Problem 55 - Intermediate Algebra

Question

Set up an equation and solve each problem. Charlotte's time to travel 250 miles is 1 hour more than Lorraine's time to travel 180 miles. Charlotte drove 5 miles per hour faster than Lorraine. How fast did each one travel?

Step-by-Step Solution

Answer

Lorraine traveled at 15 mph, and Charlotte traveled at 20 mph.

Step 1: Define variables

Let's denote the speed of Lorraine as \( v \) (in miles per hour). Then Charlotte's speed will be \( v + 5 \) mph because she drove 5 mph faster.

Step 2: Set up expressions for time

The time it takes for Lorraine to travel 180 miles can be expressed as \( \frac{180}{v} \) hours, and the time it takes for Charlotte to travel 250 miles can be expressed as \( \frac{250}{v + 5} \) hours.

Step 3: Create the equation

According to the problem, Charlotte's time is 1 hour more than Lorraine's. Thus, we set up the equation: \[ \frac{250}{v + 5} = \frac{180}{v} + 1 \]

Step 4: Solve the equation for v

Begin by eliminating the fractions. Do this by multiplying through by \( v(v + 5) \):\[ 250v = 180(v + 5) + v(v + 5) \] Simplify and solve the quadratic equation:\[ 250v = 180v + 900 + v^2 + 5v \] Rearranging gives:\[ v^2 - 65v + 900 = 0 \]

Step 5: Factor the quadratic equation

To solve \( v^2 - 65v + 900 = 0 \), factor the quadratic:\[ (v - 15)(v - 60) = 0 \] Thus, \( v = 15 \) or \( v = 60 \).

Step 6: Determine realistic solution

Since we are looking for Lorraine's speed, \( v = 60 \) seems unrealistic because it suggests Charlotte's speed would be 65 mph, and she travels in 1 hour more time for more distance. Thus, \( v = 15 \) mph is Lorraine's actual speed.

Step 7: Calculate Charlotte's speed

Given Lorraine's speed as 15 mph, Charlotte's speed must be \( 15 + 5 = 20 \) mph.

Key Concepts

Quadratic Equations

Quadratic equations are a vital part of algebra, frequently encountered in different types of problems. A quadratic equation has the general form: \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. Here, we'll focus on a specific problem involving quadratic equations related to speed and distance. In our problem, the equation \[v^2 - 65v + 900 = 0\]was derived from setting up a relationship between speeds and distances covered by Charlotte and Lorraine.To solve quadratic equations, we can use:

• Factoring, which breaks the equation into simpler products that can be solved separately. As seen, \((v - 15)(v - 60) = 0\) shows \( v = 15 \) or \( v = 60 \).
• The Quadratic Formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Solving these equations allows us to find meaningful solutions in word problems like these.In our example, by solving the quadratic equation, we determined Lorraine's speed, which gave us insight into Charlotte's speed as well, demonstrating the power of algebra in practical problems.

Speed and Distance Problems

Speed and distance problems often involve calculating how fast something is moving over a certain distance or the time it takes to move that distance. The general formula you will use is:\[\text{Distance} = \text{Speed} \times \text{Time}\]or its variations by rearranging the variables:

• Speed = \(\frac{\text{Distance}}{\text{Time}}\)
• Time = \(\frac{\text{Distance}}{\text{Speed}}\)

In this problem, we dealt with two people traveling different distances but at related speeds. Charlotte and Lorraine have a speed difference, impacting their travel times for different distances.Here's how they apply:- **Lorraine**: Travels 180 miles. Her speed is \( v \), thus her time is \(\frac{180}{v}\) hours.- **Charlotte**: Travels 250 miles, \( 5 \) mph faster than Lorraine, so her speed is \( v + 5 \) mph. Her time is \(\frac{250}{v+5}\) hours.Understanding these basic principles lets us set up an equation comparing their times effectively.

Variable Manipulation

Variable manipulation is key in solving any algebraic equation, especially when working with word problems. It involves rearranging and simplifying equations to isolate the variable of interest. In this exercise:Step-by-step manipulation is crucial:- **Defining Variables**: Start by defining variables clearly, as done by setting Lorraine's speed as \( v \).- **Setting up Expressions**: Convert word descriptions into mathematical terms, such as \( \frac{180}{v} \) for Lorraine's time.- **Equation Development**: Formulate your primary equation using the problem description, in this case, that Charlotte's time is one hour more.- **Eliminating Fractions**: Work to remove complex fractions by multiplying through by the lowest common denominator.Once you're able to solve the quadratic equation \( v^2 - 65v + 900 = 0 \), select the realistic solution, analyzing the context (speed and conditions) to choose the feasible variable value. This problem emphasizes the importance of careful manipulation to transition from word descriptions to solvable equations in algebra.