April buys 8 books for $44.1 paper back book costs$4 and 1 hard back book costs $8. how many of each did she buy? (use an equation to solve)
p ---> paperback books
h ---> hardcover books
use these equations to solve
p + h = 8
4p + 8h = 44
h = # of hardback books
p = # of paperback books
h + p = 8 = h = 8 - p
4p + 8h = $44
-4p - 4h = -32
4p + 8h = 44
4h = 12
- -
4 4
h = 3
Answer: 5 paperback books and 3 hardback books
To solve this problem, we will use a system of equations.
Let p represent the number of paperback books April bought, and h represent the number of hardback books she bought.
First, we can write the equation based on the given information:
p + h = 8
This equation states that the total number of books (paperback + hardback) April bought is equal to 8.
Next, we can write another equation based on the cost of the books:
4p + 8h = 44
This equation represents the total cost of the books. Since each paperback book costs $4 and each hardback book costs $8, we multiply the number of books by their respective prices and add them together to get the total cost.
Now, we have a system of equations:
p + h = 8
4p + 8h = 44
To solve this system, we can use the substitution method or elimination method. Let's use the elimination method:
Multiply the first equation by 4 to eliminate p by changing the equation's form:
4p + 4h = 32
Now, subtract this equation from the second equation:
4p + 8h - (4p + 4h) = 44 - 32
4p + 8h - 4p - 4h = 12
4h = 12
h = 12/4
h = 3
Now substitute the value of h into the first equation:
p + 3 = 8
p = 8 - 3
p = 5
Therefore, April bought 5 paperback books and 3 hardback books.