by Mackenzie
Andrew had $93.00. With this money, he could buy 3 CDs and 4 books. However, he bought only 2 CDs and 3 books and had $27.00 left. What was the cost of each CD?
Solution
This problem will lead to a system of linear equations
Let x be the cost of 1 cd
Let y be the cost of 1 book
Cost of 3 CDs + cost of 4 books must equal to 93 dollars.
3 × x + 4 × y = 93
3x + 4y = 93
When he bought 2 CDs and 3 books he had 27 dollars left.
The statement above means that
Cost of 2 CDs + cost of 3 books = 93 - 27
Cost of 2 CDs + cost of 3 books = 66
2x + 3y = 66
You now have a system of equations to solve
3x + 4y = 93 equation 1
2x + 3y = 66 equation 2
Multiply equation 1 by -2 and equation 2 by 3
We get
-6x + -8y = -186 equation 3
6x + 9y = 198 equation 4
Add the left sides and right sides of equation 3 and equation 4.
-6x + 6x + -8y + 9y = -186 + 198
0 + -8y + 9y = -186 + 198
-8y + 9y = -186 + 198
y = -186 + 198
y = 12
Use equation 2 to find y
2x + 3y = 66
2x + 3 × 12 = 66
2x + 36 = 66
2x + 36 - 36 = 66 - 36
2x = 66 - 36
2x = 30
Divide both sides by 2
2x / 2 = 30 / 2
x = 15
Jul 06, 18 12:29 PM
Learn how to solve two types of logarithmic equations
New math lessons
Your email is safe with us. We will only use it to inform you about new math lessons.
Jul 06, 18 12:29 PM
Learn how to solve two types of logarithmic equations
Our Top Pages
Formula for percentage
Compatible numbers
Basic math test
Basic math formulas
Types of angles
Math problem solver
Algebra word problems
Surface area of a cube
Finding the average
Scale drawings
Everything you need to prepare for an important exam!
K-12 tests, GED math test, basic math tests, geometry tests, algebra tests.
Real Life Math Skills
Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball.