Solving a modulus inequality in ##x##

[brotherbobby]

Homework Statement

Solve the equation for possible values in $x$ : $\mathbf{|2x+1|> x}$.

Relevant Equations

For the inequality $|x|>a$, the possible outcomes for $x$ are : (1) $x > a$ OR (2) $x <- a$.

The answer to the above problem is baffling, despite its straightforward nature. I will post the answer later, but here is my solution first.

Solution :

(1) $2x+1 > x$ : In this case, we have $2x - x > -1 \Rightarrow \boxed{x > - 1}$

(2) $2x+1 < -x$ : In this case, we have $2x+x < -1 \Rightarrow 3x < -1 \Rightarrow \boxed{x < -\frac{1}{3}}$.

To write it neatly, my solution to the above problem is the following : $\boxed{x \in (-\infty, -\frac{1}{3}) \cup (-1, +\infty)}$.

The book has a baffling answer - here it is - $\boxed{x \in (-\infty, +\infty)}$.

(In other words, according to the book, $x$ can be anything!)Any help would be welcome.

 

[Mark44]

Homework Statement:: Solve the equation for possible values in $x$ : $\mathbf{|2x+1|> x}$.
Relevant Equations:: For the inequality $|x|>a$, the possible outcomes for $x$ are : (1) $x > a$ OR (2) $x <- a$.

The answer to the above problem is baffling, despite its straightforward nature. I will post the answer later, but here is my solution first.

Solution :

(1) $2x+1 > x$ : In this case, we have $2x - x > -1 \Rightarrow \boxed{x > - 1}$

(2) $2x+1 < -x$ : In this case, we have $2x+x < -1 \Rightarrow 3x < -1 \Rightarrow \boxed{x < -\frac{1}{3}}$.

To write it neatly, my solution to the above problem is the following : $\boxed{x \in (-\infty, -\frac{1}{3}) \cup (-1, +\infty)}$.

The book has a baffling answer - here it is - $\boxed{x \in (-\infty, +\infty)}$.

(In other words, according to the book, $x$ can be anything!)Any help would be welcome.

Your solution is equivalent to the answer in the book, except that the book's solution is more succinct.
Are there any numbers that aren't in the intervals $(-\infty, -\frac 1 3)$ or $(-1, \infty)$ or both?

 

[member 587159]

$(-\infty, -1/3) \cup (-1,+\infty) = (-\infty, + \infty)$

Draw a picture if that helps.

 

[PeroK]

The book has a baffling answer - here it is - $\boxed{x \in (-\infty, +\infty)}$.

(In other words, according to the book, $x$ can be anything!)Any help would be welcome.

Gimme an $x$ for which that inequality does not hold!

 

[brotherbobby]

Thank you all for your replies. I realize now that the intervals I had found in my answer, viz. $x\in(-\infty, -1/3)\cup (-1, +\infty)$ includes all numbers. I suppose my error was in perhaps thinking that $-1>-1/3$. The problem underscores the fact however that some intervals, or parts of an interval, can include parts of another intervals. I draw the two intervals on the number line below, for this problem, to represent the answer as the book put it : $x\in (-\infty, +\infty)$ :

 

[member 587159]

Thank you all for your replies. I realize now that the intervals I had found in my answer, viz. $x\in(-\infty, -1/3)\cup (-1, +\infty)$ includes all numbers. I suppose my error was in perhaps thinking that $-1>-1/3$. The problem underscores the fact however that some intervals, or parts of an interval, can include parts of another intervals. I draw the two intervals on the number line below, for this problem, to represent the answer as the book put it : $x\in (-\infty, +\infty)$ :

View attachment 273104

Well done and nice picture!

 

[PeroK]

Alternatively, for $x < 0$ the result is trivial and for $x \ge 0$:
$$|2x + 1| = 2x + 1 > x$$

 

[brotherbobby]

Alternatively, for $x < 0$ the result is trivial and for $x \ge 0$:
$$|2x + 1| = 2x + 1 > x$$

Thank you. Yes, there is a simpler solution to mine giving all values of $x$ as possible.

