Phone problem - 2 variables
In this lesson, we learn how to solve a word problem involving different rates for cell phone use using two variables and two linear equations. We’re given Tim’s cell phone usage and charges for May and June and we’re asked to find the cost per minute during peak hours. We let X be the cost per minute during peak hours and Y be the cost per minute during non-peak hours. By setting up two equations and using the addition method, we find that the cost per minute during peak hours is 22 cents, while the cost per minute during non-peak hours is 15 cents. We also learn a trick to simplify the problem by dividing common factors.
Shows how to solve a word problem involving different rates for cell phone use using 2 variables and 2 linear equations.
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
- How do you write and solve a system of 2 equations to solve a word problem with two variables in Algebra?
- How do you check to make sure that your solution to a word problem is correct?
- If Tim used 25 minutes on his cell phone during peak hours and 35 minutes during non-peak hours in May and was charged $10.75, and he used 30 minutes during peak hours and 45 minutes during non-peak hours in June and was charged $13.35, what is the cost per minute during peak hours?
- How do you solve the system of equations 25x + 35y = 1075 and 30x + 45y = 1335?
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Staff Review
This problem explains a somewhat complicated scenario of minutes used during peak and non-peak hours for a cell phone plan, and the total cost for two months. The task required in this problem is to not only write out two equations with two variables that explain the problem, but also to solve the equations and check your answer. This is a great problem. This is a perfect example of systems of equations in real life.
