What is the area of the smaller right triangle?

So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:

$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$

Continuing your equation, we get:

$\dfrac{xy + 3x + 4y + 12 - xy}{2} = 42$

$xy + 3x + 4y + 12 - xy = 2 \cdot 42$

$3x + 4y + 12 = 84$

$\boxed{3x + 4y = 72}$

My question is, am I missing something so that I can solve the equation?

The lines are parallel, thus their slopes will be equal

$\implies \dfrac{x}{y} = \dfrac{x + 4}{y + 3}$

$(y + 3)(x) = (x + 4)(y)$

$xy + 3x = xy + 4y$

$3x = 4y$

$\color {red}{3x + 4y = 72},3x + (3x) = 72$

$x = 12, y = 9$

$\boxed{Area = \dfrac{xy}{2} = \dfrac{12 \cdot 9}{2} = \dfrac{108}{2} = 54}$

As shown above, consider $\triangle ABC$ and $\triangle ADE$. You have the common right angle of $\angle BAC$. Also, $\angle CBA = \angle EDA$ and $\angle BCA = \angle DEA$ because of the $2$ parallel lines on either side of the blue section, so the corresponding angles of intersection with the transversal are congruent, such as described in item #$3$ in https://en.wikipedia.org/wiki/Parallel_(geometry)#Euclidean_parallelism. Thus, you have that $\triangle ABC$ and $\triangle ADE$ are similar to each other. This means

$$\begin{equation}\begin{aligned} \frac{|AE|}{|AD|} & = \frac{|AC|}{|AB|} \\ \frac{x}{y} & = \frac{x + 4}{y + 3} \\ x(y + 3) & = y(x + 4) \\ xy + 3x & = xy + 4y \\ 3x & = 4y \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Using this in your determined equation gives

$$4y + 4y = 8y = 72 \implies y = 9 \tag{2}\label{eq2A}$$

This also gives from \eqref{eq1A} that

$$3x = 4(9) \implies x = 12 \tag{3}\label{eq3A}$$

These results confirm your guess was correct for the side lengths, so your answer of $A = \frac{1}{2}(9)(12) = 54$ is also correct.