Problem 1 - Algebra 2 and Trigonometry

Question

Ashley said that \(\frac{(a+2)(a-1)}{(a+3)(a-1)}=\frac{a+2}{a+3}\) for all values of \(a\) except \(a=-3 .\) Do you agree with Ashley? Explain why or why not.

Step-by-Step Solution

Answer

Ashley is incorrect; the expression is also undefined at \( a=1 \), not just \( a=-3 \).

Step 1: Evaluate the Given Expression

Consider the expression \( \frac{(a+2)(a-1)}{(a+3)(a-1)} \). Notice that both the numerator and the denominator contain the factor \((a-1)\).

Step 2: Simplify the Expression

The expression \( \frac{(a+2)(a-1)}{(a+3)(a-1)} \) can be simplified by cancelling out the common term \((a-1)\) from the numerator and denominator, giving us \( \frac{a+2}{a+3} \), with the restriction that \( a eq 1 \) because division by zero is undefined.

Step 3: Identify Undefined Values

The original expression is undefined when \( a=-3 \) or \( a=1 \), since the denominator \((a+3)(a-1)\) equals zero for these values.

Step 4: Evaluate Ashley's Statement

Ashley claims that \( \frac{(a+2)(a-1)}{(a+3)(a-1)}=\frac{a+2}{a+3} \) for all \( a eq -3 \). However, the expression is also undefined at \( a=1 \), so we cannot agree completely with Ashley's statement. The correct statement should exclude both \( a=-3 \) and \( a=1 \).

Key Concepts

Simplifying Rational Expressions

Simplifying rational expressions involves reducing the expression to its simplest form. A rational expression is a fraction where both the numerator and the denominator are polynomials. Here, the expression \( \frac{(a+2)(a-1)}{(a+3)(a-1)} \) contains a common term \((a-1)\) in both the numerator and the denominator. This allows us to cancel out the \((a-1)\) factor, as long as \(a eq 1\) because division by zero would occur if \(a\) was 1. The simplified expression therefore becomes \( \frac{a+2}{a+3} \).

It’s crucial to always identify and cancel only common factors. Make sure there are no mistaken identities with similar looking but different terms. Simplify rational expressions to make them easier to work with in further calculations and to find any values that potentially make the expression undefined.

Undefined Values in Algebra

In algebra, undefined values occur when expressions result in a form that is not mathematically valid, such as division by zero. With rational expressions, this happens when the denominator equals zero. For instance, in the expression \((a+3)(a-1)\), the expression is undefined when \(a+3 = 0\) or \(a-1 = 0\), which results in \(a = -3\) or \(a = 1\).

Recognizing undefined values is important when simplifying expressions, solving equations, or dealing with algebraic functions. Knowing these restrictions ensures that no invalid operations are performed. Avoid using these undefined values in your calculations to maintain mathematical accuracy. Always solve the denominator for zero to find these essential values.

Division by Zero in Algebra

Division by zero is a crucial concept in algebra that should be avoided at all times. When any part of a fraction's denominator equals zero, the fraction itself becomes undefined. This means you cannot divide any number by zero. For example, the term \(\frac{1}{0}\) has no valid numerical value.

In our original rational expression \(\frac{(a+2)(a-1)}{(a+3)(a-1)}\), if \(a = 1\) or \(a = -3\), the denominator becomes zero, which is mathematically invalid. - Always verify the denominator is non-zero. - When simplifying or solving, exclude ranges or points that cause division by zero.

Handling division by zero thoughtfully avoids errors and ensures that any expressions and equations you deal with are mathematically sound.