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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I often get confused trying to figure out absolute inequalities and I'm trying to develop some steps to solve them.

|2x - 3| < 5

I guess this means the value of the expression 2x - 3 is less then 5 and greater then -5. So in theory we can say:

2x - 3 < 5 and 2x -3 > -5

In the first inequality we removed the abosulte value symbols, and the second we did the same, reversed the greater then symbol and reversed the sign of the expression on the right. Now we can solve for x in both inequalities and I think that would give is the domain of x.

If we have the opposite:

|2x - 3| > 5

this should also mean that either 2x -3 is greater then 5, or less then -5.

2x - 3 > 5 and 2x - 3 < -5

Again we wrote a second inequality, reversed the inequality symbol, and the sign of the expression on the right.

My mathbook often gives absolute value inequalites to solve but like I said I find then very confusing so I'm trying to develop a system.

Does my reasoning seem correct?

Now I'm wondering if we could use this to solve equations with two or more absolute value terms.

|2x + 3| + |3x + 5| < 6

we could try eliminating one at a time:

|2x + 3| < 6 - | 3X + 5|

2x + 3 < 6 - |3x + 5| and 2x + 3 > -6 + |3x + 5|

Now again isolate the absolute value term:

|3x + 5| < -2x + 3 and |3x + 5| < 2x + 9

Oh boy.... here come four inequalities:

3x + 5 < -2x + 3 and 3x + 5 > 2x - 3

and

3x + 5 < 2x + 9 and 3x + 5 > -2x - 9

Hope I didn't mess up on any of those. :-/

Solved x < -0.4 and x > -8 and x < 4 and x > -2.8

This restricts us to values of x between -0.4 and -2.8.

-0.4 < x < -2.8.

If you input -0.4 or -2.8 into the original expression, |2x + 3| + |3x +5| they both produce 6. Anywhere between them is less then 6 so the solution appears to be true and the process I suppose is also true.

I worked this out as I was typing so there may be some errors in my reasoning or calculations as I haven't checked it theroughly.

Two possibilities exist at this point. Either I just invented "Absolute Value Inequality Algebra" or Avia. Or I just reinvented absolute value inequality algebra. I assume its either the later or I'm the only person on the planet not smart enough to do it in my head.

A logarithm is just a misspelled algorithm.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

The best way of solving tricky inequalities like that is to sketch it on a graph, or buy a graphical calculator and make it draw the graph for you. That way, you can see whether |f(x)| is meant to be f(x) or -f(x) and so some of the algebra can be avoided and you only need to solve the appropriate equations.

Why did the vector cross the road?

It wanted to be normal.

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