I have the following example:
$ pa_1=a_0\\ pa_0+qa_1+pa_2=a_1\\qa_0+qa_2=a_2 $
where p+q=1.
I can see how to get $a_1=(1/p)a_0 $ but from there they say from the third equation they produce $a_2 = (q/p)a_0$. Then the values are substituted into the normalising equation of $a_0 + a_1 + a_2 =1$. With a final answer of $a_0 =p/2 \ , a_1=1/2 \ , \ a_2=q/2$
Update: I can see how they get $a_2 = (q/p)a_0 $
$qa_0+qa_2=a_2 \ -> qa_0=a_2 - qa_2\ -> qa_0=a_2(1-q)\ −> a_2 = qa_0/(1-q) \ $ and as p+q=1 p=1-q hence $a_2=(q/p)a_0$
$ p\ a_1=a_0\\ p\ a_0+q\ a_1+p\ a_2=a_1\\q\ a_0+q\ a_2=a_2 $
Now replacing $a_1$ from the first equation gives:
$$p\left[ \left(1+\frac{q-1}{p^2}\right)a_0+a_2\right]=0\\ q\ a_0=\left(1-q\right)a_2$$
Using the fact that $p=1-q$, the third equation gives(which is the same as what you might have got from the second equation):
$$a_2=\frac{q}{p}a_0$$
Now replacing $a_1$ and $a_2$ in the relation$a_0+a_1+a_2=1$, we get:
$$a_0+a_1+a_2=a_0+\frac{1}{p}a_0+\frac{q}{p}a_0=\frac{a_0}{p}\left(p+1+q\right)=\frac{2}{p}a_0=1\\ \Rightarrow a_0=\frac{p}{2},a_1=\frac{1}{2},a_2=\frac{q}{2}$$