## College Algebra (6th Edition)

$(-∞,\frac{1}{2})∪ (\frac{7}{5}, ∞)$
Consider the Rational Inequality as follows: $\frac{x+4}{2x-1}<3$ Here are the steps required for Solving Rational Inequalities: Step 1: One side must be zero and the other side can have only one fraction, so we simplify the fractions if there is more than one fraction. Let us subtract 3 from both sides to obtain zero on the right. $\frac{x+4}{2x-1}<3$ $\frac{x+4}{2x-1}-3<0$ $\frac{x+4-3(2x-1)}{2x-1}<0$ $\frac{-5x+7}{2x-1}<0$ Step 2: Critical or Key Values are first evaluated. In order to this, set the numerator and denominator of the fraction equal to zero and then simplified rational inequality is solved. $-5x+7 = 0$ This implies $x =7/5$ And $2x-1=0$ This implies $x = 1/2$ These solutions are used as boundary points on a number line. Step 3: Locate the boundary points on a number line found in Step 2 to divide the number line into intervals. The boundary points divide the number line into three intervals: $(-∞,\frac{1}{2}), (\frac{1}{2}, \frac{7}{5}), (\frac{7}{5}, ∞)$ Step 4: Now, one test value within each interval is chosen and $f$ is evaluated at that number. Intervals: $(-∞,\frac{1}{2}), (\frac{1}{2}, \frac{7}{5}), (\frac{7}{5}, ∞)$ Test value: $0$ $1$ $1.5$ Sign Change: Negative Positive Negative $f (x)< 0?$: T F T Step 5: Write the solution set, selecting the interval or intervals that satisfy the given inequality. Solve $f (x)< 0$. Based on our work done in Step 4, we see that $f (x)< 0$ for all x in $(-∞,\frac{1}{2})$or$(\frac{7}{5}, ∞)$ Conclusion: The interval notation of the given inequality is $(-∞,\frac{1}{2})∪ (\frac{7}{5}, ∞)$ and the graph of the solution set on a number line is shown as follows: