Problem 17 - Algebra 1: Concepts and Skills

Question

Find the missing numerator. $$ \frac{8}{5}=\frac{?}{15 y^{2}} $$

Step-by-Step Solution

Answer

The missing numerator in the proportion is \(x = \frac{8 * 15y^2}{5}\).

Step 1: Set up the proportion

Firstly, we recognize that the problem involves ratios or proportions. The equation provided, \(\frac{8}{5} = \frac{?}{15y^2}\), can be viewed as two ratios that are equal to each other. So the problem here is to find the numerator of the second ratio (call it x) which makes the two ratios equivalent.

Step 2: Apply cross multiplication

To find the unknown numerator (x), we can use the technique called cross multiplication which states that the product of the means equals the product of the extremes in a proportion. Here, we have 8 and \(15y^2\) as the extremes, and 5 and x as the means. Cross multiplying, we get \(8 * 15y^2 = 5x\).

Step 3: Solve for x

The question asks for the numerator (x) of the second ratio. So we will make x the subject of the formula from the equation obtained in step 2. Dividing both sides of \(8 * 15y^2 = 5x\) equation by 5, we get \(x = \frac{8 * 15y^2}{5}\).

Key Concepts

Cross Multiplication

In algebra, especially when dealing with proportions, cross multiplication is a powerful tool. It helps you solve for the unknown variable in a set of equal ratios.
To understand cross multiplication, think of it as a way of "crisscrossing" the terms. You multiply the numerator of one ratio by the denominator of the other ratio, and vice versa. For example:

• If you have the proportion \(\frac{a}{b} = \frac{c}{d}\), cross multiplication gives you \(a \times d = b \times c\).

This technique is particularly useful because it turns an equation into a more straightforward linear form that you can solve easily.
In our problem, applying cross multiplication to \(\frac{8}{5} = \frac{x}{15y^2}\), we multiply the extremes: 8 and \(15y^2\), and the means: 5 and \(x\). This results in the equation \(8 \times 15y^2 = 5x\).
By simplifying this equation, it leads us to the solution for \(x\), illustrating the power of cross multiplication in solving proportions.

Numerator and Denominator

The terms "numerator" and "denominator" are fundamental when dealing with fractions and proportions. A fraction is composed of two parts:

• The **numerator** is the top number in a fraction, representing how many parts are being considered.
• The **denominator** is the bottom number, indicating the total number of equal parts in the whole.

When you have a fraction like \(\frac{8}{5}\), 8 is the numerator and 5 is the denominator. These positions are vital because they dictate how the fraction should be interpreted or compared in algebraic operations.
In our given problem, the task is to find the missing numerator in the equation \(\frac{8}{5} = \frac{x}{15y^2}\). Recognizing that \(x\) is the unknown numerator helps us utilize tools like cross multiplication to solve the equation effectively.
Understanding the role of numerators and denominators will make working with fractions much more intuitive in both basic arithmetic and complex algebraic expressions.

Equivalent Ratios

Equivalent ratios are two or more ratios that express the same relationship between numbers. If you multiply or divide both the numerator and denominator of a fraction by the same number, the ratio stays equivalent.
For example, the ratios \(\frac{2}{3}\) and \(\frac{4}{6}\) are equivalent because \(2 \times 2 = 4\) and \(3 \times 2 = 6\). This simple scaling helps preserve the proportion represented by the original fraction.
In the context of our exercise, recognizing equivalent ratios supports the solution process. The problem \(\frac{8}{5} = \frac{x}{15y^2}\) requires identifying a value for \(x\) that maintains equivalency between the given and unknown ratios.
By using cross multiplication, we find \(x = \frac{8\times 15y^2}{5}\), ensuring the two sides of the equation represent equivalent ratios. This process confirms the importance of recognizing and accurately balancing ratios to solve algebraic problems effectively.