Problem 91 - Elementary Algebra

Question

For the following problems, convert the given rational expressions to rational expressions having the same denominators. $$ \frac{x-2}{2 x^{2}+5 x-3}, \frac{x-1}{5 x^{2}+16 x+3} $$

Step-by-Step Solution

Answer

$$ The equivalent rational expressions with the same denominator are: $$ \frac{(x-2)(5x+1)}{(2x-1)(x+3)(5x+1)}, \frac{(x-1)(2x-1)}{(2x-1)(x+3)(5x+1)} $$

Step 1: Factor the denominators

We will start by factoring the denominators of the given rational expressions: $$ 2x^2 + 5x - 3 = (2x - 1)(x + 3) $$ $$ 5x^2 + 16x + 3 = (5x + 1)(x + 3) $$ So, the given rational expressions can be written as: $$ \frac{x-2}{(2x-1)(x+3)}, \frac{x-1}{(5x+1)(x+3)} $$

Step 2: Find the least common denominator (LCD)

The factors in the denominators are \((2x-1)\), \((x+3)\), and \((5x+1)\). Thus, the least common denominator (LCD) is the product of all distinct factors: $$ \text{LCD} = (2x-1)(x+3)(5x+1) $$

Step 3: Convert the rational expressions to have the same denominator

Now, we will convert each rational expression to an equivalent one with the same denominator (LCD). To do this, we must multiply both numerator and denominator of each fraction by the appropriate factors: For the first rational expression: $$ \frac{x - 2}{(2x - 1)(x + 3)} \times \frac{(5x + 1)}{(5x + 1)} = \frac{(x - 2)(5x + 1)}{(2x - 1)(x + 3)(5x + 1)} $$ For the second rational expression: $$ \frac{x - 1}{(5x + 1)(x + 3)} \times \frac{(2x - 1)}{(2x - 1)} = \frac{(x - 1)(2x - 1)}{(2x - 1)(x + 3)(5x + 1)} $$

Step 4: Verify the result

The rational expressions are now: $$ \frac{(x-2)(5x+1)}{(2x-1)(x+3)(5x+1)}, \frac{(x-1)(2x-1)}{(2x-1)(x+3)(5x+1)} $$ Both expressions have the same denominator, which is \((2x-1)(x+3)(5x+1)\). Therefore, we have successfully converted the given rational expressions to rational expressions having the same denominators.

Key Concepts

Factoring Denominators

Factoring denominators is an essential step in simplifying rational expressions and finding common denominators. Whenever you have complicated expressions in the denominator, your first instinct should be to factor them as completely as possible.
For example, consider the polynomial denominators in the expressions given:

• For the first expression: \(2x^2 + 5x - 3\). We factor this as \((2x - 1)(x + 3)\).
• For the second expression: \(5x^2 + 16x + 3\). This factors into \((5x + 1)(x + 3)\).

When factoring, we look for patterns or ways to break a polynomial down into simpler pieces. Common methods include:

• Identifying and factoring out common factors.
• Recognizing special products, like the difference of squares.
• Using the quadratic formula if necessary for more complex trinomials.

Successful factoring simplifies further operations, like finding a common denominator.

Least Common Denominator

Once we have factored the denominators, the next goal is to identify the Least Common Denominator (LCD). The LCD is the simplest expression containing all distinct factors from the denominators of the rational expressions. It is crucial because it allows us to combine rational expressions effectively.
In our case, the factored forms were \((2x - 1)(x + 3)\) and \((5x + 1)(x + 3)\). Thus, the least common denominator must include each distinct factor:

• \((2x - 1)\), from the first rational expression.
• \((5x + 1)\), from the second rational expression.
• \((x + 3)\), present in both expressions, but only needed once in the LCD.

So, the complete LCD is \((2x - 1)(x + 3)(5x + 1)\).
The LCD allows us to rewrite each rational expression with the same denominator, paving the way for addition, subtraction, or comparison of the rational expressions.

Equivalent Fractions

Creating equivalent fractions involves modifying the numerator and denominator by the same factor without altering the overall value of the fraction. This principle is key in adjusting fractions so that their denominators match—important for operations like addition or subtraction.
To convert the rational expressions to equivalent fractions with the least common denominator:

• For \(\frac{x - 2}{(2x - 1)(x + 3)}\), multiply by \(\frac{(5x + 1)}{(5x + 1)}\).
• For \(\frac{x - 1}{(5x + 1)(x + 3)}\), multiply by \(\frac{(2x - 1)}{(2x - 1)}\).

By applying these modifications, you get:

• \(\frac{(x - 2)(5x + 1)}{(2x - 1)(x + 3)(5x + 1)}\) for the first expression.
• \(\frac{(x - 1)(2x - 1)}{(2x - 1)(x + 3)(5x + 1)}\) for the second expression.

These fractions are now equivalent to the original ones but have the handy feature of a shared denominator, allowing further manipulation if necessary.