Problem 31 - 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

Solution: \( \left( \frac{3}{2}, \frac{7}{3} \right) \).

Step 1: Isolate the Inequality

The inequality given is \( \frac{x+1}{2x-3} > 2 \). Begin by re-writing the inequality in a form that allows isolating \( x \) terms. Multiply both sides by \( 2x - 3 \) to eliminate the fraction, but remember this depends on the sign of \( 2x - 3 \). We consider the cases where \( 2x - 3 > 0 \) and \( 2x - 3 < 0 \):Case 1: \( 2x - 3 > 0 \) implies \( 2x > 3 \) or \( x > \frac{3}{2} \).Case 2: \( 2x - 3 < 0 \) implies \( 2x < 3 \) or \( x < \frac{3}{2} \).

Step 2: Solve for Case 1

In this case, assume \( 2x - 3 > 0 \). The inequality becomes: \[ x + 1 > 2(2x - 3) \]Distribute on the right side:\[ x + 1 > 4x - 6 \]Subtract \( x \) from both sides:\[ 1 > 3x - 6 \]Add 6 to both sides:\[ 7 > 3x \]Divide by 3:\[ x < \frac{7}{3} \]For case 1, the solution must satisfy both \( x > \frac{3}{2} \) and \( x < \frac{7}{3} \). Thus, \( \frac{3}{2} < x < \frac{7}{3} \).

Step 3: Solve for Case 2

For \( 2x - 3 < 0 \), the inequality becomes:\[ x + 1 < 2(2x-3) \]After distributing, it transforms into:\[ x + 1 < 4x - 6 \]Subtract \( x \) from both sides, giving:\[ 1 < 3x - 6 \]Add 6 to each side:\[ 7 < 3x \]Divide by 3:\[ x > \frac{7}{3} \]This implies no interval solution since \( 2x - 3 < 0 \) already assumes \( x < \frac{3}{2} \).

Step 4: Express the Solution in Interval Notation

Based on the valid case from Step 2, the solution to the inequality is the interval: \( \left( \frac{3}{2}, \frac{7}{3} \right) \).

Key Concepts

Interval Notation

Interval notation is a way of representing a set of numbers on a number line. We use it mainly to express the solution set of inequalities clearly.
In interval notation, we can use:

• Round brackets \( ( ) \) to show that an endpoint is not included (known as an open interval),
• Square brackets \( [ ] \) to include an endpoint (known as a closed interval).

For example, the solution \( \left( \frac{3}{2}, \frac{7}{3} \right) \) means any number greater than \( \frac{3}{2} \) and less than \( \frac{7}{3} \,\) excluding those endpoints.
This notation is particularly useful because it gives a quick and concise way to describe a range of values, which is often seen in mathematics to represent solutions to inequalities. Understanding how to read and write interval notation is crucial for solving and expressing solutions of inequalities.

Compound Inequalities

Compound inequalities are formed when two inequalities are combined into one statement by the words "and" or "or".
This can be thought of as a way to show conditions where more than one constraint applies simultaneously. Let's look at the different types:

• “And” compound inequalities: These require both conditions to be satisfied at the same time. For example, \( a < x < b \) means that \( x \) must be greater than \( a \) and less than \( b \). In our exercise, \( \frac{3}{2} < x < \frac{7}{3} \) is an "and" statement.
• “Or” compound inequalities: These indicate that at least one of the conditions must be true. You will often write them as \( x < a \) or \( x > b \).

Understanding compound inequalities is essential for solving inequalities that have more than one part or condition.
It allows us to find which values can satisfy all given conditions.

Rational Inequalities

Rational inequalities involve expressions in the form of fractions where the numerator or the denominator is a variable expression. They can be a bit tricky because of the restrictions that variables can create on the domain of the function.
For example, in solving rational inequalities like \( \frac{x+1}{2x-3} > 2 \, \) several steps are necessary:

• Determine critical values: These are points that make the denominator zero or the expression undefined. For \( 2x - 3 = 0 \, \) we find \( x = \frac{3}{2} \). This indicates the value where the inequality might change.
• Set up test intervals: Use the critical values to create intervals on the number line and test each one to see which satisfy the inequality.
• Solve each interval: Analyze and solve for each part separately as shown in steps with cases, then combine valid solutions.

Rational inequalities can result in a range of solutions, but they often require careful handling of cases where the denominator approaches zero.
By following these careful steps, rational inequalities become more manageable and understandable.