Solve a system of equations using elimination: word problems

Question

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. Pruitt purchased 22 new hardcover books and 29 new paperback books, which cost a total of $402. He also purchased 57 new hardcover books and 67 new paperback books for the adult fiction section, spending a total of $1,009. What is the special price for each type of book?

Answer 1

Let H = price of hardcover books and P = price for paperbacks

 

Children's: 22H + 29P = 402

 

Adult section: 57H + 67P = 1009

 

Elimination: It looks as if eliminating the H will be easiest...but still a nuisance.  There are no common factors to 22 and 57.

 

Multiply the children's equation by 57 and the adult equation by 22 to get

 

1254H + 1653P = 22914

1254H + 1474P = 22198

 

Subtract the bottom equation from the top to get 

179P = 716 to get P = 4

 

I will leave the rest to you: plug P = 4 into either of the original equations to get H

 

There are NO easy ways to solve this system of equations...the numbers require onerous arithmetic no matter what way you elect.  As I think about it, probably this elimination method is actually the least troublesome.