Show that it is impossible to find $a,b \in \mathbb{Q}$ such that $$ \frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7} $$
We have $\frac{1}{20}=a^{2}+\frac{2 \sqrt{3}}{7} a b+b^{2} \frac{3}{49}=\left(a^{2}+b^{2} \frac{3}{49}\right)+\frac{2 \sqrt{3}}{7} a b$ with $a \neq 0, b \neq 0$.
Suppose $a=0$, then $$ \frac{1}{20}=b^{2} \frac{3}{49} \Leftrightarrow 3 b^{2}= \frac{49}{20} $$ (here I don't know how to conclude)
If $b=0$, then $$ \sqrt{5}=10 a $$ which is impossible because $\sqrt{5} \neq \mathbb{Q}$.
Alternative approach.
Lemma 1
$\sqrt{15}$ is not a rational number.
$\underline{\text{Proof}}$
Suppose that $~\displaystyle \sqrt{15} = \frac{p}{q} ~:$
$~p,q~$ relatively prime integers, and $~q > 0.$
Then $q\sqrt{15} = p \implies 15q^2 = p^2.$
However, since $p,q$ relatively prime, this implies that $15$ divides $p^2$. However, by the fundamental theorem of arithmetic, this implies that the prime factorization of $p$ contains both $3$ and $5$.
This implies that $15$ divides $p$.
This implies that $(15)^2$ divides $p^2.$
This implies that $(15)$ divides $q^2,~$ since $15q^2 = p^2.$
This is a contradiction, because $p,q$ are relatively prime.
Lemma 2
If $~k \in \Bbb{Z_{\neq 0}},~$ then $k\sqrt{15}$ is not an integer.
$\underline{\text{Proof}}$
Suppose $k\sqrt{15} = n \in \Bbb{Z}.$
Then $~\displaystyle ~\sqrt{15} = \frac{n}{k}.$
This contradicts Lemma 1.
Lemma 3
If $R,S \in \Bbb{Z_{\neq 0}}$,
then $R\sqrt{3} + S\sqrt{5}$ is not an integer.
$\underline{\text{Proof}}$
Suppose $~R\sqrt{3} + S\sqrt{5} = n \in \Bbb{Z}.$
Then $3R^2 + 5S^2 + 2RS\sqrt{15} = n^2,$ which is also an integer.
This implies that $2RS\sqrt{15}$ is an integer.
This contradicts Lemma 2.
Suppose that $~\displaystyle a,b = \frac{p_a}{q_a}, \frac{p_b}{q_b}~$ respectively,
where $p_a,q_a$ are relatively prime integers,
where $p_b,q_b$ are relatively prime integers,
and $q_a, q_b$ are both positive.
Then
$$\frac{p_a}{q_a} + \left(\frac{p_b}{q_b}\times \frac{\sqrt{3}}{7}\right) - \frac{\sqrt{5}}{10} = 0.\tag1 $$
In (1) above, you can ensure that each term has the common denominator of $(q_a \times q_b \times 7 \times 10).$
This implies that
$$\frac{70p_aq_b + 10q_ap_b\sqrt{3} - 7q_aq_b\sqrt{5}}{q_a \times q_b \times 7 \times 10} = 0. $$
This implies that
$$70p_aq_b + 10q_ap_b\sqrt{3} - 7q_aq_b\sqrt{5} = 0. $$
This implies that
$$\left(10q_ap_b\sqrt{3} - 7q_aq_b\sqrt{5}\right) ~~\text{is an integer}. $$
This is a contradiction, by Lemma 3.
Edit
I overlooked that you could have $p_b = 0$.
However, this would result in $7q_aq_b\sqrt{5}$ being an integer.
The exact same analysis used in Lemmas 1 through 3 may be repeated on $\sqrt{5}$ to show that that would be impossible.