Problem 57 - Algebra 2

Question

An airplane is traveling at the rate \(r\) of 500 miles per hour for a time \(t\) of \((6+x)\) hours. A second airplane travels at the rate \(r\) of \((540+90 x)\) miles per hour for a time \(t\) of 6 hours. Simplify the rational expression. What does this expression tell you about the distances traveled of the two airplanes?

Step-by-Step Solution

Answer

The distance of the first plane is \(\frac{5}{9}\) of the second plane's distance.

Step 1: Calculate the Distance of the First Airplane

The distance traveled by the first airplane can be calculated using the formula for distance: \[ \text{Distance} = \text{Rate} \times \text{Time} \]Substituting the given values:\[ \text{Distance} = 500 \times (6+x) = 3000 + 500x \]

Step 2: Calculate the Distance of the Second Airplane

Similarly, calculate the distance traveled by the second airplane:\[ \text{Distance} = \text{Rate} \times \text{Time} \]Given values:\[ \text{Rate} = 540 + 90x, \quad \text{Time} = 6 \]Substituting these:\[ \text{Distance} = (540 + 90x) \times 6 = 3240 + 540x \]

Step 3: Form the Rational Expression

Now, form the rational expression for the comparison of distances:\[ \frac{3000 + 500x}{3240 + 540x} \]

Step 4: Simplify the Rational Expression

To simplify the rational expression, find the common factor in the numerator and the denominator:\[ \frac{3000 + 500x}{3240 + 540x} = \frac{100(30 + 5x)}{180(18 + 3x)} \]Simplifying this:\[ \frac{30 + 5x}{54 + 9x} = \frac{5(6 + x)}{9(6 + x)} \]Cancel the common factor \((6 + x)\) (assuming \(6 + x eq 0\)):\[ \frac{5}{9} \]

Step 5: Interpret the Simplified Expression

The simplified expression \(\frac{5}{9}\) indicates that the distance traveled by the first airplane is \(\frac{5}{9}\) times the distance traveled by the second airplane. This reflects the relative distances traveled without depending on the variable \(x\).

Key Concepts

Algebraic Simplification

Algebraic simplification is a process of making mathematical expressions more manageable and understandable. It involves reducing expressions to a simpler form while maintaining the original relationships between variables.
When simplifying, it's crucial to identify common factors and operations such as factoring, distributing, and canceling terms. This can help to streamline calculations and reveal underlying patterns.
In the problem, the simplification involved expressing the distances traveled by airplanes as a rational expression and reducing it:

• Identify shared components in the numerators and denominators.
• Factor these components out to simplify the expression.
• Cancel common factors (in this case, the term \(6 + x\)).

By canceling common factors, the resultant expression \(\frac{5}{9}\) becomes more intuitive. It provides a clearer comparison of the relative distances of the two airplanes.

Distance Formula

The distance formula provides a way to calculate the total distance an object travels. It's a fundamental concept in physics and engineering and is given by multiplying the rate of travel by the time of travel.
In algebra, it is expressed as:\[ \text{Distance} = \text{Rate} \times \text{Time} \]For the first airplane, the formula calculates the distance it traveled:

• Rate = 500 miles per hour
• Time = \(6 + x\) hours
• Distance = \(500 \times (6 + x) = 3000 + 500x\)

Similarly, for the second airplane, using its specific rate and time, we find:

• Rate = \(540 + 90x\) miles per hour
• Time = 6 hours
• Distance = \( (540 + 90x) \times 6 = 3240 + 540x \)

The distance formula allows us to quantify and compare how far each airplane has traveled. This forms the basis for analyzing the problem and understanding the relationship between the airplanes' journeys.

Rate and Time Calculations

Understanding the relationship between rate and time is crucial for solving travel-related problems. Rate represents how fast an object is moving, whereas time indicates how long it travels at that rate.
Combining rate and time gives us the total distance traveled, as seen in the distance formula above. In this exercise, two different airplanes travel at different rates and possibly different times. Being able to dissect these components is key to forming accurate expressions and making informed comparisons.

• First airplane: fixed rate of 500 mph with a variable time \(6 + x\).
• Second airplane: a variable rate \(540 + 90x\) mph, with a fixed time of 6 hours.

By manipulating these rates and times, we can derive the distances traveled and form expressions that compare the two flights. This understanding is critical in simplifying the expression, highlighting the role of rate and time in travel scenarios.

Ratio and Proportion Analysis

Ratio and proportion are mathematical concepts that compare values, essential in solving the problem at hand. A ratio indicates how much one quantity is relative to another, providing straightforward comparisons.
In this context, the simplified expression \(\frac{5}{9}\) shows the ratio of the distances traveled by the two airplanes:

• The numerator (5) represents the distance relative to the first airplane's journey.
• The denominator (9) represents the distance relative to the second airplane's journey.

When the rational expression reduces to \(\frac{5}{9}\), it means for every 9 units of distance the second airplane travels, the first airplane travels 5 units. Understanding ratios and proportions thus provides insight into the relative distances, helping to interpret results consistently without getting bogged down with complex variables. This ratio is independent of the variable \(x\), emphasizing a constant comparative relationship between the two flights.