Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks. $\newline$ A 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 $24$ new hardcover books and $37$ new paperback books, which cost a total of $\$303$ . He also purchased $24$ new hardcover books and $96$ new paperback books for the adult fiction section, spending a total of $\$480$ . What is the special price for each type of book? $\newline$ The special price is $\$\_$ for hardcover books and $\$\_$ for paperback books.

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Algebra 2

Solve a system of equations using elimination: word problems

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Q. Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.

\newline

A 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

24

new hardcover books and

37

new paperback books, which cost a total of

\$303

. He also purchased

24

new hardcover books and

96

new paperback books for the adult fiction section, spending a total of

\$480

. What is the special price for each type of book?

\newline

The special price is

\$\_

for hardcover books and

\$\_

for paperback books.

Define Prices: Let's denote the price of each hardcover book as $h$ and the price of each paperback book as $p$ . Mr. Hatfield purchased $24$ hardcover books and $37$ paperback books for the children's section, which cost a total of $\$303$ . This gives us the equation $24h + 37p = 303$ .
Children's Section Purchase: For the adult fiction section, he purchased $24$ hardcover books and $96$ paperback books, spending a total of $\$480$ . This gives us the equation $24h + 96p = 480$ .
Eliminate Variable: We now have a system of two equations. We need to eliminate one of the variables, $h$ or $p$ . We choose to eliminate $h$ because it has the same coefficient in both equations.
Solve for Paperback Price: To eliminate $h$ , we subtract the first equation from the second equation. This gives us $59p = 177$ .
Substitute and Solve: We now divide both sides of the equation by $59$ to solve for $p$ . This gives us $p = \frac{177}{59}$ , which simplifies to $p = 3$ .
Final Hardcover Price: We substitute $p = 3$ into the first equation and solve for $h$ . This gives us $24h + 37(3) = 303$ . Simplifying the equation, we get $24h + 111 = 303$ .
Final Hardcover Price: We substitute $p = 3$ into the first equation and solve for $h$ . This gives us $24h + 37(3) = 303$ . Simplifying the equation, we get $24h + 111 = 303$ . Subtracting $111$ from both sides of the equation, we get $24h = 303 - 111$ , which simplifies to $24h = 192$ .
Final Hardcover Price: We substitute $p = 3$ into the first equation and solve for $h$ . This gives us $24h + 37(3) = 303$ . Simplifying the equation, we get $24h + 111 = 303$ . Subtracting $111$ from both sides of the equation, we get $24h = 303 - 111$ , which simplifies to $24h = 192$ . We now divide both sides of the equation by $24$ to solve for $h$ . This gives us $h = \frac{192}{24}$ , which simplifies to $h$ $0$ .