April buys eight books for $44. Paper back books cost $4 and hard back books cost $8. How many of

each book did she buy?

Hmm -- that's not right.

4 * 4 = 16
4 * 8 = 32

16 + 32 = 48.

Ooops!

For this kind of problem, I usually start off with all paperbacks, thus 8*4=$32.

Since I am short 44-32=$12, and changing from a paperback to a hard-cover costs 8-4=$4, So I must have bought 3 hard-covers and 8-3=5 paperbacks.
Check: 3*8+5*4=24+20=$44.

And then there's the LONG way of doing it.

H = number of hard covers.
P = number of paper covers.
===========================
H + P = 8
8H + 4P = 44.
================
Two equations in two unknowns. Solve for H and P.

To determine how many of each book April bought, we can set up a system of equations based on the given information.

Let's assume that April bought x paperback books and y hardback books.

According to the information given, the total number of books April bought is 8. Therefore, we can write the first equation as:

x + y = 8

The total cost of the book purchase is $44. Since paperback books cost $4 and hardback books cost $8, we can write the second equation as:

4x + 8y = 44

Now, we have a system of two equations with two variables. We can solve this system to find the values of x and y.

To solve this system of equations, we can use substitution or elimination method. Let's use the latter:

Multiply the first equation by 4 to make the coefficients of x in both equations equal:

4(x + y) = 4(8)
4x + 4y = 32

Now, we can subtract this equation from the second equation to eliminate x:

4x + 8y - (4x + 4y) = 44 - 32
4y = 12
y = 12/4
y = 3

Substituting the value of y back into the first equation:

x + 3 = 8
x = 8 - 3
x = 5

Therefore, April bought 5 paperback books and 3 hardback books.

4 paper backs and 4 hard cover