i am asked to prove with $\epsilon$-$\delta$-argument that $x\rightarrow |-2x+3|$ is continuous
my steps: Definition of $\epsilon-\delta$-argument:
$\forall \epsilon >0 \exists \delta>0$ with $|x-x_0|<\delta \Longrightarrow |f(x)-f(x_0)|<\epsilon$
so: $|f(x)-f(x_0)|=||-2x+3|-|-2x_0+3|| = |(2x+3)-(2x_0+3)| = |2x-2x_0|=\ldots\text{help}\ldots<\epsilon$
i am stuck there again