Question: I am having a hard time factoring this equation to find the equilibrium quantity and price.

> 
> $11p+3x-66 =0$ and $2p^2+p-x=10$
> 
> 
> 

I have gotten this far.

$11p+3x-66=2p^2+p-x-10$

$11p+3x-66-2p^2-p+x+10=0$

$-2p^2-10p-56+4x=0$

$-2p^2-10p-56=-4x$

Then I get stuck here. I don't know if I'm doing this right.

Answer: 
Clearly, $$12^{39}+14^{39}$$ is divisible by $4$

Again, $$12^{39}=(13-1)^{39}\equiv-1+\binom{39}113\pmod{169}\equiv-1$$

$$14^{39}=(13+1)^{39}\equiv1+\binom{39}113\pmod{169}\equiv1$$ 

More generally if $\displaystyle a\equiv b\pmod p, a^p\equiv b^p\pmod{p^2}$

Here $\displaystyle p=13,12\equiv-1\pmod{13}\implies12^3\equiv(-1)^3\equiv-1$

Similarly, $\displaystyle14\equiv1\pmod{13}\implies14^3\equiv(1)^3$
$$\left((39-12)^{39}+(39-14)^{39}\right) \div 26 \times 26$$
so $$\left(27^{39}+25^{39}\right) \div 26$$
so $$1^{39}+(-1)^{39}$$