 

[haruspex]

You got the right answer, but your method was invalid.

(1) 2x+1>x : In this case,

This is for the case 2x+1>0. For that condition to be met, you need x>-1/2. So when you arrived at x > -1, you had not established that x > -1 ensures |2x+1|>x, only that x>-1/2 ensures |2x+1|>x.
Likewise, your case 2 was for x<-1/2, so when you got x<-1/3 what you had managed to show was that x<-1/2 ensures |2x+1|>x, not that x<-1/3 ensures it.
So your two cases abutted, but did not overlap.

 

[brotherbobby]

This is for the case 2x+1>0.

To be honest, I did not understand your response. Let me explain.

Please note that my first case is $2x+1 > x$, and not $2x+1>0$. Solving I get $x > -1$. Why should I try to test whether $2x+1 > 0$?

Likewise, for the second case, $2x+1 < -x$ and not $-(2x+1) < 0$. That solves to $x < -1/3$. Again, I would like you to explain why should I try for the case $-(2x+1) < 0$.

Thank you for your time and interest.

 

[Mark44]

To be honest, I did not understand your response. Let me explain.

Please note that my first case is $2x+1 > x$, and not $2x+1>0$. Solving I get $x > -1$. Why should I try to test whether $2x+1 > 0$?

No, your reasoning is incorrect. To solve an inequality with an absolute value, you have to convert the inequality into two separate inequalities -- one in which the expression in absolute value signs is assumed to be positive, and another in which that same expression is assumed to be negative.
In other words, since |2x + 1| is the only part in absolute values, that's the part that we treat in two cases, not the x term on the right side.

You need to look at the two possible values of |2x + 1| and what each implies about 2x + 1.
If 2x + 1 >= 0, then |2x + 1| = 2x + 1, and the inequality to work with is 2x + 1 > x. From this inequality we get x > -1.
If 2x + 1 < 0, then |2x + 1| = -(2x + 1), so the inequality this time is -(2x + 1) > x, or equivalently, x < - 1/3.

Likewise, for the second case, $2x+1 < -x$ and not $-(2x+1) < 0$. That solves to $x < -1/3$. Again, I would like you to explain why should I try for the case $-(2x+1) < 0$.

See above.

 

[brotherbobby]

No, your reasoning is incorrect. To solve an inequality with an absolute value, you have to convert the inequality into two separate inequalities -- one in which the expression in absolute value signs is assumed to be positive, and another in which that same expression is assumed to be negative.
In other words, since |2x + 1| is the only part in absolute values, that's the part that we treat in two cases, not the x term on the right side.

You need to look at the two possible values of |2x + 1| and what each implies about 2x + 1.
If 2x + 1 >= 0, then |2x + 1| = 2x + 1, and the inequality to work with is 2x + 1 > x. From this inequality we get x > -1.

Thank you very much. I should do the entire problem again. However, I have a small matter to clarify.

If $|x|> a$, isn't it correct to say that : (1) $x > a$, or (2) $x < -a$?

Likewise, if I have an inequality involving a function of $x$, $|f(x)| > a$, can we say : (1) $f(x) > a$ or (2) $f(x) < -a$?

In each of the (two) cases, do we have to compare whether $f(x) > 0$ or not?

Thank you for your time.

 

[Office_Shredder]

I think what you wrote is right, but if a is negative you get the situation in the original post where you construct overlapping intervals.

 

[haruspex]

If |x|>a, isn't it correct to say that : (1) x>a, or (2) x<−a?

Yes, but I thought there would be a problem with reversibility. For 2x+1>x you correctly deduce x>-1, but what you need is the reverse implication: that for a certain range of x,|2x+1|>x.
However, I can't find a case where your method fails, so I have to concede.
Interesting that @Mark44 had the same concern.

