Problem 79 - Elementary Algebra

Question

Set up an algebraic equation and then solve. Bill and Ted are racing across the country. Bill leaves 1 hour earlier than Ted and travels at an average rate of 60 miles per hour. If Ted intends to catch up at a rate of 70 miles per hour, then how long will it take?

Step-by-Step Solution

Answer

It takes Ted 6 hours to catch up to Bill.

Step 1: Define the Variables

Let's define the variables to use in setting up the equation. Let \( t \) be the time in hours that Ted travels until he catches up with Bill. Since Bill leaves an hour earlier, he has been traveling for \( t + 1 \) hours when Ted catches up.

Step 2: Write the Equations for Distance

The distance Bill travels can be expressed as \( 60(t + 1) \) miles, since he travels at 60 mph for \( t + 1 \) hours. The distance Ted travels is \( 70t \) miles since he travels at 70 mph for \( t \) hours.

Step 3: Set Up the Equation

Since Ted catches up with Bill, they cover the same distance by the time Ted meets Bill. Therefore, we can set their distance equations equal to each other: \[ 60(t + 1) = 70t \]

Step 4: Solve the Equation

Expand and simplify the equation: 1. Distribute 60 on the left side: \[ 60t + 60 = 70t \]2. Move all terms involving \( t \) to one side: \[ 60 = 70t - 60t \]3. Simplify the equation: \[ 60 = 10t \]4. Solve for \( t \): \[ t = \frac{60}{10} = 6 \]

Step 5: Check the Solution

To verify, calculate the distances and see if they're equal. Bill travels for 7 hours (since \( t+1 = 6+1 \)) at 60 mph: \[ 60 \times 7 = 420 \text{ miles} \]Ted travels for 6 hours at 70 mph: \[ 70 \times 6 = 420 \text{ miles} \]The distances are equal, confirming the solution.

Key Concepts

Solving Equations

When solving equations, the goal is to find the value of the variable that makes the equation true. In this exercise, our variable is called \( t \), representing the time in hours Ted travels until he catches up with Bill.

To solve, we first set up equations based on the problem's context. Here, we used the following steps:

• Define the variable that represents the unknown quantity.
• Write equations based on known relationships, like distance or speed.
• Set the equations equal, usually because two quantities are the same, like distances in our problem.
• Solve for the variable by simplifying the equation through steps such as distributing, combining like terms, and isolating the variable.

These steps help us find that \( t = 6 \) is the solution, meaning Ted takes 6 hours to catch up with Bill.

To ensure accuracy, it's important to verify by substituting the variable's value back into the original equations to check if both sides equal.

Distance Problems

Distance problems in algebra often involve finding how far something travels over a period of time. They typically use the formula \( \text{Distance} = \text{Rate} \times \text{Time} \).

In our example:

• Bill's distance is calculated as \( 60(t + 1) \) because he travels 60 miles each hour, starting an hour earlier than Ted.
• Ted's distance is \( 70t \) because he travels at 70 miles per hour.

The heart of a distance problem usually involves finding these expressions and understanding how variables affect them. Once we find expressions for the distances both travelers cover, we can set these equal because Ted catches up with Bill, which means he covers the same distance as Bill by that time.

Distance problems often require setting up and solving equations based on such relationships to find the solution.

Rate Problems

Rate problems are closely tied to distance and time. These problems often deal with finding the speed of an object, the time taken, or the distance covered, using relationships in the form \( \text{Rate} = \frac{\text{Distance}}{\text{Time}} \).

In the problem at hand, the rates for Bill and Ted are given:

• Bill moves at 60 mph.
• Ted moves faster at 70 mph.

Understanding each person's rate is crucial in determining how quickly they cover distance and, eventually, how Ted catches up with Bill. Given that Ted's speed is faster, the equation \( 60(t + 1) = 70t \) shows us that he catches up because he travels further in the same length of time compared to Bill.

By calculating and understanding rates algebraically, we can predict and solve for various scenarios depicted in word problems like this, enabling us to find the time it takes for one object to reach another.