Problem 204 - Elementary Algebra

Question

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball?

Step-by-Step Solution

Answer

Drew burned 1000 calories per hour playing basketball.

Step 1 - Define Variables

Let the number of calories burned per hour playing basketball be denoted as B, and the number of calories burned per hour canoeing be denoted as C.

Step 2 - Set Up Equations

From the information given: On Friday: B + 2C = 1800 On Saturday: 2B + 3C = 3200

Step 3 - System of Equations

We now have a system of linear equations to solve: 1. B + 2C = 1800 2. 2B + 3C = 3200

Step 4 - Solve for One Variable

Solve the first equation for B: B = 1800 - 2C

Step 5 - Substitute into Second Equation

Substitute B in the second equation with the expression from Step 4: 2(1800 - 2C) + 3C = 3200 3600 - 4C + 3C = 3200 3600 - C = 3200

Step 6 - Solve for C

Solve for C: 3600 - C = 3200 C = 400

Step 7 - Solve for B

Substitute C back into the equation B = 1800 - 2C: B = 1800 - 2(400) = 1800 - 800 B = 1000

Step 8 - Verify

Verify by substituting back into the original equations: B + 2C = 1800: 1000 + 2(400) = 1000 + 800 = 1800 2B + 3C = 3200: 2(1000) + 3(400) = 2000 + 1200 = 3200

Key Concepts

Algebraic Equations

An algebraic equation is a mathematical statement that shows the equality of two expressions. In this problem, we used algebraic equations to represent the total calories burned by Drew. The equations must accurately reflect the relationships described in the problem.

We wrote two equations based on the information provided:

• Friday: B + 2C = 1800
• Saturday: 2B + 3C = 3200

Here, B represents the calories burned per hour playing basketball, and C represents the calories burned per hour canoeing. These equations help us to solve for the unknowns, B and C. Understanding how to form and manipulate these equations is crucial for solving real-life problems using algebra.

Linear Equations

A linear equation is an equation between two variables that gives a straight line when plotted on a graph. In the Drew's calorie burning rate problem, we used linear equations to solve for the unknown rates of calorie burning per hour (B for basketball and C for canoeing).

The equations we set up are linear because each equation is of the first degree, meaning the highest exponent of the variables (B and C) is one. For instance:

• B + 2C = 1800
• 2B + 3C = 3200

These straight-line equations intersect at a point that represents the solution to the system. Solving linear equations often involves methods like substitution or elimination, as used in this problem.

Problem-Solving Steps

Solving algebraic word problems can be straightforward if you follow systematic problem-solving steps. Here’s a breakdown based on Drew's calorie problem:

1. Define Variables: Identify what you need to find and assign letters (variables) to these unknowns. In this case, let B stand for the number of calories burned per hour playing basketball and C for canoeing.
2. Set Up Equations: Write equations based on the problem’s information. This problem provided two key time and calorie data points for both activities on different days.
3. System of Equations: Form a system of equations from the relationships described.
4. Solve for One Variable: Isolate one variable in one equation. We solved for B in the first equation.
5. Substitute and Solve: Substitute the isolated variable’s value into the other equation to solve for the second variable.
6. Back-Substitute: Use the value found to solve for the first variable again, if necessary.
7. Verify: Check the solution by substituting back into the original equations to ensure correctness.

Keeping a systematic approach helps in solving complex algebraic problems easily and accurately.

Variable Definition

Defining variables is a fundamental first step in solving word problems. Variables represent unknown quantities and can be used to form algebraic expressions. In Drew’s calorie burning scenario, we defined the variables as follows:

• B: The number of calories burned per hour playing basketball
• C: The number of calories burned per hour canoeing

By clearly defining these variables, we translated the problem’s words into mathematical terms. This allows us to write equations that represent the scenario accurately. Proper variable definition ensures that each step of the solution is logical and traceable back to the problem’s original context, making it easier to solve complex problems systematically.