Problem 43 - Algebra and Trigonometry

Question

Solve each rational inequality in Exercises \(43-60\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x-4}{x+3}>0 $$

Step-by-Step Solution

Answer

The solution to the given inequality is \(x \in (-3,4)\)

Step 1: Identify Critical Points

Critical points are the values that either makes the inequality 0 or undefined. Roots of the numerator, as well as the values that make denominator 0 are the critical points. For \(\frac{x-4}{x+3}>0\), the critical points are \(x=4\) (root of the numerator) and \(x=-3\) (which makes denominator equal to 0).

Step 2: Define the Intervals

Form intervals based on critical points in ascending order. Here, our critical points divide the number line into three intervals: \(-\infty,-3\), \(-3,4\), and \(4,\infty\).

Step 3: Test Each Interval

For an interval, take a test point and substitute it in the inequality. Note that an interval may satisfy the inequality, not satisfy the inequality or undefined. Testing these intervals we get: \nFor \(-\infty,-3\), test point is \(-4\), \(\frac{-4-4}{-4+3}<0\), which is FALSE. \nFor \(-3,4\), test point is 0, \(\frac{0-4}{0+3}<0\), which is TRUE. \nFor \(4,\infty\), test point is 5, \(\frac{5-4}{5+3}>0\), which is FALSE.

Step 4: Interpret the Results

Since we are looking for \(\frac{x-4}{x+3}>0\), and the interval \(-3'.

Step 5: Express the Solution in Interval Notation and Plot

The solution to the inequality in interval notation is \((-3,4)\). To plot this on a number line, mark an open circle at -3 and 4, and shade the region between them indicating all the numbers between -3 and 4 are the solutions to the given inequality.