Problem 288 - Elementary Algebra

Question

In the following exercises, solve. Charlie and Violet met for lunch at a restaurant between Memphis and New Orleans. Charlie had left Memphis and drove 4.8 hours towards New Orleans. Violet had left New Orleans and drove 2 hours towards Memphis, at a speed 10 miles per hour faster than Charlie's speed. The distance between Memphis and New Orleans is 394 miles. Find the speed of the two drivers.

Step-by-Step Solution

Answer

Charlie drives at 55 mph and Violet drives at 65 mph.

Step 1 - Define variables

Let Charlie's speed be denoted by \( c \) (in miles per hour). Since Violet's speed is 10 miles per hour faster, Violet's speed can be denoted by \( c + 10 \) mph.

Step 2 - Calculate distances traveled

Charlie drove for 4.8 hours, so the distance Charlie traveled is \( 4.8c \). Violet drove for 2 hours, so the distance Violet traveled is \( 2(c + 10) \).

Step 3 - Establish the equation

Since the combined distance Charlie and Violet traveled equals the distance between Memphis and New Orleans, set up the equation: \( 4.8c + 2(c + 10) = 394 \).

Step 4 - Solve for Charlie's speed

Expand the equation: \( 4.8c + 2c + 20 = 394 \)Combine like terms: \( 6.8c + 20 = 394 \)Subtract 20 from both sides: \( 6.8c = 374 \)Divide both sides by 6.8: \( c = \frac{374}{6.8} \)\( c = 55 \). Therefore, Charlie drives at 55 mph.

Step 5 - Solve for Violet's speed

Since Violet's speed is 10 mph faster than Charlie's: \( c + 10 = 55 + 10 \)\( c + 10 = 65 \). Therefore, Violet drives at 65 mph.

Key Concepts

linear equations

In algebra, linear equations are fundamental and very useful, especially for solving real-world problems. They are mathematical statements that show that two expressions are equal using constants and versatile variables. A linear equation with one variable typically looks like this: \( ax + b = c \). Here:

• \( a \) is a coefficient that multiplies the variable \( x \)
• \( b \) is a constant term that either adds to or subtracts from \( a \)
• \( c \) is the result of the equation.

To solve for the variable, you perform the same operations on both sides of the equation to isolate the variable. This usually involves:

• Combining like terms
• Adding or subtracting terms from both sides
• Multiplying or dividing both sides by coefficients or constants

speed and distance problems

Solving speed and distance problems is about understanding the relationship between distance, speed, and time. These three elements are connected by the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \]In such problems, you will often know some of these variables and need to solve for another one. Here's a breakdown:

• If you know speed and time, multiply them to find the distance.
• If you know distance and speed, divide the distance by the speed to find the time.
• If you know distance and time, divide the distance by time to find the speed.

In the exercise, Charlie's and Violet's driving times were given: 4.8 hours and 2 hours respectively. By multiplying these times by their respective speeds, we calculated how far each drove. Understanding these relationships helps in setting up accurate equations to solve complex problems, like finding the speeds of Charlie and Violet.

variable substitution

Variable substitution is a powerful technique in algebra. It involves replacing variables with numerical values or other variables to simplify equations and solve them efficiently.Consider our problem where Violet's speed is 10 mph faster than Charlie's speed. We can denote Charlie's speed with \( c \), thus Violet's speed becomes \( c + 10 \). When you have multiple variables linked by relationships, variable substitution helps in forming single-variable equations that are easier to tackle.In this exercise, substituting Violet's speed as \( c + 10 \) and calculating the distances they both traveled led us to form the equation \( 4.8c + 2(c + 10) = 394 \). By solving this single-variable equation, we first found Charlie's speed, then used substitution again to find Violet's speed.Understanding how and when to substitute variables can make solving algebraic problems more straightforward and less daunting.

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