Find the smallest positive integer for which x mod 3 = 2 and x mod 4 = 3 I came across a question on a piece of homework and my solution yields no possible solutions.

Find the smallest positive integer for which
x mod 3=2 and x mod 4=3 (and then find the next one)

So I figure (by the Quotient-Remainder Theorem)

n mod 3 = 2 => n = 3q + 2
n mod 4 = 3 => n = 4q + 3

No problems thus far and everything works out fine logically so I set the equations equal to each other and find that q = -1.
Wait, it can't be negative. So I put it on a graph to check my comprehension and like I thought that is the only point at which they meet is outside my possible domain.
Could someone point out the error in my logic here? Or perhaps I've misinterpreted the question? Apologies if the answer is simple and I've missed it.
 A: If $x\pmod4\equiv3$, then $x=4k+3$ for some integer $k$.  Considering this equation modulo $3$, we get
$$x\equiv k\equiv2\pmod3$$
So for some integer $a,k=3a+2$.  Combining these, we get
$$x=4k+3=4(3a+2)+3=12a+11$$
A: Let me illustrate it in this way:

A number is a valid solution to your problem when there is both blue and red square next to it. However, your equations impose an additional condition: that the white numbers have to be the same. As you can observe, there is a solution where the white numbers happen to be equal (the one you have found), but it is not the only one.
In fact, your method might fail to find a solution, consider $x \bmod 3 = 2$ and $x \bmod 5 = 3$. One of the solutions could be $x = 8$, but with your method we get $q=-\frac{1}{2}$, so $x = \frac{1}{2}$. It's easy to see why it does not work on the diagram below – the lower white $(-1)$ (red background) is left of the higher $(-1)$ (blue background), while for $0$ it is the other way around, so the place where the white numbers would meet is actually "in-between", namely $q = -\frac{1}{2}$.

I hope this helps $\ddot\smile$
A: You have $x = 3p+2 = 4q+3$ with $p, q \in \mathbb{N}$. Thus $3p-4q = 1\implies p = \dfrac{4q+1}{3}=q+\dfrac{q+1}{3}$. Thus $q+1 = 3 \implies q = 2\implies x = 11$ is the smallest natural number you sought.
A: well, guy, you can solve your problem using modular arithmetic and I think this is the best way for this problem.
pay attention:
x≡2 in base 3
In base 4    x≡3
We can say:
 x≡-1  in base 3  and   x≡ -1 in base 4 
According to modular arithmetic rules we have:
x≡ -1 in base 12
As a result, x can be written in form of (x =  12k – 1)
If k = 1, then x is 11 which is the smallest integer with the properties mentioned in the problem.
By giving positive integers to k, we can create more numbers:
(k = 1) x = 11, (k = 2) x = 23, (k = 3) x = 35 and …
