math mixture

[151AND8TH]

I know its a simple problem and I know the answer is
4 paperback and 2 hardcover books... My problem is
figuring out the problem with an algebraic equation...


at a book sale, the books are prices at $.60 for the paperback and $1.00 for hardcover. John spent $4.40. How many of each kind of book did he buy?

I thought ; p(.6) + h(1)=4.40 i tried a couple more ways and couldnt come out with an equation

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[domonas]

I know its a simple problem and I know the answer is
4 paperback and 2 hardcover books... My problem is
figuring out the problem with an algebraic equation...


at a book sale, the books are prices at $.60 for the paperback and $1.00 for hardcover. John spent $4.40. How many of each kind of book did he buy?

I thought ; p(.6) + h(1)=4.40 i tried a couple more ways and couldnt come out with an equation

I wouldn't even use an equation. since the total is $4.40, how many 60 cent books can John buy to equal something with 40 cents? 4 (4x60 cents = $2.40), which means he bought 2 hardcover books.

 

[doc toothache]

I know its a simple problem and I know the answer is
4 paperback and 2 hardcover books... My problem is
figuring out the problem with an algebraic equation...


at a book sale, the books are prices at $.60 for the paperback and $1.00 for hardcover. John spent $4.40. How many of each kind of book did he buy?

I thought ; p(.6) + h(1)=4.40 i tried a couple more ways and couldnt come out with an equation

With only the information given you cannot solve an algebraic equation with two unknowns. Additional information, such as the total number of books bought would be needed. In the absence, your only choice is to solve the problem as suggested by domonas by calculating for 1-6 hardcover books and dividing the remaining amount by .6 until you get a whole number.

 

[151AND8TH]

Thanks guys