Nine books are to be bought by a student. Some cost $6.00 each and the remainder cost $6.50 each. If the total amount spent was $56, how many of each book are bought?
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To solve the given problem, let's assign variables to the unknowns. Let's say the number of books purchased at $6.00 each is "x" and the number of books purchased at $6.50 each is "y."
According to the problem, the total number of books purchased is nine. So, we have the equation:
x + y = 9 ----(1)
Now, let's consider the total amount spent. The number of books purchased at $6.00 each costs $6.00 multiplied by the quantity x. The number of books purchased at $6.50 each costs $6.50 multiplied by the quantity y. So, we have the equation:
6.00x + 6.50y = 56 ----(2)
We now have a system of equations consisting of equations (1) and (2). We can solve this system of equations to find the values of x and y.
There are different methods to solve the system of equations, such as substitution or elimination. Here, we will use the elimination method.
Multiplying equation (1) by 6.50 and equation (2) by -1.00 yields:
6.50x + 6.50y = 58.50 ----(3)
-6.00x - 6.50y = -56.00 ----(4)
Adding equations (3) and (4) eliminates the variable "y":
(6.50x + (-6.00x)) + (6.50y + (-6.50y)) = (58.50 + (-56.00))
Simplifying, we get:
0.50x = 2.50
Dividing both sides of the equation by 0.50, we find:
x = 5
Now, substituting this value of x into equation (1), we can solve for y:
5 + y = 9
y = 9 - 5
y = 4
Therefore, the student bought 5 books priced at $6.00 each and 4 books priced at $6.50 each.