Solving a Word Problem Using Algebra
Introduction
Algebra is a powerful tool in solving real-life problems, such as financial ones. This article will guide you through solving a word problem related to the purchase of cannolis and pies. Understanding these types of problems can be greatly beneficial for SEO, especially when you're dealing with content related to math, finance, or educational resources.
The Problem
Lakan bought a total of 40 cannolis and pies. Each cannoli costs $2.75, and each pie costs $5.20. The pies cost $9.25 more than the cannoli. How many of each did Lakan buy?
Solution 1: Substitution Method
Let's start with a simple method that involves substitution:
Total Quantities:Let c be the number of cannolis and p be the number of pies. The total number of items is:
c p 40
Cost Calculation:The cost of the cannolis is 2.75c, and the cost of the pies is 5.20p. The cost of the pies is $9.25 more than the cannolis:
5.20p 2.75c 9.25
Substitution:Since c 40 - p, we can substitute this into the second equation:
5.20p 2.75(40 - p) 9.25
Algebraic Manipulation:Solving for p gives:
5.20p 110 - 2.75p 9.25
5.20p 2.75p 119.25
7.95p 119.25
p 15
Conclude:c 40 - 15 25
Solution 2: Equation Method
In this method, we set up and solve a system of equations:
Total Quantities:c p 40
Cost Calculation:5.20p 2.75c 9.25
Substitute and Solve:We start with the substitution of c 40 - p into the second equation:
5.20p 2.75(40 - p) 9.25
Algebraic Manipulation:Multiplying and combining like terms gives:
5.20p 110 - 2.75p 9.25
5.20p 2.75p 119.25
7.95p 119.25
p 15
Conclude:c 40 - 15 25
Solution 3: Direct Equation Method
This method uses a more straightforward equation setup:
Define Variables:Let X be the number of cannolis, so the number of pies is 40 - X.
Set Up Equation:The difference in cost is given as:
(40 - X) * 5.20 - X * 2.75 9.25
Algebraic Manipulation:Expanding and solving for X gives:
208 - 5.2 - 2.75X 9.25
-7.95X -198.75
X 25
Conclusion:40 - X 15
Conclusion
Your calculations and methods will be cleaner and more efficient if you choose one approach over the others. Practice using these methods to solve similar problems for better performance in SEO and general problem-solving.
Keywords
Word problem, algebraic equation, cost calculation, price difference, Google SEO