Solving an equality with absolute values

[brotherbobby]

Homework Statement

Solve $\boldsymbol{\mid x-3\mid+\mid x-2\mid=1}$

Relevant Equations

Given $|x-a| = x-a$ if $x\ge a$ but $|x-a| = a-x$ if $x<a$.

Problem statement : Let me copy and paste the problem to the right as it appears in the text.

Solution attempt (mine) : There are mainly three cases to consider.

(1) $\boldsymbol{x\ge 3\; :}$ Using the relevant equations given above, the problem statement reduces to $$x-3+x-2 = 1\Rightarrow 2x=6\Rightarrow \underline{x =3}$$. We accept this solution as it lies in the given range for this case ($x \ge 3$).

(2) $\boldsymbol{2\le x<3\; :}$ Using the relavant equations, the problem statement reduces to $$3-x+x-2=1\Rightarrow 1 = 1,$$ which is always true. Hene the whole domain is valid as an answer : $\underline{2\le x<3}$.

(3) $\boldsymbol{x<2\; :}$ Using the relevant equations, the problem statement reduces to $$3-x+2-x = 1\Rightarrow 5-2x=1\Rightarrow 2x = 4\Rightarrow x = 2$$, which falls out of the current range and hence is rejected as a solution.

From the underlined answers above, we can say that the solution to the given problem is $\boxed{2\le x\le 3\Rightarrow x \in [2,3]}$.

Issue : The text agrees with my answer, but solves the problem in a way which should result in a slightly different answer and not the correct one.

Solution (text) : I copy and paste the solution as it appears in the text below, annotated in places by me where I believe the author is incorrect.

Question : Is the text mistaken? An answer would be welcome.

 

[Mark44]

Homework Statement:: Solve $\boldsymbol{\mid x-3\mid+\mid x-2\mid=1}$
Relevant Equations:: Given $|x-a| = x-a$ if $x\ge a$ but $|x-a| = a-x$ if $x<a$.

View attachment 314803Problem statement : Let me copy and paste the problem to the right as it appears in the text.

Solution attempt (mine) : There are mainly three cases to consider.

(1) $\boldsymbol{x\ge 3\; :}$ Using the relevant equations given above, the problem statement reduces to $$x-3+x-2 = 1\Rightarrow 2x=6\Rightarrow \underline{x =3}$$. We accept this solution as it lies in the given range for this case ($x \ge 3$).

(2) $\boldsymbol{2\le x<3\; :}$ Using the relavant equations, the problem statement reduces to $$3-x+x-2=1\Rightarrow 1 = 1,$$ which is always true. Hene the whole domain is valid as an answer : $\underline{2\le x<3}$.

(3) $\boldsymbol{x<2\; :}$ Using the relevant equations, the problem statement reduces to $$3-x+2-x = 1\Rightarrow 5-2x=1\Rightarrow 2x = 4\Rightarrow x = 2$$, which falls out of the current range and hence is rejected as a solution.

From the underlined answers above, we can say that the solution to the given problem is $\boxed{2\le x\le 3\Rightarrow x \in [2,3]}$.

Issue : The text agrees with my answer, but solves the problem in a way which should result in a slightly different answer and not the correct one.

Solution (text) : I copy and paste the solution as it appears in the text below, annotated in places by me where I believe the author is incorrect.

View attachment 314804

Question : Is the text mistaken? An answer would be welcome.

$|x - 3| = x - 3$ if $x \ge 3$, not merely as x > 3 as you wrote. Where I have a problem in the author's solution is in the line you noted as "nice step," in which he doesn't qualify by the range of values of x.

 

[Steve4Physics]

$|x - 3| = x - 3$ if $x \ge 3$, not merely as x > 3 as you wrote. Where I have a problem in the author's solution is in the line you noted as "nice step," in which he doesn't qualify by the range of values of x.

The 'nice step' is simply using the identity $(3-x) + (x-2)=1$ which is valid for all values of $x$.

 

[brotherbobby]

$|x - 3| = x - 3$ if $x \ge 3$, not merely as x > 3 as you wrote. Where I have a problem in the author's solution is in the line you noted as "nice step," in which he doesn't qualify by the range of values of x.

The author writes $|x-3|=3-x$ when $x\le 3$. Is this correct?
I think $|x-3| = 3-x$ when $x<3$. The $\text{equal to}$ option does not apply here.

 

[PeroK]

I would look at it graphically. The distance from $x$ to $2$ plus the distance from $x$ to $3$ equals $1$. We see immediately that $x$ cannot be greater than $3$ or less than $2$. Then we see that it must hold for any point between $2$ and $3$.

 

[Mark44]

The author writes $|x-3|=3-x$ when $x\le 3$. Is this correct?

Yes. Is the equation a true statement when x = 3? What does the equation reduce to in this case?

 

[brotherbobby]

Yes. Is the equation a true statement when x = 3? What does the equation reduce to in this case?

When $x=3$, both sides of $|x-3|=3-x$ reduce to zero, so yes the statement is true.
However, this contradicts something basic. We are told that $|x-a|=x-a$ when $x\ge a$ and $\boldsymbol{|x-a|=-(x-a)=a-x}$ when $\boldsymbol{x<a}$. Note the second set of statements in bold. We have $x<a$, not $x\le a$, and yet this is what the author is doing. I agree he is right. But does it mean we should modify our basic understanding of absolute values and put $\ge$ and $\le$ for both cases?

 

[PeroK]

When $x=3$, both sides of $|x-3|=3-x$ reduce to zero, so yes the statement is true.
However, this contradicts something basic. We are told that $|x-a|=x-a$ when $x\ge a$ and $\boldsymbol{|x-a|=-(x-a)=a-x}$ when $\boldsymbol{x<a}$. Note the second set of statements in bold. We have $x<a$, not $x\le a$, and yet this is what the author is doing. I agree he is right. But does it mean we should modify our basic understanding of absolute values and put $\ge$ and $\le$ for both cases?

I see no contradiction.$$|x| = x \ \ (x \ge 0)$$$$|x| = -x \ \ (x \le 0)$$There is no contradiction at $x = 0$.

 

[Mark44]

However, this contradicts something basic.

Not at all. The only difference between $|x - 3| = 3 - x$ when x < 3, and between $|x - 3| = 3 - x$ when $x \le 3$ is that the former doesn't include 3 while the latter does.

 

[brotherbobby]

Thank you, to both @PeroK and @Mark44. I will henceforth change my understanding of modulus slightly to include the "equal to" for both cases. Thus, $$|x-a| = x-a\; \text{when}\; \boxed{x\ge a}\; \text{and}\; |x-a|=-(x-a)\;\text{when}\; \boxed{x\le a}.$$

 

[Mark44]

Thank you, to both @PeroK and @Mark44. I will henceforth change my understanding of modulus slightly to include the "equal to" for both cases. Thus, $$|x-a| = x-a\; \text{when}\; \boxed{x\ge a}\; \text{and}\; |x-a|=-(x-a)\;\text{when}\; \boxed{x\le a}.$$

It's really a minor point. Any of the following three would serve as the definition of |x - a|
|x - a| = x - a, if $x \ge a$; |x - a| = a - x if $x \le a$ (a is included in both sets)
|x - a| = x - a, if $x \ge a$; |x - a| = a - x if $x < a$ (a is included only in the first set)
|x - a| = x - a, if $x > a$; |x - a| = a - x if $x \le a$ (a is included only in the second