Question

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.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. Whitaker purchased 78 new hardcover books and 46 new paperback books, which cost a total of $840. He also purchased 78 new hardcover books and 14 new paperback books for the adult fiction section, spending a total of $744. What is the special price for each type of book?

Answer 1

Ans

Step-by-step explanation:

Let's call x the price for the hardcover books and y the price for the paperback book.

If he purchased 78 hardcover books and 46 new paperback books for $840, we can write the following equation

78x + 46y = 840

In the same way, if he purchased 78 hardcover books and 14 paperback books for $744, we get:

78x + 14y = 744

So, the system of equations is

78x + 46y = 840

78x + 14y = 744

To solve the system, we need to multiply the second equation by -1

-78x - 14y = -744

And then add this to the first equation

78x + 46y = 840

-78x - 14y = -744

32y = 96

Solving for y, we get:

32y/32 = 96/32

y = 3

Finally, we can replace y by 3 and solve for x

78x + 46y = 840

78x + 46(3) = 840

78x + 138 = 840

78x + 138 - 138 = 840 - 138

78x = 702

78x/78 = 702/78

x = 9

Therefore, the price for the hardcover books is $9 and the price for the paperback books is $3