# Solving an inequality involving absolute values

• brotherbobby
In summary: This may be summed up by saying that the function stretches from the origin to the point ##x##.]3. If both conditions in (1) and (2) are met, then the function ##f(x)\ge a##. [This may be summed up by saying that the function goes from the origin to the point ##x##.]
brotherbobby
Homework Statement
Solve ##\boldsymbol{\left|\dfrac{x-3}{x+1}\right|\le 1}##
Relevant Equations
Given ##|f(x)|\le a\Rightarrow -a\le f(x)\le +a##
Problem Statement :
I copy and paste the problem as it appeared in the text to the right.

Attempt (mine) : I copy and paste my attempt using Autodesk Sketchbook##^{\circledR}## below. I hope the writing is legible.

My answer : I have three answers and confused as to which of them hold.
Crucially, when we say that ##|f(x)|\le a\Rightarrow -a\le f(x)\le a##, do we mean that ##f(x) \ge -a\; \textbf{(and?)}\; f(x)\le a## or do we mean that ##f(x) \ge -a\; \textbf{(or?)}\; f(x)\le a##?.

I suspect it is the answer OR. (The text disagrees, as you will see in the text solutions soon.)

My three solutions : ##x>-1\; \textbf{OR}\; x\ge 1\; \textbf{OR}\; x<-1##

These solutions overlap (giving the condition OR) to yield the solution that ##\boxed{x\in (-\infty,+\infty)}-\{-1\}##. In other words, ##x## can be any real number but ##-1##.

Check (my answer) : I am aware that my answer is wrong. Given ##\boldsymbol{\left|\frac{x-3}{x+1}\right|\le 1}##. To take one number from my solution, ##x = -5##. But clearly, ##\left|\frac{-5-3}{-5+1}\right|=2\nleq 1##!

Text's Solution : I copy and paste the solution in the text. I must admit that, despite my failure to understand why it took the crucial condition above to be AND, the answer matches the inequality given in the problem upon taking a few representative examples.

A hint or a suggestion would be welcome.

brotherbobby said:
A hint or a suggestion would be welcome.
I’m not sure I fully understand the difficulty but see if this helps.

Consider the number line:
… -3 -2 -1 0 +1 +2 +3 …

When we say (for example) ##|y| \le 1##, this is equivalent to saying that y is between -1 and +1 (inclusive) on the number line.

This means two conditions have to be met at the same time:
##y \ge -1## AND ##y \le 1##

If meeting only one condition were sufficient, then OR would be used. But that’s not the case here.

Steve4Physics said:
If meeting only one condition were sufficient, then OR would be used. But that’s not the case here.
I did not understand this. ##|y|\le 1## implies both conditions : ##y\ge -1## and ##y\le 1##. Where is the scope for only one of these conditions to apply?

brotherbobby said:
I did not understand this. ##|y|\le 1## implies both conditions : ##y\ge -1## and ##y\le 1##. Where is the scope for only one of these conditions to apply?
We are looking for all allowed values of ##y## which make ##|y| \le 1## true.

Condition 1: ##y \ge -1##
Condition 2: ##y \le +1##

Condition 1 by itself allows values such as y = +100 (for example).
Condition 2 by itself allows values such as y = -100 (for example).

There is no possibility for only one condition, by itself, to always be true and still satisfy ##|y| \le 1##.

Both conditions must simultaneously be met in order to satisfy ##|y| \le 1##. Which is why we must ’AND’ the two conditions.

Sorry, I can’t think of another way to explain it.

EDIT. Some signs and tyop's corrected.

Steve4Physics said:
Condition 1: y≥−1
Steve4Physics said:
Condition 2: y≤+1

Condition 1 by itself allows values such as y = -100 (for example).
Condition 2 by itself allows values such as y = +100 (for example).
I suppose you meant that Condition 2 by itself allows y = -100 but Condition 1 will forbid it. Likewise, condition 1 by itself will allow y = 100 but condition 2 will forbid.

Thank you.

May I ask you what would you say if it was given that ##|f(x)|\ge 1##? Surely, there are two conditions here too : (1) ##f(x)\ge 1## and ##f(x)\le -1##. Is the word and correct here? Should it be replaced by or?

