Question

Solve the given inequality algebraically.

$\frac{x+1}{x-1}>0$

Short answer

Required solution set is$(-\infty ,-1)\cup (1,\infty )$

Step-by-step solution

Step 1. Given information 

The given inequality is :

$\frac{x+1}{x-1}>0$

We have to solve for x.

Step 2. Finding zeroes 

Zeroes of the inequality $f\left(x\right)=\frac{x+1}{x-1}>0$are

$\frac{x+1}{x-1}=0$

$f\left(x\right)$is undefined for $x=1$

Therefore$x=-1$

Step 3. Dividing real number line into 3 intervals 

Now we use the zeroes to separate the real number line into intervals$(-\infty ,-1),(-1,1),(1,\infty )$

Step 4. Selecting a test number in each interval  

Now we select a test number in each interval found in Step 3 and evaluate at each number to determine if

$f\left(x\right)=\frac{x+1}{x-1}=0$is positive or negative

In the interval $(-\infty ,-1)$, we chose -2 where f is positive

In the interval $(-1,1)$, we chose 0 where f is negative

In the interval $(1,\infty )$, we chose 2 where f is positive

We know that our required inequality is $f\left(x\right)>0$

Here the inequality is not strict $(\ge or\le )$so we have to exclude the solutions of $f\left(x\right)=0$in the solution set.

So we want the interval where f is positive.

So required solution set is$(-\infty ,-1)\cup (1,\infty )$