## Intermediate Algebra (6th Edition)

$\color{blue}{(-3.5, -1)}$
Solve the equation $\dfrac{5}{x+1}=-2$ to obtain: \begin{array}{ccc} \require{cancel} &\dfrac{5}{x+1}&=&-2 \\&(x+1)\cdot \dfrac{5}{x+1}&=&-2(x+1) \\&\cancel{(x+1)}\cdot \dfrac{5}{\cancel{x+1}}&=&-2x-2 \\&5&=&-2x-2 \\&5+2&=&-2x-2+2 \\&7&=&-2x \\&\frac{7}{-2} &= &\frac{-2x}{-2} \\&-\frac{7}{2} &= &x \\&-3.5&=&x \end{array} Find the number that will make the $\dfrac{5}{x+1}$ undefined. Note that when $x=-1$ the denominator becomes $0$, making the expression undefined.. Thus, the numebrs $-3.5$ and $-1$ are the critical points of the given inequality. These numbers divide the number line into three parts: $(-\infty, -3.5)$, $(-3.5, -1)$, and $(-1, +\infty)$ Pick a test point from each interval/part, and test if it satisfies the given inequality. For $(-\infty, -3.5)$, use the test point $-4$ and substitute it into the given inequality: $\dfrac{5}{x+1}\lt -2 \\\dfrac{5}{-4+1} \lt -2 \\\dfrac{5}{-3} \lt -2 \\-1.\overline{6} \not \lt -2$ Thus, the numbers in this interval/part are not solutions to the given inequality. For $(-3.5, -1)$, use the test point $-2$ and substitute it into the given inequality: $\dfrac{5}{x+1}\lt -2 \\\dfrac{5}{-2+1}\lt -2 \\\dfrac{5}{-1} \lt -2 \\-5 \lt -2$ This statement is true therefore the numbers within this interval are solutions to the given inequality. For $(-1, +\infty)$, use the test point $0$ and substitute it into the given inequality: $\dfrac{5}{x+1} \lt -2 \\\dfrac{5}{0+1} \lt -2 \\\dfrac{5}{1}\lt -2 \\5 \not \lt -2$ Thus, the numbers in this interval are not solutions to the given inequality. $-3.5$ is not a solution since the inequality involves strictly less than. $-1$ is not a solution because it makes an expression undefined. Therefore, the solution to the given inequality is: $\color{blue}{(-3.5, -1)}$.