Question

Solve each inequality algebraically.

$\frac{x-1}{x+2}\ge -2$

Short answer

Solution set and the interval is:
$$\begin{gathered} \left\{x\right|x<-2or1\le x\le \infty \left\} \\ \right(-\infty ,-2)\cup [1,\infty ) \end{gathered}$$

Step-by-step solution

Step 1. Change the inequality to equal to zero. 

Change the inequality equal to zero to make it easier and then get the value of x
First arrange the inequality,

$\begin{array}{r}\frac{x-1}{x+2}\ge -2\\ \frac{x-1}{x+2}+2\ge 0\\ \frac{x-1}{x+2}+2\cdot \frac{x+2}{x+2}\ge 0\\ \frac{x-1+2x+4}{x+2}\ge 0\\ \frac{3x-3}{x+2}\ge 0\end{array}$

Make it equal to zero.

$\begin{array}{r}\frac{3x-3}{x+2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ (3x-3)(x+2)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ 3x-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0;x+2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ 3x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3;x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-2\\ x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{3}{3}\\ x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1\end{array}$

Step 2. Form the intervals

From the obtained values of x, we can form the interval,

So the interval we have is:

$(-\infty ,-2)\cup (-2,1)\cup (1,\infty )$

Step 3. Form the table

Since, $f\left(x\right)\ge 0$, so is positive or zero.

check a number in each interval and evaluate the function to see whether it is satisfying the function.

Interval	Number chosen	Resultant
$(-\infty ,-2)$	-3	$12\ge 0$
$(-2,1)$	0	$-\frac{3}{2}\ge 0$
$(1,\infty )$	2	$\frac{3}{4}\ge 0$

Step 4. Solution set and interval

Since, $12\ge 0,\frac{3}{4}\ge 0$satisfy $f$. Thus,

role="math" localid="1646208744522" $$\begin{gathered} \left\{x\right|x<-2or1\le x\le \infty \left\} \\ \right(-\infty ,-2)\cup [1,\infty ) \end{gathered}$$