Inequality question

[tomcenjerrym]

Does anyone can solve the following equation?

|x − 1| = 1 − x

Thanks

Tom

[BSMSMSTMSPHD]

Yes I can. Thanks for inquiring!

[BSMSMSTMSPHD]

Sorry, that's unfair - it's your first post. Can you show us the work you have so far? Do you understand what absolute value means?

[Kummer]

Does anyone can solve the following equation?

|x − 1| = 1 − x

Thanks

Tom
Square both sides. And see what you get.

[symbolipoint]

Square both sides. And see what you get. A more introductory way is to consider (x -1) as separately positive, and negative.

If x - 1 is positive, then x - 1 = 1 - x;
If x - 1 is negative, then x - 1 = -(1 - x )

[symbolipoint]

Does anyone can solve the following equation?

|x − 1| = 1 − x

Thanks

Tom

Actually, that is not an inequality question; but an absolute value equation.

[nicktacik]

If you have |a| = -a, what is the only way this can be true?

[tomcenjerrym]

Sorry, that's unfair - it's your first post. Can you show us the work you have so far? Do you understand what absolute value means?

This is the solutions which is done by myself:

|x − 1| = 1 − x
x − 1 = ±(1 − x)

First
x − 1 = +(1 − x)
= 1 − x

Second
x − 1 = −(1 − x)
= −1 + x
= x − 1

By |x| = ±x properties.

Actually, that is not an inequality question; but an absolute value equation.

That's absolutely correct. I mean ABSOLUTE VALUE QUESTION. Thanks for remind me about that.

Square both sides. And see what you get.

Why should I square-ing both sides of them? This is the most question I want to know in the ABSOLUTE VALUE. As I can read on Calculus book, there is no rules about SQUARE on absolute value except the |x| = √x^2 one. Can you explain me further about this?

[symbolipoint]

tomcenjerrym - Your first condition yields x=0 as a solution; and your second condition allows ALL real numbers as solutions. All real numbers will satisfy the equation.

[d_leet]

tomcenjerrym - Your first condition yields x=0 as a solution; and your second condition allows ALL real numbers as solutions. All real numbers will satisfy the equation.

No they won't. Take x=5 as an example |5-1|=|4|=4, but 1-5=-4, so in this case |x-1| does not equal 1-x, hence it is obviously not true for all real numbers. There is, however, a subset of the real numbers (with more than a single element) that satisfies the above equation.

[symbolipoint]

No they won't. Take x=5 as an example |5-1|=|4|=4, but 1-5=-4, so in this case |x-1| does not equal 1-x, hence it is obviously not true for all real numbers. There is, however, a subset of the real numbers (with more than a single element) that satisfies the above equation.
You're correct. I was not careful enough when I solved the problem. We must watch around the critical point. The first part indicates x=1. When we check a value less than 1, we find equality; when we check a point greater than 1, we do not find equality.

The solution seems to be x<=1

[tomcenjerrym]

If you have |a| = -a, what is the only way this can be true?

I like nicktacik way to solve the problem. It’s simple and easy to understand. Thank you very much nicktacik and thank you to everyone too.

[Nesk]

The title "Inequality Question" is thus not entirely wrong, since the solution is an inequality.

[cks]

#12
nicktacik,

ya, his answer was kind of making me suddenly awake!