B Solving Absolute Value Inequalities: How to Define Cases

[blueblast]

Hi there,

I'm having trouble understanding this math problem:

|x| + |x-2| = 2

The answer says its: 0<=x<=2

I understand you need different "cases" in order to solve this. For example, cases for when x is less than 0, when x-2 is less than 0, etc.

Thanks,

blueblast

 

[Doc Al]

For each of those cases, write an equation in terms of x (without the absolute value signs) and try to solve it. For example, when x < 0, how would you write |x| ?

 

[blueblast]

Yes, so far, this is what I got:

There are a total of four cases:

1. x + x-2 = 2
2. x + -(x-2) = 2
3. -x + x-2 = 2
4. -x + -(x-2) = 2

Simplified, this is:

1. x = 2
2. 2 = 2 (all reals)
3. -2 = 2 (no solution)
4. x = 0

Not sure where to go from there.

 

[S.G. Janssens]

It could help to make a drawing of the real line. On that line, indicate the points $0$ and $2$.

Now note that $|x|$ equals the distance from $x$ to $0$ and, likewise, $|x - 2|$ equals the distance from $x$ to $2$. Apparently the sum of these two distances should equal $2$, for $x$ to be a solution.

So, could any $x < 0$ be a solution of your equation? Could any $x > 2$ be a solution?
You have now narrowed down the possible solutions to the interval $[0,2]$. For $x \in [0,2]$ your equation becomes
$$
|x| + |x - 2| = x + (2 - x) = 2
$$
For which $x \in [0,2]$ is this equation satisfied?

 

[Doc Al]

There are a total of four cases:

You'll need to define those cases. (Think in terms of ranges.) For example, for x < 0, your equation can be written as:
-x + 2 - x = 2
For which the solution is x = 0, which violates x < 0. Thus we can exclude the range x < 0.

And so on...