Shane and Kate are preparing to go on long car trips with their families. Shane bought 8 books and 2 CDs for $73.60. Kate bought 3 books and 4 CDs for $69.85. Both totals were withour tax. All of the books were the same price and all of the CDs cost the same. How much did a book cost?
8B+2C=73.60
3B+4C=69.95
I would multiply the first equation by 2, then subtract equation 2 from the new 1.
4.50
To find out how much a book costs, we'll need to use algebra to solve the system of equations created by the given information.
Let's define the cost of a book as 'b' and the cost of a CD as 'c'.
From the problem, we can write two equations:
1) 8b + 2c = 73.60 (equation 1)
2) 3b + 4c = 69.85 (equation 2)
To solve these equations, we can use either substitution or elimination method. Let's use the elimination method to eliminate the 'c' variable.
Multiply equation 1 by 2, so that the coefficient of 'c' in both equations will be the same:
2(8b + 2c) = 2(73.60)
16b + 4c = 147.20 (equation 3)
Now, subtract equation 2 from equation 3:
(16b + 4c) - (3b + 4c) = 147.20 - 69.85
16b + 4c - 3b - 4c = 77.35
Simplifying the equation:
13b = 77.35
b = 77.35 / 13
b = 5.95
Therefore, the cost of a book is $5.95.