As soon as you can represent three consecutive integers as $3x+5y$, you can represent them all by just adding a $3$ to the previous representations. Since $8=3+5,9=3+3+3$ and $10=5+5$, all the integers $\geq 8$ can be represented.
Another way to prove this is to consider that:
$$ r(n)=|\{(a,b)\in\mathbb{N}^2:3a+5b=n\}|$$
is the coefficient of $x^n$ in the product:
$$ (1+x^3+x^6+x^9+\ldots)(1+x^5+x^{10}+\ldots),$$
hence:
$$\begin{eqnarray*}r(n)&=&[z^{n}]\frac{1}{(1-z^3)(1-z^5)}\\&=&[z^n]\left(\frac{1}{15(1-z)^2}-\frac{1}{5(1-z)}+h(z)\right)\end{eqnarray*}\tag{1}$$
where:
$$h(z) = \sum_{\xi\in Z}\frac{\operatorname{Res}\left(\frac{1}{(1-z^3)(1-z^5)},z=\xi\right)}{\xi-z}$$
and $Z=\left\{\exp\frac{2\pi i}{3},\exp\frac{4\pi i}{3},\exp\frac{2\pi i}{5},\exp\frac{4\pi i}{5},\exp\frac{6\pi i}{5},\exp\frac{8\pi i}{5}\right\}$.
Since the sum of the residues is $0$, the contribute to the coefficients given by the residues in $Z$ can never exceed $\frac{|Z|}{5}=\frac{6}{5}$. Hence we just need to prove that for any $n\geq N$,
$$[z^n]\left(\frac{1}{15(1-z)^2}-\frac{1}{5(1-z)}\right)>\frac{6}{5}$$
holds, in order to prove that any $n\geq N$ can be represented as $3x+5y$. However:
$$[z^n]\left(\frac{1}{15(1-z)^2}-\frac{1}{5(1-z)}\right)=\frac{n+1}{15}-\frac{1}{5},$$
hence we can take $N=21$ and fill the remaining cases ($n\in[8,20]$) by hand.