absolute values

**psolaki** · December 23rd, 2011, 07:35 AM

In solving an equation for e.g : 2x-7=9 we start by using some axioms or thoerem ,like adding or subtracting or multiplying both sides of the equation the equation will not change e.t.c

But in solving the equation : |x+2|+|x-3| =2 what axioms or definitions e.t.c can we use to start solving the equation??

**Siron** · December 23rd, 2011, 07:48 AM

Use the definition of the absolut value. You should consider 4 cases.

**MarkFL2** · December 23rd, 2011, 08:09 AM

I only find 3 cases to consider:

$x<-2$

$-2\le x<3$

$3\le x$

What am I missing?

**Siron** · December 23rd, 2011, 08:24 AM

Originally Posted by MarkFL2

I only find 3 cases to consider:

$x<-2$

$-2\le x<3$

$3\le x$

What am I missing?

Yes, you're right because one case isn't possible, but my intention was that psolaki would notice that.

**MarkFL2** · December 23rd, 2011, 08:33 AM

Ah, now I get what you mean...I thought I was neglecting something...my apologies for being obtuse and spoiling your intention.

**psolaki** · December 23rd, 2011, 01:04 PM

Since $x+2\geq \vee x+2<0$ and $x-3\geq 0\vee x-3<0$ by the axiom of trichotomy,the 4 cases we should consider are the following:

1) $x+2\geq 0\wedge x-3\geq 0$

2) $x+2\geq 0\wedge x-3<0$

3) $x+2<0\wedge x-3\geq 0$

4) $x+2<0\wedge x-3<0$

Correct??

**Siron** · December 23rd, 2011, 01:26 PM

Yes, indeed! But there's one case which is not possible. Which one? Can you solve the equation now? What's your conclusion?

**Plato** · December 23rd, 2011, 01:31 PM

Originally Posted by psolaki

But in solving the equation : |x+2|+|x-3| =2 what axioms or definitions e.t.c can we use to start solving the equation??

The equation $|x+2|+|x-3| =2$ has no solution.
It is a very good example of showing that knowing the basics is the best way to do mathematics.

The expression $|x-y|$ is the distance from $x\text{ to }y$ .

So look at that equation again.
The distance from $x$ to $-2$ plus the distance from $x$ to $3$ equals $2$ .

Now draw a number line and mark off $-2~\&~3$ .
They are five units apart: $|(-2)-(3)|=5$ .
So can there be an $x$ such that $|x-(-2)|+|x-(3)|=2~?$

**psolaki** · December 23rd, 2011, 04:38 PM

o.k

1) $x+2\geq\wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}$

2) $x+2\geq 0\wedge x-3<0\Longrightarrow x+2-x+3=2\Longrightarrow 5=2$

3) $x+2<0\wedge x-3\geq 0\Longrightarrow -x-2+x-3=2\Longrightarrow -5=2$

4) $x+2<0\wedge x-3<0\Longrightarrow -x-2-x+3=2\Longrightarrow x=-\frac{1}{2}$

So the solutions are: x=3/2 or x= -1/2

Correct??

**Prove It** · December 23rd, 2011, 04:41 PM

Originally Posted by psolaki

o.k

1) $x+2\geq\wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}$

2) $x+2\geq 0\wedge x-3<0\Longrightarrow x+2-x+3=2\Longrightarrow 5=2$

3) $x+2<0\wedge x-3\geq 0\Longrightarrow -x-2+x-3=2\Longrightarrow -5=2$

4) $x+2<0\wedge x-3<0\Longrightarrow -x-2-x+3=2\Longrightarrow x=-\frac{1}{2}$

So the solutions are: x=3/2 or x= -1/2

Correct??

Did you read Plato's post directly above yours?

**psolaki** · December 23rd, 2011, 05:51 PM

Siron in guiding me thru the solution of the equation asked me to show calculations,or must accept platos way of solving the problem as a general way of solving equations of this type??

I am confused

**MarkFL2** · December 23rd, 2011, 07:20 PM

I believe Plato wanted you to recognize that the sum of the distances from any point x on the number line to the points -2 and 3 is a minimum when:

$-2\le x\le 3$ and that sum is 5.

Look at the 2 solutions you obtained...do they violate your stated conditions?

**ILikeSerena** · December 23rd, 2011, 07:36 PM

Let's start with:

Originally Posted by psolaki

1) $x+2\geq 0 \wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}$

You found $x=\frac{3}{2}$ which is good.
However, this can only be a solution if it satisfies the conditions $x+2\geq 0 \wedge x-3\geq 0$ .
Does it?

**Deveno** · December 23rd, 2011, 08:01 PM

Originally Posted by psolaki

Siron in guiding me thru the solution of the equation asked me to show calculations,or must accept platos way of solving the problem as a general way of solving equations of this type??

I am confused

your calculations are ok, but you haven't gone far enough.

x+2 ≥ 0 and x-3 ≥ 0, means that x ≥ -2 and x ≥ 3, that is, x ≥ 3. however, if x ≥ 3, then that is inconsistent with x = 3/2. so that's a contradiction.

now examine your other three cases.

**HallsofIvy** · December 24th, 2011, 07:52 AM

Originally Posted by psolaki

o.k

1) $x+2\geq\wedge x-3\geq 0\Longrightarrow x+2+x-3=2\Longrightarrow x=\frac{3}{2}$

If x= 3/2, then x- 3= -3/2 which is NOT greater than or equal to 0

2) $x+2\geq 0\wedge x-3<0\Longrightarrow x+2-x+3=2\Longrightarrow 5=2$

3) $x+2<0\wedge x-3\geq 0\Longrightarrow -x-2+x-3=2\Longrightarrow -5=2$

4) $x+2<0\wedge x-3<0\Longrightarrow -x-2-x+3=2\Longrightarrow x=-\frac{1}{2}$

If x=-1/2 then x+ 2= 3/2 which is NOT less than 0.

So the solutions are: x=3/2 or x= -1/2

Correct??

No. You did not check to see if your values for x met the required conditions.

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