Problem 23 - Prealgebra

Question

Write a proportion that could be used to solve for each variable. Then solve. \(s\) inches in 0.54 hour 4.55 inches in 1.89 hours

Step-by-Step Solution

Answer

\(s \approx 1.3\) inches.

Step 1: Understand the Problem

We are given two sets of data describing two instances of a situation. Each instance provides a number of inches covered over a span of hours. We need to find a proportion that will relate these two instances using the given information.

Step 2: Set Up a Proportion

To create a proportion, we set up a ratio of inches to hours for each scenario. We'll have \( \frac{s}{0.54} \) for the first scenario and \( \frac{4.55}{1.89} \) for the second. We equate these two ratios to reflect that they are proportionate:\[ \frac{s}{0.54} = \frac{4.55}{1.89} \]

Step 3: Cross-Multiply to Solve for \(s\)

Using the property of cross-multiplication, we can solve the proportion. This means we multiply across the equal sign in a diagonal: \( s \times 1.89 = 0.54 \times 4.55 \). Simplifying this gives:\[ 1.89s = 2.457 \]

Step 4: Isolate \(s\)

To solve for \(s\), divide both sides by 1.89:\[ s = \frac{2.457}{1.89} \]Calculating the division gives the value of \(s\).

Step 5: Calculate the Value of \(s\)

Perform the division:\( s = \frac{2.457}{1.89} \approx 1.3 \). Thus, \(s\) is approximately equal to 1.3 inches.

Key Concepts

Understanding Ratios

A ratio is a way to compare two quantities by expressing how many times one value contains or is contained within the other. In simpler terms, it's a comparison of two numbers describing their relative size. For instance, if you have a scenario involving distance traveled and time taken, you could express these as a ratio like "miles per hour."Here are a few key points about ratios:

• Ratios can be represented in several forms; such as fractions (\(\frac{a}{b}\)), using a colon (a:b), or even in words ("a to b").
• Proportions occur when two ratios are set equal to each other.
• Most everyday problems involving comparison or scaling deal with ratios.

Understanding how and why to use ratios in real-world applications is essential. For example, using the problem above, the first ratio of inches to hours is \(\frac{s}{0.54}\) and the second ratio is \(\frac{4.55}{1.89}\). These ratios provide a way to compare different scenarios under similar conditions, allowing for problem-solving by setting them equal in a proportion.

Cross-Multiplication Technique

Cross-multiplication is an invaluable tool in solving equations involving proportions. It involves multiplying across the equal sign of a proportion, allowing you to eliminate the fraction and solve for the unknown variable directly.Here's how it works:

• When you have a proportion \(\frac{a}{b} = \frac{c}{d}\), you multiply across the diagonals to get \(a \times d = b \times c\).
• This step removes the fractions, simplifying the equation into a form that's easier to solve.

In our specific problem, we had the proportion \(\frac{s}{0.54} = \frac{4.55}{1.89}\). By applying cross-multiplication, it transforms into the equation \(s \times 1.89 = 0.54 \times 4.55\), which can be simplified further. This method is particularly helpful as it transitions you smoothly from understanding the relationship between ratios to performing a calculation.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. Once you've simplified the problem using techniques like cross-multiplication, it's time to solve for the variable, in this case, \(s\).The steps to solve an equation effectively include:

• Isolating the variable on one side of the equation.
• Performing algebraic operations like addition, subtraction, multiplication, or division as necessary.
• Checking your answer to ensure it's reasonable and fits the context of the problem.

In our example, after cross-multiplying, we had \(1.89s = 2.457\). Isolating \(s\) involves dividing both sides by \(1.89\), yielding \(s = \frac{2.457}{1.89}\). By performing this division, we obtain \(s \approx 1.3\). Thus, solving equations is about logical manipulation until the variable is isolated, allowing us to find its precise value.