 

[brotherbobby]

Dear user Haruspex, I would really have to do the same problem again. Best way out really, since my original method was very questionable. (I have been writing $|2x+1|= 2x+1$ without looking to see under what conditions does that hold. It does if $2x+1 > 0\Rightarrow x > -1/2$). I would like to ask you something very different at the moment, if you're aware.

The mathjax engine which generates math equations on this forum is no longer working with the command \color. If I am right, do you know a way around it?

 

[brotherbobby]

You need to look at the two possible values of |2x + 1| and what each implies about 2x + 1.
If 2x + 1 >= 0, then |2x + 1| = 2x + 1, and the inequality to work with is 2x + 1 > x. From this inequality we get x > -1.
If 2x + 1 < 0, then |2x + 1| = -(2x + 1), so the inequality this time is -(2x + 1) > x, or equivalently, x < - 1/3.

I do the problem again. Thank you for your help. It would be best if I mention the problem statement to start with.

Problem Statement : Solve the inequality for the possible values for $x$ : $\mathbf{|2x+1| > x}$.

Solution : The inequality can have two outcomes.

(1) $|2x+1| = 2x+1$. This is possible if $2x+1 \geq 0\Rightarrow x \geq -1/2$. Going to the inequality, we have $2x+1 > x \Rightarrow x > - 1$. Since we have $-1/2 > -1$, the former answer ($x \geq -1/2$) is more constraining than $x > -1$. Hence, for this case, the possibilities for $x$ are : $x\geq -1/2 \Rightarrow \boxed{x \in [-1/2, +\infty)}$.

(2) $|2x+1| = -(2x+1)$. This is possible if $2x+1 < 0 \Rightarrow x < -1/2$. For the given inequality, we have $-2x - 1 > x \Rightarrow x < -1/3$. The number $-1/3 > -1/2$, the former inequality ($x < -1/2$) is more constraining than $x< -1/3$. Hence, for this case, the possibilities for $x$ are : $x < -1/2 \Rightarrow \boxed{x \in (-\infty, -1/2)}$.

Putting the two cases together, we get the answer : $\boxed{x \in (-\infty, -1/2) \cup [-1/2, +\infty) \Rightarrow \boldsymbol{\mathbf{x \in (-\infty, +\infty)}}}$.

It matches the answer in the book, but I wonder if my answers for the two cases are right.

 

[haruspex]

I do the problem again. Thank you for your help. It would be best if I mention the problem statement to start with.

Problem Statement : Solve the inequality for the possible values for $x$ : $\mathbf{|2x+1| > x}$.

Solution : The inequality can have two outcomes.

(1) $|2x+1| = 2x+1$. This is possible if $2x+1 \geq 0\Rightarrow x \geq -1/2$. Going to the inequality, we have $2x+1 > x \Rightarrow x > - 1$. Since we have $-1/2 > -1$, the former answer ($x \geq -1/2$) is more constraining than $x > -1$. Hence, for this case, the possibilities for $x$ are : $x\geq -1/2 \Rightarrow \boxed{x \in [-1/2, +\infty)}$.

(2) $|2x+1| = -(2x+1)$. This is possible if $2x+1 < 0 \Rightarrow x < -1/2$. For the given inequality, we have $-2x - 1 > x \Rightarrow x < -1/3$. The number $-1/3 > -1/2$, the former inequality ($x < -1/2$) is more constraining than $x< -1/3$. Hence, for this case, the possibilities for $x$ are : $x < -1/2 \Rightarrow \boxed{x \in (-\infty, -1/2)}$.

Putting the two cases together, we get the answer : $\boxed{x \in (-\infty, -1/2) \cup [-1/2, +\infty) \Rightarrow \boldsymbol{\mathbf{x \in (-\infty, +\infty)}}}$.

It matches the answer in the book, but I wonder if my answers for the two cases are right.

Yes, that's the method @Mark44 and I see as natural and convincing.