System of equations modulo primes Is it generally possible to solve a set of linear equations modulo some prime numbers $\{p,q,r\}$. For example if I have the following congruences:

$$ 
xa_p + yb_p \equiv d \pmod {p}\\
xa_q + yb_q \equiv d \pmod {q}\\
xa_r + yb_r \equiv d \pmod {r}\\
$$

for some known $a_i$ and $b_i$ with $i \in \{p,q,r\}$ can I determine $d$ (assuming a solution exists and $d&lt;p,q,r$)?

Thanks in advance! (and sorry if this is trivial.)

EDIT: Additionally, $a_i$ and $b_i$ are congruences of some unknown (and large) integers $a,b$. (i.e. $a\equiv a_i \pmod{i}$ and $b\equiv b_i \pmod{i}$ )

Any pointers are be welcome! Also knowing that there is no "good" algorithm for this problem would also help me a lot.
Since the problem is finite, a brute force approach will either find a solution or prove none exist. In more detail, if $x,y$ is a solution, then $x+kpqr,y+spqr$ is a solution for all $k,s\in \mathbb Z$ (since $pqr=0$ modulo each of $p,q,r$). It thus follows (by using the division algorithm) that if a solution exists then a solution exists with $x,y\in \{0,1,\cdots ,pqr-1\}$. So, the brute force approach will be to iterate over all $0\le x,y&lt; pqr$ and check for solutions. If one is found then you found one, otherwise none exist.
x+y ≡ d (mod p)

d ≡ (x (mod p)) + (y (mod p)) (mod p)

Example:

a ≡ 1 mod 3,
b ≡ 2 mod 3,
a + b ≡ 3 mod 3 ≡ 0,

Also:

xa ≡ d (mod p),
d = c (mod p),

c = a (mod p) iff (a (mod p) > x (mod p) or a (mod p) = x (mod p) OR a (mod p) is = 0)
c = x (mod p) iff (x (mod p) > a (mod p) or x (mod p) = a (mod p) OR a (mod p) is = 0)