Many SAT word problems have two unknown quantities, such as the numbers of adult and child tickets sold. Each quantity gets its own variable, and one equation ties them to a total: $12a + 8c = 480$.

Read each term as rate × quantity. In $12a$, 12 dollars for each adult ticket is multiplied by $a$, the number of adult tickets, so $12a$ is the dollars that came from adult tickets. Each variable stands for an amount of something: a number of tickets here, but in other problems a time, a distance or a weight. Its coefficient is the value for each unit of that amount (12 is the price of one adult ticket in dollars, and in a distance problem a speed is the miles for each hour). The number on the other side is the total (480 dollars in all).

If you know one quantity, substitute it and solve for the other. If 20 adult tickets were sold, $12(20) + 8c = 480$, so $240 + 8c = 480$, $8c = 240$ and $c = 30$ child tickets.

Intercepts have meanings too. Setting $c = 0$ gives $12a = 480$, so $a = 40$: if no child tickets were sold, 40 adult tickets would bring in 480 dollars. Setting $a = 0$ gives $c = 60$.

For which equation represents this situation, pair each rate with its own quantity and keep the units matched: a per-hour speed multiplies a time in hours. To check a choice, make up one case you can work out in your head, with two different, nonzero amounts, such as 10 of the first item and 5 of the second. The right equation's left side gives the total that case really produces. Choosing different amounts matters: with equal amounts, a choice that swaps the two rates gives the same total and can't be ruled out.