Dgdrxrt

Consider two dissimilar surds $sqrt p$ and $sqrt q$. Then the problem asks to find rational numbers $a,b,c$ and $d$ such that for $x=sqrt p + sqrt q$ we can write,
$$
sqrt p - sqrt pq + q = frac ax+bcx+d
$$
Initially, my attempt was straightforward rationalisation:
$$
frac ax+bcx+d=frac a(sqrt p + sqrt q)+bc(sqrt p + sqrt q)+d
$$
$$
=frac (ad-bc)sqrt p + (ad+bc)sqrt q + ac(q-p)+bd2cdsqrt q + c^2(q-p)+d^2
$$
$$
=frac Asqrt p +Bsqrt q + CDsqrt q +E
$$
$$
=frac -AEsqrt p-BEsqrt q+ADsqrt pq+(BDq-CE)D^2q-E^2
$$
Now it appears quite easy to equate the coefficients of surds in the LHS and the RHS, put in the values of the reduced constants $A,B,C,D$ and $E$ so as to get 4 equations in 4 variables $a,b,c$ and $d$ and finally solve them. Trust me, this is an absolutely ridiculous idea. Is there any other way to solve this problem, perhaps a simpler shortcut? Any help would be appreciated.

I may have made a mistake, because I find that it is generally not possible. But I'll write what I have, and then hopefully someone can correct it.

Since $sqrt p$ and $sqrt q$ are dissimilar, any number of the form $k_1 + k_2sqrt p + k_3 sqrt q + k_4sqrtpq$, where $k_iinmathbb Q$, is uniquely determined by the coefficients $k_i$. Using this we see that $c=0$ is impossible, so we can assume without loss of generality that $c=1$ (by dividing through).

Going on Berci's suggestion in the comment, we get:

$$
asqrt p + asqrt q + b
= (sqrt p -sqrt pq + q)(sqrt p + sqrt q + d)
$$
$$
= dsqrt p + (q-p)sqrt q + (1-d)sqrtpq + (p+qd)
$$

We get that $1-d=0$, so $d=1$. But then we get $q-p=a=d=1$, so we only have a solution in the special case $q=p+1$. In this case the solution is

$$
fracx+(p+q)x+1 = fracx+(2p+1)x+1
$$

So, what do you guys think? I hope I haven't simply made an arithmetic error. But since we get four equations in three variables, it seems reasonable that we don't get a solution in general.