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:

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