Problem 18 - Algebra A Combined Function

Question

Perform each indicated operation. Simplify if possible. \(\frac{-9}{25 x^{2}-1}+\frac{7}{1-25 x^{2}}\)

Step-by-Step Solution

Answer

\(\frac{-2}{(5x - 1)(5x + 1)}\)

Step 1: Factor the Denominators

To simplify the expression, start by factoring the denominators. Notice that both denominators appear to be in the form of a difference of squares. Rewrite them:\[25x^2 - 1 = (5x - 1)(5x + 1)\]\[1 - 25x^2 = -(25x^2 - 1) = -(5x - 1)(5x + 1)\]Thus, the expression becomes:\[\frac{-9}{(5x - 1)(5x + 1)} + \frac{7}{-(5x - 1)(5x + 1)}\]

Step 2: Combine the Fractions

Since the denominators are now the same (apart from a factor of -1 in the second fraction), we can combine the fractions. The common denominator is \[(5x - 1)(5x + 1)\].Combine the numerators by addition:\[\frac{-9}{(5x - 1)(5x + 1)} + \frac{7}{-(5x - 1)(5x + 1)} = \frac{-9 - 7(-1)}{(5x - 1)(5x + 1)}\]Simplify the numerator:\[-9 + 7 = -9 - (-7) = -9 + 7 = -2\]So,\[\frac{-2}{(5x - 1)(5x + 1)}\]

Step 3: Simplify the Expression

There appears to be no common factors between the numerator and the denominator other than 1. Therefore, the expression\[\frac{-2}{(5x - 1)(5x + 1)}\]is already simplified.

Key Concepts

Difference of Squares

In the world of rational expressions, recognizing the pattern of the _difference of squares_ can be a powerful tool. This pattern appearswhen two perfect squares are subtracted. The standard form of a difference of squares is:\[ a^2 - b^2 = (a - b)(a + b) \]This simple formula allows us to factor expressions into two binomials,which is a crucial step in simplifying complex fractions.In our example exercise, the denominators \( 25x^2 - 1 \) and \( 1 - 25x^2 \) are rewrittenusing the difference of squares. Note that:- \( 25x^2 \) is the square of \( 5x \) because \( (5x)^2 = 25x^2 \)- \( 1 \) is the square of 1 because \( 1^2 = 1 \)Thus, both expressions are factored as:- \( 25x^2 - 1 = (5x - 1)(5x + 1) \)- \( 1 - 25x^2 = -(5x - 1)(5x + 1) \)Understanding this technique is essential, as it enables you to rewrite and factor expressions swiftly and accurately.

Factoring Denominators

Factoring denominators is a critical step when dealing with rational expressions. It not only simplifies expressions but also aids in identifying common denominators.When approaching an expression that involves a fraction, always check if the denominator can be factored. In the case of our exercise, bothdenominators are in the form of a difference of squares, making the factorization process straightforward.It's important to remember:- A factorable denominator can reveal a common factor numerator might share, simplifying your work later.- Factorization often reduces complicated expressions into more recognizable forms, making other arithmetic operations easier.In the original exercise, recognizing and factoring the difference of squares immediately set us up for success in subsequentsteps. The result of the factoring was:- \( (5x - 1)(5x + 1) \)which appeared in the denominators of both fractions. This insightmakes combining and further simplification more manageable.

Combining Fractions

Combining fractions may sound daunting at first, but with practice, it becomes a straightforward process. It's an action that involves finding a common denominator and then adding or subtracting numerators.Once you factor both denominators, as shown using the difference of squares, you will likely notice common factors. In our exercise:- Both denominators were factored into \((5x - 1)(5x + 1)\), with the second fraction having an additional -1 factorWith identical or similar denominators achieved, the next step is tocombine the numerators, keeping the common denominator:- Numerators \(-9\) and \(7\) (from the second fraction \(7\times(-1)\)) were combined over the common denominator to simplify the expression:\[ \frac{-9}{(5x - 1)(5x + 1)} + \frac{7}{-(5x - 1)(5x + 1)} = \frac{-9 + 7}{(5x - 1)(5x + 1)} \]This step allows you to manage and simplify expressions effectively,leading to a clean, simplified result. Understanding these steps buildsa strong foundation for working with more complex rational expressions as well.