Problem 51 - Elementary Algebra

Question

Simplify each algebraic fraction. $$\frac{y^{2}-6 y-72}{y^{2}-8 y-84}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{y - 12}{y - 14}\).

Step 1: Factor the Numerator

The numerator is given as \(y^2 - 6y - 72\). We need to factor it. Look for two numbers that multiply to -72 and add to -6. These two numbers are 6 and -12. Thus, the numerator can be factored as:\[y^2 - 6y - 72 = (y + 6)(y - 12)\]

Step 2: Factor the Denominator

The denominator is \(y^2 - 8y - 84\). We need to factor it similarly to the numerator. Look for two numbers that multiply to -84 and add to -8. The numbers are 6 and -14. Thus, the denominator factors to:\[y^2 - 8y - 84 = (y + 6)(y - 14)\]

Step 3: Simplify the Fraction

Now that both the numerator and denominator have been factored, the fraction is:\[\frac{(y + 6)(y - 12)}{(y + 6)(y - 14)}\]Notice that the factor \((y + 6)\) appears in both the numerator and the denominator. We can cancel it out, provided \(y eq -6\) to avoid division by zero. This simplifies the fraction to:\[\frac{y - 12}{y - 14}\]

Step 4: State the final simplified expression

After canceling the common factors, the fraction \(\frac{y^2 - 6y - 72}{y^2 - 8y - 84}\) simplifies to \(\frac{y - 12}{y - 14}\). Remember the restrictions on \(y\) are that \(y eq 6\), \(y eq 14\), and \(y eq -6\) because these would make the original expressions undefined due to division by zero.

Key Concepts

Factoring Trinomials

Factoring trinomials involves breaking down a quadratic expression into a product of two binomials. For a trinomial like \(ax^2 + bx + c\), the goal is to find two numbers that multiply to \(ac\) and add up to \(b\). This can sometimes feel like solving a puzzle, but once you get the hang of it, it becomes a straightforward process.

In our example, we focus on two trinomials: \(y^2 - 6y - 72\) and \(y^2 - 8y - 84\). We need to identify pairs of numbers:

• For \(y^2 - 6y - 72\), we find numbers 6 and -12 since they multiply to -72 and add to -6. The expression factors to \((y + 6)(y - 12)\).
• For \(y^2 - 8y - 84\), the numbers are 6 and -14, which multiply to -84 and sum to -8. The expression factors to \((y + 6)(y - 14)\).

Factoring is a critical step because it reveals the component parts of the expression, allowing for further simplification.

Simplifying Expressions

Simplifying expressions means making them as straightforward as possible, often by canceling out common factors. It's like decluttering a room, removing any unnecessary items for clarity. In the context of algebraic fractions, once we factor both the numerator and denominator, we can "cancel" terms that are present in both.

In our example:

• We have the factored form: \(\frac{(y + 6)(y - 12)}{(y + 6)(y - 14)}\).
• The \((y + 6)\) term appears in both the numerator and the denominator, so it can be canceled out.

This results in the simpler expression: \(\frac{y - 12}{y - 14}\), making it easier to understand and work with in subsequent calculations.
Remember, simplifying doesn't change the value of the expression, but it does make it easier to interpret and work with for future calculations.

Division by Zero

Division by zero is a fundamental concept in algebra and mathematics in general. It is important because dividing by zero is undefined, which can "break" mathematical operations.

When simplifying algebraic fractions, it's crucial to identify values that make the denominator zero, as these are restrictions on the expression. In our example:

• The initial fraction \(\frac{(y + 6)(y - 12)}{(y + 6)(y - 14)}\) becomes undefined if \(y = -6\), \(y = 14\), or \(y = 6\).
• Therefore, it's necessary to state that \(y\) cannot equal these values, ensuring the expression doesn't invoke division by zero.

Without these restrictions, we'd risk encountering an undefined situation, which is something always to avoid in algebraic manipulation.