## College Algebra (6th Edition)

$(-∞, -5) ∪ (-3, ∞)$
Consider the Rational Inequality as follows: $\frac{x+1}{x+3}<2$ 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 2 from both sides to obtain zero on the right. $\frac{x+1}{x+3}<2$ $\frac{(x+1)}{x+3}-2<0$ $\frac{-x-5}{x+3}<0$ 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. $-x-5 = 0$ This implies $x =-5$ and $x+3=0$ This implies $x =-3$ 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: $(-∞, -5), (-5, -3), (-3, ∞)$ Step 4. Now, one test value within each interval is chosen and $f$ is evaluated at that number. Intervals: $(-∞, -5), (-5, -3), (-3, ∞)$ Test value: $-6$ $-4$ $-2$ 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. We are interested in solving $f (x) < 0$. Based on our work done in Step 4, we see that $f (x) < 0$ for all x in is $(-∞, -5)$ or $(-3, ∞)$ . Conclusion: Thus, the interval notation of the given inequality is $(-∞, -5) ∪ (-3, ∞)$ and the graph of the solution set on a number line is shown as follows: