Chapter 4.4, Problem 34AYU

To determine

To calculate:

The solution of following inequality equation:

  $\frac{\left(x-3\right)}{\left(x+1\right)}>0$

Answer to Problem 34AYU

The solution of the inequality equation $\frac{\left(x-3\right)}{\left(x+1\right)}>0$ is $x=3$ .

Explanation of Solution

Given information:

  $\frac{\left(x-3\right)}{\left(x+1\right)}>0$

Calculation:

The given equation is:

  $\begin{array}{l}\frac{\left(x-3\right)}{\left(x+1\right)}>0\\ x-3=0OR\text{}x+1=0\\ x=3x=-1\end{array}$

Need the inequality value which is less than zero, from the above solution $x>0,3>0$ .

Hence, the inequality is solved for $x=-1,3$ .