Problem 67 - Algebra and Trigonometry

Question

Solve each inequality in Exercises \(65-70\) and graph the solution set on a real number line. $$ \frac{3}{x+3}>\frac{3}{x-2} $$

Step-by-Step Solution

Answer

The solution set of the inequality is (-∞, -3) ∪ (2, ∞)

Step 1: Simplify the inequality

We can start by subtracting the expression \(\frac{3}{x-2}\) from both sides of the inequality to combine the fractions on one side. The inequality then becomes: \(\frac{3}{x+3} - \frac{3}{x-2} > 0\)

Step 2: Find a common denominator

Given the two fractions have different denominators, it's necessary to find a common denominator. The least common denominator between \(x+3\) and \(x-2\) is \((x+3)(x-2)\). Multiply both fractions by the common denominator for simplification: \(\frac{3(x - 2)}{(x+3)(x-2)} - \frac{3(x + 3)}{(x+3)(x-2)} > 0\)

Step 3: Simplify the fractions

After multiplying, simplify the fractions further: \(\frac{3x - 6 - 3x - 9}{(x+3)(x-2)} > 0. Simplifying again, -15/(x^2 + x - 6) > 0\)

Step 4: Factor the denominator and solve for critical points

The denominator can be factored into \((x-2)(x+3)\). Set the fractions equal to zero and solve for x, which will give the critical points x = -3, 2. These values divide the number line into three intervals: (-∞, -3), (-3, 2), and (2, ∞) to be tested

Step 5: Test the intervals

Choose any number within each interval to substitute into the inequality. \nFor interval (-∞, -3), let x = -4: -15/((-4 - 3)(-4 + 2)) < 0, which is true; \nFor interval (-3, 2), let x = 0: -15/((0 - 3)(0 + 2)) < 0, which is false; \nFor interval (2, ∞), let x = 3: -15/((3 -3)(3 +2)) < 0, which is true.

Step 6: Graph the solution and write in interval notation

Plot (-∞, -3) and (2, ∞) on a real number line. Since the inequality doesn't include 'or equal to', use open circles at x = -3 and x = 2. Write the final solution in interval notation as (-∞, -3) ∪ (2, ∞)