Be ready to die because of your bragging on your latex

Let's approach this problem by turning the last two congruences into equations.

$$x=9+20a$$
$$x=4+45b$$
Substituting:
$$9+20a=4+45b$$
We can start applying the Chinese Remainder theorem by finding $b$ first.
$$20a=45b-5$$
We can take $\pmod{20}$ to both sides of this equation
$$0=(45b-5)\pmod{20}$$
$$5\pmod{20}=5b$$
$$b=1\pmod{20}$$

We can now plug in $b$ into the second equation

$$x=4+45(1\pmod{20})$$
And since we want this to remain as an equation, we can just set another variable
$$x=4+45(1+20c)$$
$$x=4+45+900c$$
$$x=49+900c$$
Let's take $\pmod6$ on both sides now:
$$x\pmod6=1$$


Now let's solve for $a$
$$9+20a=4+45b$$
$$45b=20a+5$$
Let's take $\pmod{45}$ of both sides
$$0\pmod{45}\equiv (20a+5)\pmod{45}$$
$$-5\pmod{45}\equiv 20a\pmod{45}$$
$$40\pmod{40}\equiv 20a\pmod{45}$$
$$a\equiv 8\pmod{45}$$
Plugging it back into the original equation:
$$x=20(45c+8)+9$$
$$x=900c+160+9$$
$$x=900c+169$$
We can take $\pmod6$ of both sides to eliminate $c$
$$x\equiv 1\pmod6$$

Since both of the ending results are $x\equiv 1\pmod6$, the only solution is $\boxed{\bold{1}}$