Problem 232 - Elementary Algebra

Question

A small jet can fly 1,435 miles in 5 hours with a tailwind but only 1215 miles in 5 hours into \(a\) headwind. Find the speed of the jet in still air and the speed of the wind.

Step-by-Step Solution

Answer

The speed of the jet in still air is 265 mph, and the speed of the wind is 22 mph.

Step 1: Define Variables

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

Step 2: Set Up Equations

Using the given information, we can set up two equations based on the concept of speed, distance, and time. With a tailwind, the speed of the jet becomes \( j + w \). With a headwind, it becomes \( j - w \). Thus, based on the distance formula \( \text{Distance} = \text{Speed} \times \text{Time} \), we have:1. \( 1435 = (j + w) \times 5 \)2. \( 1215 = (j - w) \times 5 \)

Step 3: Simplify Equations

Divide both sides of each equation by 5 to simplify:1. \( 287 = j + w \)2. \( 243 = j - w \)

Step 4: Solve the System of Equations

Add the two simplified equations to eliminate \( w \):\( 287 + 243 = j + w + j - w \)\( 530 = 2j \)\( j = 265 \)Now, substitute \( j = 265 \) back into either equation to find \( w \). Using \( 287 = j + w \):\( 287 = 265 + w \)\( w = 22 \)

Step 5: Verify the Solution

Substitute the values of \( j \) and \( w \) back into the original equations to check:\( 1435 = (265 + 22) \times 5 \)\( 1435 = 287 \times 5 \)\( 1435 = 1435 \) (True)\( 1215 = (265 - 22) \times 5 \)\( 1215 = 243 \times 5 \)\( 1215 = 1215 \) (True)

Key Concepts

speed distance time relationship

The relationship between speed, distance, and time is fundamental in solving many algebra word problems. Understanding this relationship can help you break down complex problems into manageable steps. The basic formula connecting these three variables is: \[ \text{Distance} = \text{Speed} \times \text{Time} \] In this problem, we deal with a small jet flying different distances depending on the direction of the wind. When there's a tailwind, the wind helps the jet move faster, increasing its speed. Conversely, a headwind slows the jet down. Here's how it works:

• With a tailwind, the jet's speed is increased by the wind's speed.
• With a headwind, the jet's speed is decreased by the wind's speed.

So, we have two scenarios:

• Tailwind: Speed of jet = speed of jet in still air (j) + speed of wind (w).
• Headwind: Speed of jet = speed of jet in still air (j) - speed of wind (w).

This understanding allows us to set up crucial equations to solve the problem.

system of equations

A system of equations is a set of two or more equations with the same variables. In this problem, we have two equations involving the speed of the jet in still air and the speed of the wind. Here are the steps we followed: First, define the variables:

• Let \( j \) be the speed of the jet in still air (in mph).
• Let \( w \) be the speed of the wind (in mph).

For the tailwind scenario: \( 1435 = (j + w) \times 5 \) For the headwind scenario: \( 1215 = (j - w) \times 5 \) We simplify these equations by dividing both sides by 5:

• \( 287 = j + w \)
• \( 243 = j - w \)

Now, we can solve this system of linear equations using either addition or subtraction to eliminate a variable, which in this case is \( w \). Adding the two equations results in:

• \( 287 + 243 = j + w + j - w \)
• \( 530 = 2j \)

Thus, \( j = 265 \) Substitute \( j \) back into one of the simplified equations to find \( w \):

• \( 287 = 265 + w \)
• \( w = 22 \)

We've successfully solved the system of equations!

problem-solving in algebra

Problem-solving in algebra often involves breaking down a problem into smaller, manageable parts. Here's how we tackled this word problem step by step:

• First, we defined our variables: speed of the jet in still air \( j \) and speed of the wind \( w \).
• Next, we set up equations based on the given scenarios (tailwind and headwind).
• We simplified the equations to make them easier to solve.
• We used the method of addition to solve the system of equations, eliminating one variable.
• We found the value of \( j \) first and then used it to find \( w \).
• Finally, we verified our solution by plugging the values back into the original equations.

Verification is critical in algebra to ensure our results are correct. It instills confidence and confirms the logical steps we've taken. In any algebra problem, it's essential to stay organized and methodical, verifying each step along the way.