How can I solve |3x-5| - |2x+3| >0 using interval division method?

[thomas49th]

Homework Statement

Find the solution set of:

$$|x+1| + |x-2| \leq 5$$

Homework Equations

I'm going to rearrange this to

$$|x+1| \leq 5 - |x-2|$$

The Attempt at a Solution

Well i sketched a graph of it. The line 5-|x-2| cross the y-axis at 3 and it 'pongs' back of the x-axis at -3 (is there a proper name for this value of x). The line |x+1| at cross y at 1 and x at -1.

The graph lines seem to cross between -3 and -1. One of them is a pongy line (reflected up from the x-axis due to the modulus symbol) and the other is the original line.

5 - |x-2| = -|x+1|
|x-2| -5 = |x+1|

but how do i solve for that?

another concern i have is whether or the lines are going to cross again higher up (if you see what i mean). Is there a sound way of checking it.

Thanks

 

[dirk_mec1]

Divide the entire real axis in several intervals corresponding to the several absolute values. Then look at each interval and determine for each term the correct form.

You can then solve the equation by looking at each interval.

 

[m_s_a]

http://www.a7bk-a-up.com/pic/Ek226065.jpg

 

[phymatter]

hi!
pl. help me out with this:

|3x-5| - |2x+3| >0
How can i solve this by applying your method :

Divide the entire real axis in several intervals corresponding to the several absolute values. Then look at each interval and determine for each term the correct form.

You can then solve the equation by looking at each interval.

thanks in advance