How can we solve an equation with two variables, when only given one equation?

$7w+3d=44 \implies d = \dfrac{44-7w}{3}=\dfrac{42-6w}{3}+\dfrac{2-w}{3} =14-2w+\dfrac{2-w}{3}$

The numbers that give integer solutions are of the form $w=3x+2$.

Letting $w=3x+2$, we get $d = 10-7x$

$x < 0$ makes $w$ negative. $x>1$ makes $d$ negative.

$x=0$ gives $(w,d)=(2,10)$ and $x=1$ gives $(w,d)=(5, 3)$.

Response to comment.

I chose $d = \dfrac{44-7w}{3}$ because you only have to try $w=0,1,2$ to find a value of $d$ that makes $w$ an integer. If you felt like writing

$$w=\dfrac{44-3d}{7}= 6 - \dfrac{3d-2}{7}$$ then $d=3$ gives us $w=5$.

For small numbers, this works just fine. If you have to, the best way is to use modular arithmetic.

For example, to search for an initial value of $d$,

\begin{align} 7w+3d=44 &\implies 3d \equiv 44 \pmod 7 \\ &\implies 3d \equiv 2 \pmod 7 \\ &\implies 5 \cdot 3d \equiv 5 \cdot 2 \pmod 7 \\ & \implies d \equiv 3 \pmod 7 \end{align}

You can then use $d=3+7t$ and you will get all integer values for $w$ and $d$.

If $7w+3d=44$, then $7w+3d-44$ is divisible by $3$,

so $7w+3d-44-3(2w+d-14)=w-2 $ is,

so $w\in\{2,5,8,11,14,17,...\}$,

but in that set only $2$ and $5$ can be multiplied by $7$ to get a result not more than $44$.