Problem 49 - 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. $$ \frac{4-2 x}{3 x+4} \leq 0 $$

Step-by-Step Solution

Answer

The solution to the inequality \(\frac{4-2x}{3x+4} \leq 0\) is \((-∞,-4/3) \cup (-4/3,2]\) in interval notation.

Step 1: Identify the Critical Points

The critical points of a rational inequality are the values of x that make the numerator 0 and the values of x that make the denominator 0 since at these points the inequality can change from true to false. To identify these, set the numerator and the denominator equal to zero separately and solve for x. Solving \(4 - 2x = 0\) gives \(x = 2\). Solving \(3x + 4 = 0\) gives \(x = -4/3\). These are the critical points.

Step 2: Test Values in each Interval

We divide the number line into intervals using the critical points as boundaries. The intervals are \((-∞,-4/3)\), \((-4/3,2)\) and \((2,∞)\). Test a point in each of these intervals in the original inequality to determine whether the interval is in the solution set. For example, testing \(x=-2\) in \((-∞,-4/3)\) , \(x=0\) in \((-4/3, 2)\), and \(x=3\) in \((2,∞)\).

Step 3: Determine the Solution

If the result is true, then the interval is part of the solution set. For example, if we substitute \(-2\) into the inequality, we get \(\frac{4-2*(-2)}{3*(-2)+4} \leq 0\), which simplifies to \(\frac{8}{-2} \leq 0\). As \(-4 \leq 0\) is true, the interval \((-∞,-4/3)\) is part of the solution. Using the same procedure, we find that the interval \((-4/3, 2)\) also belongs to the solution set while \((2,∞)\) does not. Thus, the solution is \((-∞,2]\) in interval notation. Note that the 2 is included because the original inequality was \(\leq 0\). Therefore, if the value of x is such that the inequality exactly equals zero, it is included in the solution set.