Show a $\delta$-$\epsilon$ proof that $\lim_{x\to a} (\pi f(x)-10g(x)) = 2\pi -90$

Question:

Let $a \in R$ . If $\lim_{x\to a} f(x) = 2$ and $\lim_{x\to a} g(x) = 9$, show a $\delta$-$\epsilon$ proof that $\lim_{x\to a} (\pi f(x)-10g(x)) = 2\pi -90$.

My attempt at it:

Let $\epsilon > 0, \exists \delta_1,\delta_2,\delta_3>0$ such that $$0<|x-a|<\delta_1 \implies |f(x)-2|<\epsilon$$ $$0<|x-a|<\delta_2 \implies |g(x)-9|<\epsilon$$ $$0<|x-a|<\delta_3 \implies |\pi f(x)-10g(x) - 2\pi -90|<\epsilon$$

Consider:

$$|\pi f(x)-10g(x) - 2\pi -90| = |\pi f(x)-10g(x) - 0 + 0 - 2\pi -90| =$$ $$|\pi (f(x)-2)- 0 + 0 - 10(g(x)-9)|$$ $$\le |\pi||f(x)-2| + |10||g(x)-9|$$

At this point I'm not sure how to continue. Do I just assume $|f(x)-2|<\epsilon$ and $|g(x)-9|<\epsilon$ and say that $$\le |\pi||f(x)-2| + |10||g(x)-9|$$ $$< |\pi|\epsilon + |10|\epsilon ?$$

Or do I need to figure out what $\delta_1$ and $\delta_2$ is to continue, but if that is the case how would I do that?

If anyone could help me with this that would be really appreciated.

You would be fine if you let $\delta$ be the smaller of $\delta_1$ and $\delta_2$, and note that for all $x$, $|x - a| < \delta$ implies $|x - a| < \delta_1$ and $|x - a| < \delta_2$, which implies $|f(x) - 2| < \epsilon$ and $|g(x) - 9| < \epsilon$; hence $$|\pi f(x) - 10 g(x) - (2\pi - 90)| = |\pi(f(x) - 2) - 10(g(x) - 9)| \le \pi|f(x) - 2| + 10|g(x) - 9| < (\pi + 10)\epsilon.$$ As $\epsilon$ was arbitrary, the result follows.