A bag contains 8 red marbles, 9 yellow marbles, and 7 green marbles. How many additional red marbles must be added to the 24 marbles already in the bag so that the probability of randomly drawing a red marble is $\dfrac{3}{5}$ ?

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One option is to simply try out the answer choices. For example, using choice B,

$$P(\text{red}) = \frac{\text{red}}{\text{total}}$$ $$P(\text{red}) = \frac{16+8}{16+24}$$ $$= \frac{24}{40} = \frac{3}{5} \checkmark$$

We can write an equation, using $x$ for the unknown amount of red marbles to be added:

$$P(\text{red}) = \frac{\text{red}}{\text{total}}$$ $$\frac{3}{5} = \frac{8+x}{24+x}$$ $$3(24+x)=5(8+x)$$ $$72+3x=40+5x$$ $$32=2x$$ $$x=\boxed{16}$$