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&amp;=16\\
2x+3y&amp;=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?
If everybody had bicycles, there would be $16 \times 2 = 32$ wheels.  There are actually $38$, so that's $6$ additional wheels.  Each one must be on a different tricycle.
So there are $6$ tricycles, and the other $10$ students have bicycles.
Take variable $x$ for bicycles and $(16-x)$ for tricycles.

Now, consider the equation: 

$$ 2x+3(16-x)=38$$

On solving this, we get $x=10.$ So, $10$ bicycles and 6 tricycles.