Problem 25 - Algebra A Combined Function

Question

Solve. Write the solution set in interval notation. $$ \frac{x+2}{x-3}<1 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, 3)\).

Step 1: Consider the Inequality

The given inequality is \( \frac{x+2}{x-3} < 1 \). Start by rewriting the inequality such that one side is zero. To do this, subtract 1 from both sides: \[ \frac{x+2}{x-3} - 1 < 0 \].

Step 2: Create a Common Denominator

To subtract 1, you'll need a common denominator. The current denominator is \( (x-3) \). Therefore, rewrite 1 as \( \frac{x-3}{x-3} \) to get: \[ \frac{x+2 - (x-3)}{x-3} < 0 \].

Step 3: Simplify the Numerator

Simplify the expression in the numerator: \[ x+2 - (x-3) = x+2-x+3 = 5 \]. Thus, the inequality becomes: \[ \frac{5}{x-3} < 0 \].

Step 4: Determine Where the Expression is Negative

The fraction \( \frac{5}{x-3} \) is negative when the denominator is negative. This happens when \( x-3 < 0 \). Solving this, we find \( x < 3 \).

Step 5: Identify Invalid Values

The expression is undefined when \( x = 3 \) as it causes division by zero. Thus, \( x eq 3 \).

Step 6: Write the Solution in Interval Notation

Combine the results of the previous steps: the solution is all \( x < 3 \), excluding \( x = 3 \). Therefore, the solution set in interval notation is \((-\infty, 3)\).

Key Concepts

Interval Notation

Interval notation is a shorthand way of describing a range of values. It is particularly useful in expressing solutions to inequalities without having to write lengthy verbal explanations. In interval notation, intervals come in two types: open and closed.

For an open interval, the endpoints are not included in the set, and they are denoted with parentheses. For example, \( (a, b) \) means all numbers between \( a \) and \( b \), but not \( a \) or \( b \) themselves. Conversely, a closed interval, which includes both endpoints, uses square brackets, \( [a, b] \).

You can also have half-open intervals such as \( [a, b) \) where \( a \) is included but \( b \) is not, or \( (a, b] \) where \( b \) is included but \( a \) is not. In this exercise, the solution set for the inequality is \( (-\infty, 3) \).

Note how we use a parenthesis at \( 3 \) to exclude it from the interval, as the expression is undefined at that point.

Common Denominator

Finding a common denominator is an essential step when dealing with the subtraction or addition of fractions. A common denominator allows you to combine fractions effectively by aligning them under a single system.

Let's say you have two fractions, \( a/b \) and \( c/d \). To add or subtract them, you must express both fractions with the same denominator. This often involves multiplying the numerator and denominator of each fraction by an appropriate factor.

In the original problem, \( \frac{x+2}{x-3} - 1 < 0 \), the fraction \( 1 \) is rewritten with the common denominator \( x-3 \), as \( \frac{x-3}{x-3} \). This transformation allows both terms to be combined under a single fraction, simplifying calculation.

Undefined Expression

An expression becomes undefined when any operation in it leads to an impossible calculation. Among such situations, division by zero is a common roadblock that makes an expression undefined.

For example, in the inequality you worked with \( \frac{5}{x-3} < 0 \), the fraction becomes undefined if its denominator equals zero. This happens when \( x = 3 \). At \( x = 3 \), the equation causes division by zero, which is undefined.

This is why it's crucial to exclude such values from your solutions. By identifying these values first, you can both avoid errors and accurately determine the range of legitimate solutions. This exclusion is represented in interval notation by using round brackets, indicating the point where the expression is undefined is not part of the solution set.