Problem 69 - Algebra and Trigonometry

Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{x+2}{x+3}<\frac{x-1}{x-2} $$

Step-by-Step Solution

Answer

\((-\frac{1}{2}, -3) \cup (-3, 2) \cup (2, \infty)\) is the solution.

Step 1: Eliminate Fractions by Cross-Multiplying

To solve the inequality \( \frac{x+2}{x+3} < \frac{x-1}{x-2} \), first eliminate the fractions by cross-multiplying, assuming \( x+3 eq 0 \) and \( x-2 eq 0 \). This results in:\[(x+2)(x-2) < (x-1)(x+3)\].

Step 2: Expand and Simplify Both Sides

Expand both sides of the inequality:- LHS: \( (x+2)(x-2) = x^2 - 4 \)- RHS: \( (x-1)(x+3) = x^2 + 2x - 3 \)The inequality becomes:\[x^2 - 4 < x^2 + 2x - 3\].

Step 3: Simplify the Inequality by Canceling Terms

Cancel out \(x^2\) on both sides since they appear identical:\[-4 < 2x - 3\].

Step 4: Solve the Linear Inequality

Solve the inequality \(-4 < 2x - 3\) for \(x\). Add 3 to both sides:\[-1 < 2x\].Divide both sides by 2:\[-\frac{1}{2} < x\].

Step 5: Consider Domain Restrictions

Remember the restrictions \(x eq -3\) and \(x eq 2\) due to the original denominators. These restrictions ensure that\(x\) cannot equal \(-3\) or \(2\).

Step 6: Write the Solution in Interval Notation

Combine the results from Steps 4 and 5 to express the solution in interval notation, excluding \(-3\) and \(2\):The solution set is \((-\frac{1}{2}, -3) \cup (-3, 2) \cup (2, \infty)\).

Step 7: Graph the Solution Set

On the number line, represent the solution set by an interval:- Draw an open circle at \(-\frac{1}{2}\), \(-3\), and \(2\). - Shade the regions from \(-\frac{1}{2}\) to \(-3\), from \(-3\) to \(2\), and from \(2\) to infinity.

Key Concepts

Interval Notation

Interval notation is a concise way to represent a range of values, especially useful when dealing with inequalities. Rather than listing all possible values in a range, interval notation shows the endpoints and the type of interval.

- A round bracket, "]" or "(", indicates that the endpoint is not included, also known as an open interval.
- A square bracket, "]" or "(", indicates that the endpoint is included, also known as a closed interval.

For example, the solution \((-\frac{1}{2}, -3) \cup (-3, 2) \cup (2, \infty)\) includes all numbers between \-\frac{1}{2}\ and \-3\ and between \-3\ and \2\, excluding these points, as well as all numbers greater than \2\. This is vividly depicted using open brackets, highlighting the exclusion of endpoints \-3\ and \2\ due to domain restrictions. The union symbol \cup\ is used to show a combination of these intervals.

Cross-Multiplying

Cross-multiplying is a technique for solving equations involving fractions. By cross-multiplying, you remove the fractions by multiplying the numerator of one side by the denominator of the other, and vice-versa.

For the inequality \(\frac{x+2}{x+3} < \frac{x-1}{x-2}\), cross-multiplication involves multiplying \(x+2\) by \(x-2\) and \(x-1\) by \(x+3\). This results in a new inequality without fractions: \((x+2)(x-2) < (x-1)(x+3)\).

This method is beneficial for simplifying the problem, making it easier to isolate the variable. Nevertheless, it's crucial to keep in mind the domain restrictions caused by denominators to avoid invalid results.

Domain Restrictions

Domain restrictions arise from the conditions that make a function or equation undefined, such as dividing by zero. In the original inequality \(\frac{x+2}{x+3} < \frac{x-1}{x-2}\), the denominators \(x+3\) and \(x-2\) cannot be zero.

- Thus, \(x eq -3\) and \(x eq 2\), as these values would result in division by zero, making them inadmissible for the inequality.

When presenting solutions in interval notation, these restrictions must be considered to ensure only valid values of \(x\) are included. In the given solution, \-3\ and \2\ are excluded, altering the intervals to reflect valid solutions by using the open intervals such as \((-3, 2)\). This consideration guarantees the mathematical integrity of the solution set.