PeroK
brotherbobby said:
May I ask you what would you say if it was given that ##|f(x)|\ge 1##? Surely, there are two conditions here too : (1) ##f(x)\ge 1## and ##f(x)\le -1##. Is the word and correct here? Should it be replaced by or?
Note that ##|x| \ge 1## implies ##x \le -1## or ##x \ge +1##

Whereas ##|x| \le 1## implies ##-1 \le x \le 1##, which means ##-1 \le x## and ##x \le 1##.

PeroK said:
Note that ##|x| \ge 1## implies ##x \le -1## or ##x \ge +1##

Whereas ##|x| \le 1## implies ##-1 \le x \le 1##, which means ##-1 \le x## and ##x \le 1##.
Thank you @PeroK. I hope you are not dismissive of my query above in post# 5 on this matter, though it might appear to be trivial. I think it is an important point, recognition of which would save the student several heartaches when he proceeds to solve problems. Not to mention that I have not seen a textbook where the point is made. The text I am using has only made use of it in its examples, without once mentioning it.

To recapitulate :

1. If ##|f(x)|\le a##, we have the finction ##f(x)\le a## and ##f(x)\ge -a##. [This may be summed up by saying that ##-a\le f(x)\le a##, long as one understands that both conditions must be obeyed simultaneously].

2. If ##|f(x)|\ge a##, the function ##f(x)\ge a## or ##f(x)\le -a##.

brotherbobby said:
May I ask you what would you say if it was given that ##|f(x)|\ge 1##? Surely, there are two conditions here too : (1) ##f(x)\ge 1## and ##f(x)\le -1##. Is the word and correct here? Should it be replaced by or?
Yes. In that case 'and' is wrong and 'or' is needed.

I’ll use y rather than f(x), for neatness/clarity.

So now we are looking for all allowed values of ##y## such that ##|y| \ge 1##.

Our new conditions are:
##y \le -1## This is (-∞, -1] on the number line.
##y \ge +1## This is [+1,∞) on the number line.

Note the two conditions correspond to two non-overlapping sections of the number line. y could be on either section.

So in this case, we require either (not both) of the conditions to apply. We now require ##y \le -1## OR ##y \ge +1##.

Compare this to the original question with ##|y| \le 1## [Edit: typo'corrected] . This had two overlapping sections of the number line and ##y## had to be on the single region of overlap. That’s where the AND came from.

Steve4Physics said:
Compare this to the original question with |y|≥1.
You mean ##|y|\le 1##.

brotherbobby said:
You mean ##|y|\le 1##.
Yes. I just corrected it but you beat me to it!

Simpler approach: $$\begin{split} |x - 3| &\leq |x + 1| \\ (x - 3)^2 &\leq (x+1)^2 \\ x^2 - 6x + 9 &\leq x^2 + 2x + 1 \\ 8 &\leq 8x \\ 1 &\leq x.\end{split}$$

Mark44
Maybe you can consider a more geometric approach: |x-1| can be seen as the distance from the point x to 1. Same for |x+3|.
Now we look for the set of points whose distance to 3 are less than the distance to -1.

SammyS, Mark44 and PeroK

## What is an inequality involving absolute values?

An inequality involving absolute values is an expression that includes absolute value symbols, which indicate the distance of a number from zero on a number line. These inequalities can have one or more absolute values and may include variables.

## What are the steps for solving an inequality involving absolute values?

The steps for solving an inequality involving absolute values are:

1. Isolate the absolute value expression on one side of the inequality sign.

2. Rewrite the absolute value expression as two separate inequalities, one with a positive sign and one with a negative sign.

3. Solve each inequality separately.

4. Combine the solutions to find the final solution set.

## Can an inequality involving absolute values have more than one solution?

Yes, an inequality involving absolute values can have more than one solution. This is because absolute values represent distance, so there can be multiple numbers that satisfy the inequality.

## How can I check my solution for an inequality involving absolute values?

To check your solution for an inequality involving absolute values, plug the solution back into the original inequality and see if it is true. If the solution makes the inequality true, then it is a valid solution.

## What are some common mistakes to avoid when solving an inequality involving absolute values?

Some common mistakes to avoid when solving an inequality involving absolute values are:

- Forgetting to split the absolute value expression into two separate inequalities.

- Not correctly distributing the negative sign when rewriting the absolute value expression.

- Forgetting to consider both positive and negative solutions when solving the separate inequalities.

- Not checking the validity of the solution by plugging it back into the original inequality.

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