What is the solution to this inequality: $| 2x-3| > - | x+3|?$ By using graphical method, I am getting all real numbers..
Where am I wrong in graphical method? How to solve this using calculation?

 A: Since the absolute value of a number is always non-negative, the only way for this inequality to fail would be for both sides to be zero at the same time, which is clearly not possible since


*

*$x+3=0\implies x=-3$

*$2x-3=0\implies x=\frac32$
Your picture is fine, but usually a picture does not make a proof, it only gives a hint.
A: Because absolute values are always positive we have
$$
\lvert 2x - 3 \rvert \geq 0 \geq -\lvert x + 3 \rvert
$$
for all values of $x$. Hence the only time $\lvert 2x - 3 \rvert > -\lvert x + 3 \rvert$ could potentially not hold is when both sides equal $0$. But $2x - 3 = 0$ if and only if $x = \frac{3}{2}$ and $x + 3 = 0$ if and only if $x = -3$ and clearly these cannot happen at the same time. Thus $\lvert 2x - 3 \rvert > -\lvert x + 3 \rvert$ holds for all values of $x$.
A: Rearrange:
$$|2x-3|>-|x+3| \iff|2x-3|+|x+3|>0 \iff \\
|2x-3|\ne 0 \ \text{and} \ |x+3|\ne 0 \iff x\ne \frac32 \ \text{and} \ x\ne -3 \iff x\in(-\infty,+\infty).$$
A: 1) Let $x \not = -3. $
The right hand side of the inequality $<0$ for $x \in \mathbb{R}$ \ {$-3$}.
The left hand side $\ge 0$ for $x \in \mathbb{R}$ \ {$-3$}.
2)Let $x=-3:$ 
Right hand side = $0$, left hand side $>0$, ok.
Hence inequality is valid for all $x \in \mathbb{R}.$
A: Solution
Denote $$f(x)=|2x-3|+|x+3|.$$ We want the solution for $f(x)>0$.
In fact, 
$$f(x)=
\begin{cases}
-3x，x<-3;\\
-x+6,-3 \leq x<\dfrac{3}{2};\\
3x,x \geq \dfrac{3}{2}
\end{cases}$$

Thus, all $x \in \mathbb{R}$ are the solution for $f(x)>0$.
