Question

Solve the following equation and find the value of x?
$\dfrac{{x + 3}}{{x - 1}} > 0,x \in R$

Answer 1

Hint:To solve this type of question first we have to define the function that means we have to check where the function is defined and in that range only we have to solve. Here we have to find a range of x for which the fraction is greater than 0.

Complete step-by-step answer:

For defining this question, the denominator should not be equal to 0.
$
  x - 1 \ne 0 \\
  x \ne 1 \\
 $
We have to find a range of x for which this fraction is positive so if numerator and denominator both have the same sign then it will always be greater than 0.
Case 1. When both numerator and denominator both are positive.
$
  x + 3 > 0 \\
   = x > - 3 \\
$
And
$
  x - 1 > 0 \\
   = x > 1 \\
 $
We have to take the range of x for which both conditions are satisfied. That means intersection of both x
So, $x \in \left( {1,\infty } \right)$
Case 2. When both numerator and denominator are negative
$x + 3 < 0 \Rightarrow x < - 3$
And
$x - 1 < 0 \Rightarrow x < 1$
Now again we have to satisfy both the conditions that means take intersection of both.
$x \in \left( { - \infty , - 3} \right)$
Hence final answer of function where it is greater than 0 is
$x \in \left( { - \infty , - 3} \right) \cup \left( {1,\infty } \right)$

Note: Whenever we get this type of question the key concept of solving is we have to find a range where both numerator and denominator is positive and both is negative then we have to take intersection so that both the function should satisfy that condition. And also care about where function is defined or not.