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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineA librarian is expanding some sections of the city library. He buys books at a special price from a dealer who charges one price for any hardback book and another price for any paperback book. For the children's section, Mr. Hatfield purchased 2424 new hardcover books and 3737 new paperback books, which cost a total of $303\$303. He also purchased 2424 new hardcover books and 9696 new paperback books for the adult fiction section, spending a total of $480\$480. What is the special price for each type of book?\newlineThe special price is $_\$\_ for hardcover books and $_\$\_ for paperback books.

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.\newlineA librarian is expanding some sections of the city library. He buys books at a special price from a dealer who charges one price for any hardback book and another price for any paperback book. For the children's section, Mr. Hatfield purchased 2424 new hardcover books and 3737 new paperback books, which cost a total of $303\$303. He also purchased 2424 new hardcover books and 9696 new paperback books for the adult fiction section, spending a total of $480\$480. What is the special price for each type of book?\newlineThe special price is $_\$\_ for hardcover books and $_\$\_ for paperback books.
  1. Define Prices: Let's denote the price of each hardcover book as hh and the price of each paperback book as pp. Mr. Hatfield purchased 2424 hardcover books and 3737 paperback books for the children's section, which cost a total of $303\$303. This gives us the equation 24h+37p=30324h + 37p = 303.
  2. Children's Section Purchase: For the adult fiction section, he purchased 2424 hardcover books and 9696 paperback books, spending a total of $480\$480. This gives us the equation 24h+96p=48024h + 96p = 480.
  3. Eliminate Variable: We now have a system of two equations. We need to eliminate one of the variables, hh or pp. We choose to eliminate hh because it has the same coefficient in both equations.
  4. Solve for Paperback Price: To eliminate hh, we subtract the first equation from the second equation. This gives us 59p=17759p = 177.
  5. Substitute and Solve: We now divide both sides of the equation by 5959 to solve for pp. This gives us p=17759p = \frac{177}{59}, which simplifies to p=3p = 3.
  6. Final Hardcover Price: We substitute p=3p = 3 into the first equation and solve for hh. This gives us 24h+37(3)=30324h + 37(3) = 303. Simplifying the equation, we get 24h+111=30324h + 111 = 303.
  7. Final Hardcover Price: We substitute p=3p = 3 into the first equation and solve for hh. This gives us 24h+37(3)=30324h + 37(3) = 303. Simplifying the equation, we get 24h+111=30324h + 111 = 303. Subtracting 111111 from both sides of the equation, we get 24h=303−11124h = 303 - 111, which simplifies to 24h=19224h = 192.
  8. Final Hardcover Price: We substitute p=3p = 3 into the first equation and solve for hh. This gives us 24h+37(3)=30324h + 37(3) = 303. Simplifying the equation, we get 24h+111=30324h + 111 = 303. Subtracting 111111 from both sides of the equation, we get 24h=303−11124h = 303 - 111, which simplifies to 24h=19224h = 192. We now divide both sides of the equation by 2424 to solve for hh. This gives us h=19224h = \frac{192}{24}, which simplifies to hh00.

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