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.
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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.