Problem 203 - Elementary Algebra

Question

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Step-by-Step Solution

Answer

Shelly burns 10 calories per minute jogging and 10 calories per minute cycling.

Step 1: Define Variables

Let the number of calories burned per minute of jogging be denoted as \(J\) and the number of calories burned per minute of cycling be denoted as \(C\).

Step 2: Formulate Equations

Based on the given information, formulate two equations:1. For the first day: \(10J + 20C = 300\)2. For the second day: \(20J + 10C = 300\)

Step 3: Solve the System of Equations

First, solve one of the equations for one variable. For example, solve the first equation for \(J\):\(J = \frac{300 - 20C}{10} = 30 - 2C\)Next, substitute this expression for \(J\) into the second equation:\(20(30 - 2C) + 10C = 300\)This simplifies to:\(600 - 40C + 10C = 300\)\(600 - 30C = 300\)\(-30C = -300\)\(C = 10\)

Step 4: Find the Value of J

Substitute \(C = 10\) back into the equation \(J = 30 - 2C\):\(J = 30 - 2(10)\)This simplifies to:\(J = 30 - 20\)\(J = 10\)

Step 5: Verify the Solutions

Verify that \(J = 10\) and \(C = 10\) satisfy both original equations:For the first equation:\(10(10) + 20(10) = 100 + 200 = 300\)For the second equation:\(20(10) + 10(10) = 200 + 100 = 300\)Both equations are satisfied, confirming the solution is correct.

Key Concepts

Linear Equations

A system of linear equations consists of two or more equations with the same set of variables. In our exercise with Shelly, we are dealing with two linear equations. They represent the total calories burned based on the minutes spent jogging and cycling. Linear equations have a constant rate of change and do not involve any variables being multiplied together or raised to a power higher than one. This simplifies calculations and makes problem-solving straightforward. The key is to express the problem in algebraic terms. Once that's done, solving becomes a matter of simple arithmetic and logic.

Variables

Variables are an essential concept in algebra. They represent unknowns or quantities that can change. In the given exercise, we use variables to denote the calories burnt per minute while jogging (\(J\)) and cycling (\(C\)). Here are key points to remember:

• Define variables clearly at the beginning. This avoids confusion later.
• Use consistent symbols to represent each variable. This maintains clarity throughout your solution.

For Shelly's problem, we defined \(J\) as the calories burned per minute of jogging, and \(C\) as the calories burned per minute of cycling. By doing so, we made it easier to set up and solve our system of equations.

Problem-Solving

When tackling a problem involving systems of equations, follow a structured approach:

• Define the problem and your variables clearly.
• Formulate equations based on the given information.
• Use suitable methods to solve the equations:
• Substitution Method: Solve one equation for one variable and substitute it into the other equation, as we did in Shelly's exercise.
• Elimination Method: Combine equations to eliminate one variable, simplifying your work.

In Shelly's exercise, substitution was the chosen method. It was straightforward, given the specific equations provided. Remember, the goal is to simplify until you find values for all variables.

Calculation Verification

Verifying your solution is crucial. This ensures the accuracy of your problem-solving process. Here’s how to verify your calculations effectively:

• Substitute the obtained variable values back into the original equations.
• Check if both equations are satisfied with your solutions.

In Shelly’s problem, we found \(J = 10\) and \(C = 10\). We then substituted these values back into both original equations to confirm they worked.
For the first equation:
\(10(10) + 20(10) = 100 + 200 = 300\)
For the second equation:
\(20(10) + 10(10) = 200 + 100 = 300\)
Both equations were satisfied, which confirmed our solution was correct. Verification is a simple yet powerful step to ensure the reliability of your results.