Problem 15 - Prealgebra

Question

Find each quotient. Write in simplest form. $$\frac{x^{2}}{5} \div \frac{a x}{2}$$

Step-by-Step Solution

Answer

The quotient is \( \frac{2x}{5a} \).

Step 1: Rewrite Division as Multiplication

To solve the expression \( \frac{x^2}{5} \div \frac{ax}{2} \), we first rewrite the division as multiplication by the reciprocal. This changes our problem to: \[ \frac{x^2}{5} \times \frac{2}{ax} \].

Step 2: Multiply the Numerators and Denominators

Next, multiply the numerators together and the denominators together. The expression becomes: \[ \frac{x^2 \times 2}{5 \times ax} = \frac{2x^2}{5ax} \].

Step 3: Simplify the Expression

To simplify \( \frac{2x^2}{5ax} \), we need to cancel out any common factors in the numerator and the denominator. Both the numerator and denominator have an \( x \) term. So, we cancel one \( x \) from both: \[ \frac{2x}{5a} \].

Key Concepts

Understanding Algebraic Expressions

Algebraic expressions are fundamental building blocks of algebra. These expressions consist of variables, numbers, and operations such as addition, subtraction, multiplication, and division. Variables are symbols, often represented by letters like \( x \), \( a \), or \( y \), which can take on various values. In the exercise, the expression \( \frac{x^2}{5} \) is an algebraic expression containing a variable \( x \) raised to the power of 2.
The power of 2 indicates that \( x \) is multiplied by itself. Understanding how to manipulate algebraic expressions allows you to solve more complex mathematical problems. By rearranging and simplifying such expressions, you can find solutions to equations quickly and accurately. Remembering the order of operations and properties of exponents is crucial when working with algebraic expressions.

Division of Fractions

Dividing fractions might seem tricky at first, but there's a straightforward technique to handle it. The key is to convert the division problem into a multiplication problem by using the reciprocal of the divisor. For instance, in the exercise, we start with \( \frac{x^2}{5} \div \frac{ax}{2} \).
To solve it, we take the reciprocal of the second fraction \( \frac{ax}{2} \) to get \( \frac{2}{ax} \). By turning the division into multiplication \( \frac{x^2}{5} \times \frac{2}{ax} \), the problem becomes much simpler.
This approach ensures ease when working with fractions and is a valuable tool in simplifying complex expressions. Always remember to first transform the division into multiplication by the reciprocal to clear any confusion when tackling similar problems.

Solving Rational Equations

Rational equations are equations that involve fractions whose numerators and denominators contain polynomials. Solving rational equations requires proper manipulation of these fractions to simplify and solve the problem.
In our example, once we convert the division into multiplication, we progress by multiplying the numerators and the denominators separately: \( \frac{2x^2}{5ax} \).
To solve a rational equation, the next step is to simplify by canceling out any common factors. In our case, \( x \) appears in both the numerator and the denominator. Simplifying involves reducing these common factors to arrive at the simplest form: \( \frac{2x}{5a} \). This solution illustrates how rational expressions can be managed by strategically applying math rules.

Utilizing Simplification Techniques

Simplification techniques are essential for making complex expressions more manageable and easier to work with. The goal is to reduce expressions to their simplest form without changing their value. In the problem, we started with an expression \( \frac{2x^2}{5ax} \) and simplified it to \( \frac{2x}{5a} \).
This was achieved by identifying and canceling common factors.

• Identify: Recognize all common factors in the numerator and the denominator.
• Cancel: Reduce them by dividing both numerator and denominator by these factors.

This process of simplification not only makes expressions easier to interpret but prepares them for subsequent operations. Practicing these techniques often will make algebraic concepts more intuitive and less intimidating.