A librarian is expanding some sections of the city library. He buys book at a special price from a dealer who charges one price for any hardback book and another price for any paperback books. For the children's section, Mr. Kent purchased 12 new hardcover books and 73 new paperback books, which cost a total of $594. He also purchased 11 new hardcover books and 87 new paperback books for the adult fiction section, spending a total of $665. What is the special price for each type of book?

The special price of 1 hardcover book = $13

The special price of 1 paperback book = $6

Step-by-step explanation:

Here, let us assume the special price of 1 hardcover book = $ m

and the the special price of 1 paperback book = $ n

Now, the price of 12 hardcover books = 12 x (Price of 1 hardcover) = 12 m

and the price of 73 paperback books = 73 x (Price of 1 paperback) = 73 n

Also, The price paid for 12 hardcover and 73 paperback books is $594

⇒ 12 m + 73 n = 594 ... (1)

Similarly:

The price of 11 hardcover books = 11 x (Price of 1 hardcover) = 11 m

The price of 87 paperback books = 87 x (Price of 1 paperback) = 87 n

Also, The price paid for 11 hardcover and 87 paperback books is $665.

⇒ 11 m + 87 n = 665 ... (2)

Now, solving both equation for the value of m and n, we get:

12 m + 73 n = 594 x 11

11 m + 87 n = 665 x (-12)

we get: 132 m + 803 n = 6534

- 132 m - 1044 n = - 7980

adding both equation, we get:

132 m + 803 n - 132 m - 1044 n = - 7980 + 6534

or, - 241 n = - 1446

or, n = 1446/241 = 6, or n = 6

Now, Substituting n = 6 in 12 m + 73 n = 594,

we get: 12 m + 73 (6) = 594, m = 13

Hence, The special price of 1 hardcover book = $13

and The special price of 1 paperback book = $6