Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Sabbir****Guest**

How do I solve this equation

**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi!

The general approach to absolute value equalities is to consider where the quantities inside the

absolute values are positive, zero, or negative. Try the following:

Where both of 2x+1 and 2x-1 are positive replace the absolute values with these and solve.

Where both of 2x+1 and 2x-1 are negative replace the absolute values with the opposite of these

and solve.

Where 2x+1 is positive and 2x-1 is negative replace |2x+1| with 2x+1 and |2x-1| with -(2x-1)

and solve.

You will also find solutions where each of these quantities individually are zero.

You will find the solutions in [-1/2, 1/2].

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

Offline

**Sabbir****Guest**

Could you please show the solving?

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,054

hi Sabbir,

Welcome to the forum.

For any 'linear' expression

The critical value for x is

So, for your equation you need to consider these cases:

(i) x < - 1/2

expression becomes

Although, strictly that value is outside the range that I'm checking, a numeric check shows it is a solution.

(ii) x = -1/2 Just done that.

(iii) -1/2 < x < 1/2

expression becomes

This is true for all values of x in the range.

(iv) x = +1/2

expression becomes

So this is a solution.

(v)

expression becomes

Putting (i) to (v) together, the solution set is {x : -1/2 ≤ x ≤ 1/2} or [-1/2,1/2] as noelevans has already explained.

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

Hi Bob

The solution is correct, but the interval is wrong. It should be an open interval: (-1/2,1/2)=]-1/2,1/2[.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

- 1 / 2 is a solution, isn't it?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

True. Then the solution set should be {x: -1/2<=x<=1/2}

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,054

Thanks. In my head it said that, but, as you know, I'm getting old and doddery, so what my fingers type doesn't always square with what my brain thinks. I have edited the post.

If you look carefully at the graph, you see the end point pixels are included.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**