Prove that $\alpha f + \beta g$ is Lipschitz-continuous

Question

f, g are Lipschitz-Continuous in Intervall I. Show that $\alpha f + \beta g$ is also Lipschitz-Continuous $\forall \alpha, \beta \in \mathbb{R}$

So that gives us following prerequisites...

1. $|f(x)-f(x')| \leq L |x-x'|$
2. $|g(x)-g(x')| \leq M |x-x'|$

For $x, x' \in I \wedge L, M > 0 \quad const.$

I did following...

$\Rightarrow |(\alpha f+\beta g)(x)-(\alpha f+\beta g)(x')|=|\alpha f(x)+\beta g(x)-\alpha f(x')-\beta g(x')|$

$=|\alpha (f(x)-f(x'))+\beta (g(x)-g(x'))| \leq |\alpha (f(x)-f(x'))|+|\beta (g(x)-g(x'))|$

$=|\alpha| |(f(x)-f(x'))|+|\beta| |(g(x)-g(x'))| \leq |\alpha| L |x-x'|+|\beta| M |x-x'|$

$ =(|\alpha| L +|\beta| M) |x-x'| = N |x-x'|$

Which means, that in the end we have shown that...

$|(\alpha f+\beta g)(x)-(\alpha f+\beta g)(x')| \leq N |x-x'|$ with $N>0 \quad const.$

Is that correct? Thanks.