Problem 38 - Elementary Algebra

Question

For Problems \(33-50\), set up an equation and solve the problem. (Objective 2 ) To travel 300 miles, it takes a freight train 2 hours longer than it takes an express train to travel 280 miles. The rate of the express train is 20 miles per hour faster than the rate of the freight train. Find the rates of both trains.

Step-by-Step Solution

Answer

The freight train's rate is 50 mph, and the express train's rate is 70 mph.

Step 1: Define Variables

Let the rate of the freight train be \( r \) miles per hour. The express train travels 20 miles per hour faster, so its rate is \( r+20 \) miles per hour.

Step 2: Write Time Equations

The time it takes for the freight train to travel 300 miles is \( \frac{300}{r} \) hours, and for the express train to travel 280 miles is \( \frac{280}{r+20} \) hours. We know the freight train takes 2 hours longer: \[ \frac{300}{r} = \frac{280}{r+20} + 2. \]

Step 3: Solve for r

Clear the denominators by multiplying through by \( r(r+20) \): \[ 300(r+20) = 280r + 2r(r+20). \] Simplifying the equation: \[ 300r + 6000 = 280r + 2r^2 + 40r. \] This reduces to: \[ 2r^2 + 20r - 6000 = 0. \]

Step 4: Simplify and Solve Quadratic Equation

Simplify the equation: \[ r^2 + 10r - 3000 = 0. \] Solve using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 10, c = -3000 \). Thus, \[ r = \frac{-10 \pm \sqrt{10^2 - 4(1)(-3000)}}{2(1)}. \] \[ r = \frac{-10 \pm \sqrt{100 + 12000}}{2} = \frac{-10 \pm \sqrt{12100}}{2}. \] \[ r = \frac{-10 \pm 110}{2}. \] This gives two solutions: \( r = 50 \) or \( r = -60 \). Since rate cannot be negative, \( r = 50 \).

Step 5: Find Express Train's Rate

Since the express train's rate is 20 miles per hour faster, its rate is \( r + 20 = 50 + 20 = 70 \) miles per hour.

Key Concepts

Quadratic Equations

Quadratic equations form the backbone of many algebra problems, including those involving motion, like our trains' speeds in this exercise. An equation is termed "quadratic" when it can be expressed in the standard form: \( ax^2 + bx + c = 0 \). Here, the highest power of \( x \) is 2, indicating the presence of a square term.
When solving quadratic equations, there are several methods to choose from. In our exercise, we used the **quadratic formula**, which is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps us find the roots of the equation when it's not easily factorable. The "\( b^2 - 4ac \)" part under the square root is known as the **discriminant**. It tells us about the nature of the roots:

• If the discriminant is positive, there are two real and distinct solutions.
• If it is zero, there is one real solution, a repeated root.
• If negative, the solutions are complex numbers.

In our problem, the discriminant is positive (12100), giving us two potential values for \( r \). Finally, we selected the positive solution because speed cannot be negative.

Rate Problems

Rate problems often involve finding relationships between distance, speed, and time. The fundamental formula that guides these problems is:

• **Distance = Rate × Time**

This formula helps us set up equations based on the information given. For example, in our train problem, the freight train travels 300 miles, so its travel time can be expressed as \( \frac{300}{r} \). Similarly, the express train’s travel time is \( \frac{280}{r+20} \). These equations reflect the fundamental relationship between rate and time.
This specific problem brings in an additional layer with the express train going 20 miles per hour faster and taking less time. The key here is to determine the rate for both by finding a common time relationship, adding or subtracting time as necessary to meet the stated conditions—in this case, accounting for the 2-hour difference.

System of Equations

A system of equations is essentially a set of equations with multiple variables. Solving such systems involves finding a common solution for all the variables. In our train rate problem, the challenge is not explicitly labeled a system of equations, but it conceptually behaves like one in managing conditions and constraints:

• \( \frac{300}{r} = \frac{280}{r+20} + 2 \) was our key equation.

This single equation incorporates conditions for both the freight and the express trains. By simultaneously considering both journeys, we create a cohesive mathematical "system" to solve for the rates.
By manipulating and solving this system, we find the unknown variable, \( r \). This enables us to work out both the freight train's and the express train's rates successfully. This technique showcases how interconnected equations can help us solve multifaceted algebra problems efficiently.