannot be equa 19 Jul 2026, 03:57
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Hi, i just want to reconfirm the values of for |2x-3| so it is +ve i.e 2x+3 when x>=-3/2 and 2x-3 when x<3/2 . Am i doing the negative part calculation correct? My approach is that when x will be less than 0 then 2x will become -2x inside the mod so we'll have -2x+3 and when we'll open the mod as negative then that should be 2x-3<0 and will give x<3/ ; this is the correct way right?
Dereno

The Critical points are x= 4, -(3/2) and 0.

So, if we plot the points on the number line, we get 4 regions. Visualise it in a number line.

Region 1: (-3/2) to -ve infinity.

Region 2: Between -(3/2) and 0

Region 3: Between 0 and 4

Region 4: Between (4 and +ve infinity)

Let’s assume the value |x - 4| as A, |2x + 3| as B, and |x| as C.

For Region 1: All A,B,C are negative.

So, the modulus opens in -ve.

-(x-4) -(2x+3) -x

solving we get Option A (1-4x)

For Region 2: A is -ve and B is +ve (as it’s to the right of the critical Point (-3/2)). And C as -ve.

-(x-4)+(2x+3)-x

Solving , we get option B :7

For region 3: A is negative and B,C are both positive.

-(x-4) +2x+3+x

solving, we get Option C: 2x+7

For region 4, All are positive.

(x-4)+(2x+3)+x

Solving, we get Option D: 4x -1

So, the answer is Option E
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If x is a real number, which of the following cannot be equa 19 Jul 2026, 22:05
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Hey inciduntveniam,

The negative part is incorrect. Here is what it actually is.

In general,

| p | = +p, when p >=0
| p | = -p, when p<0

Here,

|2x+3| = 2x+3, when x>=3/2
|2x+3| = - (2x + 3) = -2x - 3 (Observe -> this is -2x - 3, not 2x - 3). when x<-3/2

Here is a solution for this question ->


---
Harsha
inciduntveniam
Hi, i just want to reconfirm the values of for |2x-3| so it is +ve i.e 2x+3 when x>=-3/2 and 2x-3 when x<3/2 . Am i doing the negative part calculation correct? My approach is that when x will be less than 0 then 2x will become -2x inside the mod so we'll have -2x+3 and when we'll open the mod as negative then that should be 2x-3<0 and will give x<3/ ; this is the correct way right?

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If x is a real number, which of the following cannot be equa 20 Jul 2026, 20:06
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Hi inciduntveniam,

I can see exactly where the wires got crossed, and it's a really common one. Let me anchor to the two pieces you wrote.

Absolute value doesn't care about the sign of x. It cares about the sign of the whole thing inside the bars - here, the whole expression 2x + 3. So the switch point is wherever 2x + 3 = 0, which is x = -3/2, not x = 0 and not x = 3/2.

So the two cases are:
- When 2x + 3 ≥ 0 (that is, x ≥ -3/2): the bars come off as-is - |2x + 3| = 2x + 3. ✅ (you had this right)
- When 2x + 3 < 0 (that is, x < -3/2): the bars flip the entire expression - |2x + 3| = -(2x + 3) = -2x - 3.

Two fixes to your negative case:

1. You don't flip just the 2x because x < 0. You flip the whole inside, so 2x + 3 becomes -2x - 3, not 2x - 3.

2. The condition isn't x < 3/2. Set the inside negative: 2x + 3 < 0 - x < -3/2.

That's why Dereno's regions split at -3/2, and why the negative branch produced -2x - 3 (part of the 1 - 4x you get in Region 1).

Quick check to lock it in - solve these the same way:
- |2x + 3| - where does it flip, and what does it become on the negative side? (Answer: flips at x = -3/2, becomes -2x - 3.)
- |3x - 6| - flip point? Negative-side form? (Set 3x - 6 < 0 - x < 2; it becomes -3x + 6.)

Same rule every time: find where the inside hits zero, and on the negative side flip the entire inside.

Answer: E

inciduntveniam
Hi, i just want to reconfirm the values of for |2x-3| so it is +ve i.e 2x+3 when x>=-3/2 and 2x-3 when x<3/2 . Am i doing the negative part calculation correct? My approach is that when x will be less than 0 then 2x will become -2x inside the mod so we'll have -2x+3 and when we'll open the mod as negative then that should be 2x-3<0 and will give x<3/ ; this is the correct way right?

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