1. **State the problem:** Mr. Roth buys hardcover and paperback books at special prices. For the children's section, he buys 17 hardcover and 21 paperback books costing a total of 288. For the adult fiction section, he buys 84 hardcover and 74 paperback books costing a total of 1304. We need to find the special price for each type of book. 2. **Define variables:** Let $h$ be the price of one hardcover book. Let $p$ be the price of one paperback book. 3. **Write the system of equations:** $$ \begin{cases} 17h + 21p = 288 \\ 84h + 74p = 1304 \end{cases} $$ 4. **Use elimination to solve:** Multiply the first equation by 4 to align the $h$ terms: $$ 4(17h + 21p) = 4(288) \implies 68h + 84p = 1152 $$ Now subtract this from the second equation: $$ (84h + 74p) - (68h + 84p) = 1304 - 1152 $$ $$ (84h - 68h) + (74p - 84p) = 152 $$ $$ 16h - 10p = 152 $$ 5. **Express $h$ in terms of $p$:** $$ 16h = 152 + 10p $$ $$ \Rightarrow h = \frac{152 + 10p}{16} $$ 6. **Substitute $h$ back into the first original equation:** $$ 17\left(\frac{152 + 10p}{16}\right) + 21p = 288 $$ Multiply both sides by 16 to clear the denominator: $$ 17(152 + 10p) + 21 \times 16 p = 288 \times 16 $$ $$ 2584 + 170p + 336p = 4608 $$ $$ 2584 + 506p = 4608 $$ 7. **Solve for $p$:** $$ 506p = 4608 - 2584 = 2024 $$ $$ p = \frac{2024}{506} = 4 $$ 8. **Find $h$ using $p=4$:** $$ h = \frac{152 + 10 \times 4}{16} = \frac{152 + 40}{16} = \frac{192}{16} = 12 $$ **Final answer:** The special price is $12 for hardcover books and $4 for paperback books.