Problem 90 - Elementary Algebra

Question

For Problems \(75-92\), perform the indicated operations and express your answers in reduced form. $$ \frac{2}{3 x^{3}}-\frac{5}{4 x} $$

Step-by-Step Solution

Answer

The reduced form is \( \frac{8 - 15x^2}{12x^3} \).

Step 1: Identify a Common Denominator

We need to find the least common denominator (LCD) for the fractions \( \frac{2}{3x^3} \) and \( \frac{5}{4x} \). The denominators are \( 3x^3 \) and \( 4x \). The LCD should contain each factor of the denominators at the highest power it appears: \( 3 \), \( 4 \), and the highest power of \( x \), which is \( x^3 \). Therefore, the LCD is \( 12x^3 \).

Step 2: Rewrite Each Fraction

Convert each fraction to have the LCD as the denominator. For \( \frac{2}{3x^3} \), multiply the numerator and the denominator by \( 4 \) to get \( \frac{8}{12x^3} \). For \( \frac{5}{4x} \), multiply the numerator and the denominator by \( 3x^2 \) to get \( \frac{15x^2}{12x^3} \).

Step 3: Combine the Fractions

Now that both fractions have the same denominator, we can combine them: \( \frac{8}{12x^3} - \frac{15x^2}{12x^3} = \frac{8 - 15x^2}{12x^3} \).

Step 4: Simplify the Result

At this stage, check if the fraction can be reduced further. The expression \( \frac{8 - 15x^2}{12x^3} \) is already simplified as its numerator and denominator have no common factors other than \( 1 \). Thus, the reduced form of the expression is \( \frac{8 - 15x^2}{12x^3} \).

Key Concepts

Common Denominator

When working with algebraic fractions, finding a common denominator is crucial. It allows us to add or subtract fractions by rewriting them with a shared base. Imagine you have two fun puzzles with different numbers of pieces and you want to compare them. You'd need them to have the same total pieces!

To find a common denominator, look at each fraction's denominator and determine the least common multiple that includes all factors. For example, if you have denominators like \(3x^3\) and \(4x\), identify which factors are at play. Here, they include the constants \(3\) and \(4\), and a variable factor \(x\) at multiple powers. The least common one will comprise the highest power of each factor: \(3\), \(4\), and \(x^3\). Therefore, the common denominator is \(12x^3\). This process lays the groundwork for successful fraction operations in algebra.

Fraction Addition

Adding fractions is like combining different colored legos to build a tower. The trick is to ensure they fit together perfectly using the same size blocks, which in math translates to the same denominator.

Once you've found a common denominator, rewrite each fraction so they share this denominator. Adjust each fraction's numerator accordingly by multiplying it with the factor used to adjust the denominator.

• For \(\frac{2}{3x^3}\), multiply the numerator and denominator by \(4\) to get \(\frac{8}{12x^3}\).
• For \(\frac{5}{4x}\), multiply the numerator and denominator by \(3x^2\) to get \(\frac{15x^2}{12x^3}\).

Now you can add or subtract the fractions easily. Just combine their numerators over the common denominator. In this exercise, it becomes \(\frac{8}{12x^3} - \frac{15x^2}{12x^3} = \frac{8 - 15x^2}{12x^3}\). Ensuring each fraction harmonizes with a common denominator is key to seamless addition and subtraction.

Simplifying Expressions

Think of simplifying an algebraic fraction as cleaning up your room by putting everything in the right place and getting rid of the unnecessary. The goal is to express the fraction in its simplest form.

After combining fractions using a common denominator, take a close look at the numerator and the denominator. Check if they share any common factors. If they do, you can divide both by that factor to make the fraction more manageable.

In our example, the expression \(\frac{8 - 15x^2}{12x^3}\) is already neat. The numerator \(8 - 15x^2\) and the denominator \(12x^3\) don't have common factors except \(1\). Sometimes a fraction might appear complex, but once you confirm no common factors exist, it’s already in its simplest form. This step cements the neatness of your solution and ensures that your math "room" is tidy and minimum.