Problem 33 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-40: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ \frac{x+1}{2 x-3}>2 $$

Step-by-Step Solution

Answer

The solution is \(\left(\frac{3}{2}, \frac{7}{3}\right)\).

Step 1: Clear the Inequality

First, we'll clear the fraction by subtracting 2 from both sides of the inequality, giving us \( \frac{x+1}{2x-3} - 2 > 0 \).

Step 2: Create a Single Fraction

Rewriting the inequality, we get a single fraction: \( \frac{x+1 - 2(2x-3)}{2x-3} > 0 \), which simplifies to \( \frac{x+1-4x+6}{2x-3} > 0 \).

Step 3: Simplify the Expression

Simplify the numerator: \( (x+1-4x+6) = -3x+7 \). So the inequality becomes \( \frac{-3x+7}{2x-3} > 0 \).

Step 4: Find the Critical Points

Set the numerator and denominator equal to zero separately to find the critical points. For the numerator: \(-3x+7=0\) gives \(x=\frac{7}{3}\). For the denominator: \(2x-3=0\) gives \(x=\frac{3}{2}\).

Step 5: Determine Intervals on Number Line

Based on critical points \(x=\frac{3}{2}\) and \(x=\frac{7}{3}\), divide the number line into three intervals: \((-\infty, \frac{3}{2})\), \((\frac{3}{2}, \frac{7}{3})\), and \((\frac{7}{3}, \infty)\).

Step 6: Test the Intervals

Choose a test point from each interval:- For \((-\infty, \frac{3}{2})\), pick \(x=0\). The result is \(\frac{-3(0)+7}{2(0)-3} = \frac{7}{-3} < 0\).- For \((\frac{3}{2}, \frac{7}{3})\), pick \(x=2\). The result is \(\frac{-3(2)+7}{2(2)-3} = \frac{1}{1} > 0\).- For \((\frac{7}{3}, \infty)\), pick \(x=3\). The result is \(\frac{-3(3)+7}{2(3)-3} = \frac{-2}{3} < 0\).

Step 7: Compare Signs and Write Solution

The inequality is satisfied where the expression is positive. This occurs in the interval \(\left(\frac{3}{2}, \frac{7}{3}\right)\). Since the numerator and denominator cannot be zero, points \(x=\frac{3}{2}\) and \(x=\frac{7}{3}\) are not included in the solution.

Key Concepts

Interval Notation

When we solve inequalities, representing the solution in interval notation is a concise way to express the range of values that satisfy the inequality. Interval notation uses parentheses and brackets to show which values are included or excluded.

For instance, if we have a solution that spans from one point to another but does not include these boundary points, we write it using parentheses. In our exercise, the solution lies between \( \frac{3}{2} \) and \( \frac{7}{3} \), and because these points make the expression undefined (the denominator is zero), we exclude them: \((\frac{3}{2}, \frac{7}{3})\). This signifies all values greater than \( \frac{3}{2} \) and less than \( \frac{7}{3} \), without including the endpoints themselves.

• Use brackets \([\,]\) if a number is included in the solution.
• Use parentheses \((\,\)) for excluded numbers.

Understanding how to properly use interval notation helps clearly communicate the solution set of an inequality.

Critical Points

Critical points are crucial when solving inequalities involving fractions or polynomials. They help us determine which intervals to test for solutions by identifying where the expression changes its sign.

In the given exercise, critical points are derived by setting the numerator and denominator of our fraction equal to zero. For the inequality \( \frac{-3x+7}{2x-3} > 0 \), the numerator \(-3x+7\) is zero at \(x=\frac{7}{3}\). The denominator \(2x-3\) is zero at \(x=\frac{3}{2}\).

These zero points divide the number line into intervals.

• The number line is split into sections around the critical points.
• Test points are chosen from each interval to determine the sign of the expression.

Thus, the critical points guide us in distinguishing the intervals to consider for further analysis.

Fraction Simplification

Simplifying fractions is essential in solving inequalities like \( \frac{x+1}{2x-3}>2 \). This step helps in transforming the original form into one that's easy to analyze and solve.

To simplify, it's best to first convert the inequality so both sides are fractions, allowing for a unified expression. Start by moving all terms to one side: \( \frac{x+1}{2x-3} - 2 > 0 \). Then, create a common denominator to combine and simplify the fractions, leading to \( \frac{-3x+7}{2x-3} > 0 \).

Simplifying further involves working with the numerator \(-3x+7\). Such simplification ensures clarity:

• Combine like terms in the numerator.
• Retain the denominator to maintain sign analysis integrity.

Each step in fraction simplification brings you closer to identifying the intervals for the solution.

Sign Analysis

Sign analysis in inequalities is the method of checking which intervals make the inequality true. By using test points from different intervals, we can determine the sign of the expression in those sections and identify where the inequality holds.

In our exercise, with critical points \( \frac{3}{2} \) and \( \frac{7}{3} \), we divide the number line into three sections: \((-fty, \frac{3}{2})\), \((\frac{3}{2}, \frac{7}{3})\), and \((\frac{7}{3}, fty)\).

For each section, choose a test point and substitute it into the inequality:

• For \((-fty, \frac{3}{2})\), a point like \(x=0\) results in a negative value.
• For \((\frac{3}{2}, \frac{7}{3})\), if \(x=2\), the result is positive, satisfying the inequality.
• For \((\frac{7}{3}, fty)\), try \(x=3\) which again gives a negative result.

With this information, only the interval \((\frac{3}{2}, \frac{7}{3})\) makes the inequality true, leading to that being the solution.