Problem 234 - Elementary Algebra

Question

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1,170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Step-by-Step Solution

Answer

Jet speed: 415 mph, Wind speed: 25 mph

Step 1 - Define Variables

Let the speed of the jet in still air be denoted as \(j\) (in miles per hour) and the speed of the wind be denoted as \(w\) (in miles per hour).

Step 2 - Set Up Equations for Tailwind

When flying with a tailwind, the effective speed of the jet is \(j + w\). Given that the jet flies 1,320 miles in 3 hours, the equation is:\[\frac{1320}{3} = j + w\]Simplifying, we get:\[440 = j + w\]

Step 3 - Set Up Equations for Headwind

When flying into a headwind, the effective speed of the jet is \(j - w\). Given that the jet flies 1,170 miles in 3 hours, the equation is:\[\frac{1170}{3} = j - w\]Simplifying, we get:\[390 = j - w\]

Step 4 - Solve the System of Equations

We now have the following system of linear equations:1. \(440 = j + w\)2. \(390 = j - w\)Add these two equations to eliminate \(w\):\[440 + 390 = (j + w) + (j - w)\]\[830 = 2j\]Solving for \(j\), we get:\[j = 415\]

Step 5 - Find the Speed of the Wind

Substitute \(j = 415\) back into the first equation to find \(w\):\[440 = 415 + w\]Solving for \(w\), we get:\[w = 25\]

Key Concepts

Understanding a System of Equations

A system of equations consists of multiple equations that share common variables. Solving such systems requires finding the values of these variables that satisfy all the equations simultaneously. For this problem, we have two equations involving the speed of the jet and the speed of the wind: one for flying with a tailwind and one for flying into a headwind. Both equations share the variables for the speed of the jet ( j ) and the speed of the wind ( w ). By solving the system, we determine the values of j and w that make both equations true.

Working with Linear Equations

Linear equations are mathematical statements that describe a straight-line relationship between two variables. They can be written in the form ax + by = c , where a , b , and c are constants. In this exercise, the equations 440 = j + w and 390 = j - w are both linear. Because they consist of straight-line relationships between j and w , we can use simple algebraic techniques to solve them. The power of linear equations lies in their simplicity and predictability, making them essential tools in algebra and real-world problem-solving.

Algebraic Solution Techniques

To solve systems of equations algebraically, we use methods such as substitution and elimination. Here, the elimination method helps us solve for one variable by removing the other. By adding the equations 440 = j + w and 390 = j - w , we eliminate w , giving us a single equation with only j . This makes it easier to solve for j . Once j is determined, we substitute this value back into one of the original equations to find w . Algebraic solution techniques like these are crucial for efficiently solving systems of equations.

Solving Word Problems in Algebra

Word problems can be challenging because they require translating verbal descriptions into mathematical equations. The key steps include:

• Reading the problem carefully to understand what is being asked.
• Identifying the variables and defining them precisely.
• Setting up relevant equations based on the given information.

In this exercise, we translate the distances and times into equations that represent the jet's speeds with a tailwind and into a headwind. By systematically converting the word problem into a system of equations, we can then apply algebraic techniques to find the solution. This method ensures clarity and helps tackle even complex algebraic word problems effectively.