Is there another simpler method to solve this elementary school math problem? I am teaching an elementary student. He has a homework as follows.

There are $16$ students who use either bicycles or tricycles. The total
  number of wheels is $38$. Find the number of students using bicycles.

I have $3$ solutions as follows.
Using a single variable.
Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times $2$ and the total number of tricycles times $3$.
$$
2\times x + 3 \times (16-x) = 38
$$
The solution is $x=10$.
Using $2$ variables.
Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively.
It implies that
\begin{align}
x+y&=16\\
2x+3y&=38
\end{align}
The solution is $x=10$ and $y=6$.
Using multiples
The multiples of $2$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$
The multiples of $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$ 
The possible wheel combinations with format $(\#\text{bicycle wheels}, \#\text{tricycle wheels})$:
$\hspace{6cm} (32,6)$ but there are $18$ students
$\hspace{6cm} (26,12)$  but there are $17$ students
$\hspace{6cm} (20,18)$ there are $16$ students
$\hspace{6cm} (14, 24)$ there are $15$ students
$\hspace{6cm} (8, 30)$ there are $14$ students
$\hspace{6cm} (2,36)$ there are $13$ students
Thus the correct combination is $10$ bicycles and $6$ tricycles.
My question
Is there any other simpler method? 

Consider the total wheel function $f$ of two variables $x$ the number of bicyclists and $y$ the number of "tricyclists". Since every time we add a bicyclist we increase the number of wheels by two, we have
$$
f_x = 2.
$$
Similarly for case of tricylists,
$$
f_y = 3.
$$
So we have the solutions
$$
f(x,y) = 2x + A(y) \\
f(x,y) = 3y + B(x)
$$
which give
$$
f(x,y) = 2x + 3y + C.
$$
However, $f(0, 0) = 0$ as we have no wheels when we have no riders, so $C = 0$.
Since the total number of cyclists is $16$, we are only interest in the portion of $f$ that lies on the line $y = 16 - x$, that is $f(x, 16 - x)$. Morever, we what to find when the total number of wheels on this line is $38$, so
$$
\begin{aligned}f(x, 16 - x) & = 38 \\
2x + 3(16-x) & = 38 \\
-x + 48 & = 38 \\
-x & = -10 \\
x & = 10.
\end{aligned}
$$
Thus the solution is the point $(x,y) = (10, 6)$.
