Problem 26 - Algebra for College Students

Question

Sharon commutes to her office by train. When she walks to the train station, it takes her 40 min. When she rides her bike, it takes her 12 min. Her average walking rate is 7 mph less than her average biking rate. Find the distance from her house to the train station. $$ \begin{array}{|l|l|l|l|} \hline & \text { Rate } & \text { Time } & \text { Distance } \\ \text { Walking } & & & \\ \text { Biking } & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The distance from Sharon’s house to the train station is 2 miles.

Step 1: Define the variables

Let the biking rate be denoted as \( x \) mph. Consequently, the walking rate is given by \( x - 7 \) mph.

Step 2: Convert time to hours

The given times in minutes need to be converted to hours for uniformity. Thus, the walking time is \( \frac{40}{60} = \frac{2}{3} \) hours, and the biking time is \( \frac{12}{60} = \frac{1}{5} \) hours.

Step 3: Set up distance equations

Using the formula \( \text{Distance} = \text{Rate} \times \text{Time} \), we write the equations for walking and biking distances. For walking: Distance = \( (x - 7) \times \frac{2}{3} \) For biking: Distance = \( x \times \frac{1}{5} \)

Step 4: Equate the distances

Since the distance to the train station remains the same, the two expressions of distance can be equated: \( (x - 7) \times \frac{2}{3} = x \times \frac{1}{5} \)

Step 5: Solve the equation

Solve the equation: \[(x - 7) \times \frac{2}{3} = x \times \frac{1}{5}\] Multiply both sides by 15 to clear the fractions: \[ 15 \times (x - 7) \times \frac{2}{3} = 15 \times x \times \frac{1}{5} \] This simplifies to: \[ 5 \times 2 \times (x - 7) = 3 \times x \] \[ 10x - 70 = 3x \] Isolate variable \( x \): \[ 10x - 3x = 70 \] \[ 7x = 70 \] \[ x = 10 \]

Step 6: Calculate the distance

Given the biking rate \( x = 10 \) mph, use either equation to find the distance. Using the biking rate equation: Distance = \( 10 \times \frac{1}{5} = 2 \) miles.

Key Concepts

rate-time-distance relationship

To solve any distance problem in algebra, we rely on the relationship between rate (speed), time, and distance. This relationship is usually expressed with the formula:

\(\text{Distance} = \text{Rate} \times \text{Time}\).

Understanding how these variables interact is crucial. Here are a few key points:

• Rate (Speed): This represents how fast something or someone is moving. It tells us the distance covered per unit of time (e.g., miles per hour).
• Time: This indicates the duration for which the movement happens. It's important to note the time units (e.g., hours, minutes).
• Distance: This is the total length of the path covered.

In the problem, Sharon's walking rate and biking rate are given. Her walking rate is 7 mph less than her biking rate, meaning these rates are directly linked. By applying the rate-time-distance formula for both walking and biking and then equating them, we can solve for the unknown rates and consequently find the distance.

solving linear equations

Solving linear equations is a core concept in algebra. These are equations that involve variables to the first power and have a constant rate of change. Let's go through the steps to solve the specific linear equation in this problem:

1. **Set up the equation**: From the problem, the two distance equations are \((x - 7) \times \frac{2}{3} = x \times \frac{1}{5}\).

2. **Clear the fractions**: To make the equation easier to solve, we remove the fractions by multiplying both sides by a common multiple, in this case, 15:
\[15 \times (x - 7) \times \frac{2}{3} = 15 \times x \times \frac{1}{5}\]

3. **Simplify** the equation: This yields: \(10(x - 7) = 3x\).

4. **Isolate the variable**: Combine like terms and solve for \(x\):
\[10x - 70 = 3x\] \[10x - 3x = 70\] \[7x = 70\] \[x = 10 \]. Once you have solved for the variable, you can use it to find other required values, such as distances.

unit conversion

Unit conversion refers to changing one kind of measurement unit to another. This is essential to ensure all variables are consistent across the equation. In this problem:

- Sharon's walking time is given in minutes. To use this in our rate-time-distance equations, we need to convert it to hours.
For example, the walking time of 40 minutes is converted as follows:

• \(40 \text{ minutes} = \frac{40}{60} \text{ hours} = \frac{2}{3} \text{ hours}\).

- Similarly, her biking time of 12 minutes is likewise converted:

• \(12 \text{ minutes} = \frac{12}{60} \text{ hours} = \frac{1}{5} \text{ hours}\).

Converting these values ensures that the time and speed units (miles per hour) are consistent, allowing us to apply the rate-time-distance relationship accurately. Ignoring proper unit conversion can lead to incorrect answers. Always double-check your units before solving the problem.