Modulo polynomials, excluding selected solution combinations Given
$$
(x-a)(x-b)(x-c) \equiv 0\ \ \pmod {p}
$$
$$
(x-d)(x-e)(x-f)\ \equiv 0\ \ \pmod {q}
$$
x: unknown variable. p,q : known primes. a,b,c,d,e,f : known values.
Are there one or more modulo equations that will exclude the combination 
$$
(x \equiv a \pmod {p})\ AND\ (x \equiv d \pmod {q})
$$
but still allow all other combinations of x solutions?
regards arthur

Edit
I tried $\left(mod\ pq\right)$ but couldn't get it to work. List all legal combinations and exclude the illegal one $\left(mod\ pq\right)$.
Let $$r_{jk} \equiv j \left(mod\ p\right)\  and\ \  r_{jk} \equiv k \left(mod\ q\right)$$
$$(x-r_{ae})(x-r_{af})(x-r_{bd})(x-r_{be})(x-r_{bf})(x-r_{cd})(x-r_{ce})(x-r_{cf}) \equiv 0 \left(mod\  pq\right)$$
But if $x = r_{ad}$ (illegal) then $x = a +k_1p$ and $x = d + k_2q$ then 
$$(x-r_{ae})\dots(x-r_{bd})\dots \equiv (a+k_1p - (a + k_3p))\dots(d+k_2q-(d+k_4q))\dots\left(mod\  pq\right)$$
$$\equiv p(k_1-k_3)\dots q(k_2-k_4)\dots \equiv 0\left(mod\  pq\right)$$
The $x\equiv r_{ad}$ (illegal) produces a $p$ and a $q$ solving the equation $\left(mod\  pq\right)$.

Edit 2
My application will work with the simplified problem:
Let
$$(x-a)(x-b) \equiv 0\ \ \pmod {p}$$
$$(x-a)(x-b) \equiv 0\ \ \pmod {q}$$
Find equations that will allow
$$x \equiv a \ \pmod {p}\ \  and \ \ x \equiv a \ \pmod {q}$$
$$or$$
$$x \equiv b \ \pmod {p}\ \ and\ \  x \equiv b \ \pmod {q}$$
but block
 $$x \equiv a \ \pmod {p}\ \  and \ \ x \equiv b \ \pmod {q}$$
$$or$$
$$x \equiv b \ \pmod {p}\ \ and\ \  x \equiv a \ \pmod {q}$$
where $p$ and $q$ are distinct.
 A: Let 
$\left(\frac{a}{p}\right)=\left(\frac{a}{q}\right)=1$ ,
$\left(\frac{b}{p}\right)=\left(\frac{b}{q}\right)=-1$
$$x^{\frac{p-1}{2}} \equiv \pm1\ (mod\ p)\ \ , \ x^{\frac{q-1}{2}} \equiv \pm1\ (mod\ q)$$
$$x^{\frac{p-1}{2}} = \pm1\ + k_1 p\ \ ,\ x^{\frac{q-1}{2}} = \pm1\ + k_2 q$$
$$qx^{\frac{p-1}{2}} = \pm q\ + k_1 pq\ \ , \ px^{\frac{q-1}{2}} = \pm p\ + k_2p q$$
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} = \pm q\  \pm p\ + (k_1 + k_2)pq$$
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} = \pm q\  \pm p\ (mod\ pq)$$
To select $\left(\frac{x}{p}\right)\left(\frac{x}{q}\right)=1$ i.e. block 
$\left(\frac{a}{p}\right)\left(\frac{b}{q}\right)=\left(\frac{b}{p}\right)\left(\frac{a}{q}\right)=-1$
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} \equiv \pm (q+p)\ (mod\ pq)$$
$$qx^{\frac{p-1}{2}} - px^{\frac{q-1}{2}} \equiv \pm (q-p)\ (mod\ pq)$$
if $d$ can be found where $d|\phi(pq)$ s.t. $a^d\equiv 1\ (mod\ pq)$ and $b^d\equiv -1\ (mod\ pq)$ then the $\pm$ sign can be removed s.t.
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} \equiv x^d (q+p)\ (mod\ pq)$$
I still need to determine if $d$ always exists.
Admittedly extra conditions were added. A more general solution would be welcome.

$$\left(x^{\frac{p-1}{2}}\right)^{odd} \equiv \pm1\ (mod\ p)\ \ , \ \left(x^{\frac{q-1}{2}}\right)^{odd} \equiv \pm1\ (mod\ q)$$
Multiply each side by itself an odd number of times preserves the sign of the $\pm1$. if $p \equiv q \equiv 3\ mod\ 4$ then $\frac{p-1}{2}$ and $\frac{q-1}{2}$ are odd.
$$x^{\frac{p-1}{2}\frac{q-1}{2}} \equiv \pm1\ (mod\ p)\ \ , \ x^{\frac{p-1}{2}\frac{q-1}{2}} \equiv \pm1\ (mod\ q)$$
$$x^{\frac{p-1}{2}\frac{q-1}{2}} \equiv \pm1\ (mod\ pq)$$
So if $p \equiv q \equiv 3\ mod\ 4$ then $d = \frac{\phi(pq)}{4}$.
A: If $f(a)≡0$ (mod$p$) and $f(a)≡0$ (mod$q$) then $f(a)≡0$ (mod $pq$).
Also if $f(a) \equiv 0$ (mod $pq$) implies $f(a) \equiv 0$ (mod $p$) and $f(a) \equiv 0$ (mod $q$)
