Problem 17 - Algebra and Trigonometry

Question

\(7-20=\) Simplify the rational expression. $$ \frac{y^{2}+y}{y^{2}-1} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{y}{y-1} \), where \( y \neq -1 \).

Step 1: Identify the Expression

Examine the given rational expression: \( \frac{y^{2} + y}{y^{2} - 1} \). Identify both the numerator and the denominator.

Step 2: Factor the Numerator

The numerator \( y^{2} + y \) can be factored by taking out the common factor \( y \), resulting in \( y(y + 1) \).

Step 3: Factor the Denominator

The denominator \( y^{2} - 1 \) is a difference of squares. It can be factored into \( (y + 1)(y - 1) \).

Step 4: Simplify the Expression

Now, the expression becomes \( \frac{y(y + 1)}{(y + 1)(y - 1)} \). Observe that \( (y+1) \) is a common factor in both the numerator and the denominator.

Step 5: Cancel the Common Factor

Cancel the common factor \( y + 1 \) in both the numerator and the denominator, simplifying the expression to \( \frac{y}{y - 1} \).

Step 6: Check for Excluded Values

Since we canceled \( y + 1 \), \( y eq -1 \) must be an excluded value because it would make the original denominator zero.

Key Concepts

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler polynomials that multiply together to give the original polynomial. In essence, it's about finding the 'pieces' that make up the whole. This process is essential in simplifying rational expressions and solving equations.

In the given expression, the numerator is a polynomial:

• Expression: \( y^2 + y \)
• Common Factor: \( y \)

By factoring out the greatest common factor, which is \( y \), we rewrite the numerator as \( y(y + 1) \).

It's important to look for other types of factoring:

• Grouping: Useful when dealing with polynomials with four or more terms.
• Quadratic: Expressible in the form \( ax^2 + bx + c \).

Each method depends on the structure of the polynomial. Recognizing which to use saves time and ensures precision.

Difference of Squares

The difference of squares is a special factoring pattern where a square number is subtracted from another square number. The pattern can be expressed as \( a^2 - b^2 = (a + b)(a - b) \). It’s an effective way to break down polynomials quickly.

In the exercise, the denominator \( y^2 - 1 \) fits this pattern:

• \( y^2 \) is a perfect square.
• 1 is a perfect square as \( 1^2 \).

We factor it as \( (y + 1)(y - 1) \). This technique helps to uncover common factors in the numerator and denominator, making the expression more straightforward to simplify.

Recognizing this pattern comes with practice, and it's handy in algebraic expressions and equations, as it allows a quick and effective reduction.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is about reducing expressions to their simplest form. It makes working with expressions much easier and often necessary before solving equations.

The current exercise involves a rational expression:

• Rational Expression: The fraction form \( \frac{y(y + 1)}{(y + 1)(y - 1)} \)
• Common Factor: \( y+1 \)

By canceling the common factor \( y+1 \) from both the numerator and the denominator, we simplify it to \( \frac{y}{y - 1} \). This reduced form is not only simpler but also easier to evaluate and use.

Be mindful of excluded values. These are specific values that can make the denominator zero or were canceled out during simplification. For this expression, \( y = -1 \) must be excluded as it leads to division by zero in the original form. Identifying and noting these values is crucial, as they define the equation's domain and correct simplification.