How Do You Solve Complex Absolute Value Inequalities?

[askor]

How to solve these two absolute value problems?

1.
$|3x - 5| > |x + 2|$

My attempt:
From what I read in my textbook, the closest properties of absolute value is the one that uses "equal" sign
$|3x - 5| = |x + 2|$
$3x - 5 = x + 2$
$3x -x = 5 + 2$
$2x = 7$
$x = \frac{7}{2}$

$|3x - 5| = |x + 2|$
$3x - 5 = -(x + 2)$
$3x - 5 = -x - 2$
$3x + x = 5 - 2$
$4x = 3$
$x = \frac{3}{4}$

However, this absolute uses ">" sign. So, how do you solve this one?

2.
|x - 3| + |2x - 8| = 5

I don't understand at all of absolute value problem like above one. Please help me.

Note: this is the absolute value properties from my textbook (please see attached file).

 

[nrqed]

How to solve these two absolute value problems?

1.
$|3x - 5| > |x + 2|$

My attempt:
From what I read in my textbook, the closest properties of absolute value is the one that uses "equal" sign
$|3x - 5| = |x + 2|$
$3x - 5 = x + 2$
$3x -x = 5 + 2$
$2x = 7$
$x = \frac{7}{2}$

$|3x - 5| = |x + 2|$
$3x - 5 = -(x + 2)$
$3x - 5 = -x - 2$
$3x + x = 5 - 2$
$4x = 3$
$x = \frac{3}{4}$

However, this absolute uses ">" sign. So, how do you solve this one?

2.
|x - 3| + |2x - 8| = 5

I don't understand at all of absolute value problem like above one. Please help me.

Note: this is the absolute value properties from my textbook (please see attached file).

Start with the simplest case: what is the solution if both $2x-5$ and $x+2$ are both greater or equal to zero? What condition(s) do you get on $x$ then?

 

[haruspex]

this absolute uses ">" sign.

The abs() function is continuous, so as x varies continuously |3x-5|-|x+2| cannot switch between >0 and <0 without passing through =0.
Thus, the solutions you found for the equals case represent the boundaries for the positive and negative ranges. It is just a matter of testing values of x between and beyond those points.

For (2), you have more cases to consider. How many?

 

[PeroK]

Following on from this, I would say the simplest approach is to draw a graph of both functions to see graphically where $|3x -5|$ is greater than $|x + 2|$.

 

[haruspex]

Following on from this, I would say the simplest approach is to draw a graph of both functions to see graphically where $|3x -5|$ is greater than $|x + 2|$.

Perhaps it comes to the same thing, but I look at the points where the individual terms become zero.
In general, we have $\Sigma a_i|x-b_i|>c$. The a_i can be signed.
The function is continuous and consists of straight lines between the points $x=b_i$. It is not hard to plot the values of the function at those points, and to see what happens as x tends to ±∞.