<s>[INST] What is the area of the smaller right triangle? 
The two diagonal lines are parallel, and the area of the region between them is $42$

What is $A,$ 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:
$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$
Solving the equation we get:
$3x + 4y = 72$
I don't know what's next here so I just assumed that $x = 12, y = 9$ which satisfies the equation and resulting to $A = 54$. My question is, am I missing something so that I can solve the equation? 
 [/INST] 
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
$\implies \dfrac{x}{y} = \dfrac{x + 4}{y + 3}$
$(y + 3)(x) = (x + 4)(y)$
$xy + 3x = xy  + 4y$
$3x = 4y$
$\color {red}{3x + 4y = 72},3x + (3x)  = 72$
$x = 12, y = 9$
$\boxed{Area = \dfrac{xy}{2} = \dfrac{12 \cdot 9}{2} = \dfrac{108}{2} = 54}$
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