Question

Solve each inequality algebraically.

$\frac{x+1}{x-3}\le 2$

Short answer

Solution set and the interval is:
$$\begin{gathered} \left\{x\right|x<3orx\ge 7\left\} \\ \right(-\infty ,3)\cup [7,\infty ) \end{gathered}$$

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

Arrange the inequality first,

$$\begin{gathered} \frac{x+1}{x-3}\le 2 \\ \frac{x+1}{x-3}-2\le 0 \\ \frac{x+1}{x-3}-2·\frac{x-3}{x-3}\le 0 \\ \frac{x+1}{x-3}-\frac{2x-6}{x-3}\le 0 \\ \frac{x+1-2x+6}{x-3}\le 0 \\ \frac{-x+7}{x-3}\le 0 \end{gathered}$$

Make it equal to zero,

$$\begin{gathered} \frac{-x+7}{x-3}=0 \\ -x+7=0;x-3=0 \\ x=7;x=3 \end{gathered}$$

Step 2. Form the intervals

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

So the interval we have is:

$(-\infty ,3)\cup (3,7)\cup (7,\infty )$

Step 3. Form the table

Since, $f\left(x\right)<0$, so is negative.

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

Interval	Number chosen	Resultant
$(-\infty ,3)$	1	$-3\le 0$
$(3,7)$	4	$3\le 0$
$(7,\infty )$	8	$-\frac{1}{5}\le 0$

Step 4. Solution set and interval

Since,

$-3\le 0,-\frac{1}{5}\le 0$satisfy $f$. Thus,

$$\begin{gathered} \left\{x\right|x<3orx\ge 7\left\} \\ \right(-\infty ,3)\cup [7,\infty ) \end{gathered}$$