Problem 57 - 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. $$ \frac{x+4}{2 x-1} \leq 3 $$

Step-by-Step Solution

Answer

The solution in interval notation for the given inequality \(\frac{x + 4}{2x - 1} \leq 3 \) is \(( -\infty, \frac{1}{2}] \cup [\frac{7}{6}, \infty )\)

Step 1: Isolate the term with variable

Subtract 3 from both sides of the inequality to isolate the rational function. After subtraction, you get \(\frac{x+ 4}{2x - 1} - 3 \leq 0 \), which can be written as \(\frac{x+4 -3(2x-1)}{2x-1} \leq 0 \). Simplification yields \(\frac{-6x +7}{2x-1} \leq 0 \).

Step 2: Factor the expression

In this case, the expression cannot be further factored. So, move to the next step.

Step 3: Set inequality to zero and solve

Solving the inequality \(\frac{-6x +7}{2x-1} \leq 0 \) by setting inequality to zero gives two critical points. Solving \( -6x +7 =0 \) gives \( x = \frac{7}{6} \) and solving \( 2x -1 =0 \) gives \( x = \frac{1}{2} \).

Step 4: Test intervals

The critical points partition the number line into three intervals. Test each by choosing a test value in the interval: less than \(\frac{1}{2}\), greater than \(\frac{1}{2}\) but less than \(\frac{7}{6}\), and greater than \(\frac{7}{6}\). Substitute these values into the inequality and determine if the result is positive or negative.

Step 5: Graph the solution on number line

To plot the graph, mark points at \( x =\frac{1}{2} \) and \( x =\frac{7}{6} \) on the number line. Then, plot the intervals for which the inequality is true (from the result of step 4). Since the inequality is \(\leq 0 \), the solutions at \( x =\frac{1}{2} \) and \( x =\frac{7}{6} \) are included in the solution set.

Step 6: Write the solution in interval notation

Based on the graph/data from previous step, write down the intervals for which the inequality holds. Be sure to check if the points need to be included or not in the interval ( use '[ ]' if point included and '( )' if not).