Question

How do you solve \[\dfrac{x+3}{x-1}\ge 0\] and write the answer as an inequality and interval notation?

Answer 1

Hint: Consider two cases for solving the above inequality. In the first condition assume that \[x+3\ge 0\] and find the condition for the denominator so that the overall inequality becomes greater than 0. Now, in the second condition assume that \[x+3\le 0\] and find the condition for the denominator so that the overall inequality becomes greater than 0. Solve for the set of values of x obtained in both the conditions and hence take their union to get the answer.

Complete step by step solution:
Here, we have been provided with the inequality \[\dfrac{x+3}{x-1}\ge 0\] and we are asked to solve this inequality and write the answer in both inequality and interval notation.
Now, for the inequality \[\dfrac{x+3}{x-1}\ge 0\] to exist the denominator cannot be 0, so we have \[x\ne 1\]. Now, we can see that the given inequality states that the expression \[\dfrac{x+3}{x-1}\] should be greater than or equal to 0. For this to happen, both the numerator and the denominator must have same sign. So, there can be two possible cases. Let us check them one – by – one.
1. Case (i): - Both numerator and denominator are greater than 0, so we have,
\[\Rightarrow x+3\ge 0\] and \[x-1>0\]
\[\Rightarrow x\ge -3\] and \[x>1\]
Since, we have to satisfy both the conditions, so we need to take the intersection of the above two sets. On taking the intersection we will get \[x>1\].
2. Case (ii): - Both numerator and denominator are smaller than 0, so we have,
\[\Rightarrow x+3\le 0\] and \[x-1<0\]
\[\Rightarrow x\le -3\] and \[x<1\]
Here, also we need to satisfy both the conditions, so we need to take the intersection of the above two sets. On taking the intersection we will get, \[x\le -3\].
Now, since both the cases are possible so either of them can be our solution that means of x obtained. So, on taking the union of the sets of values of x obtained in the above two cases, we get,
\[\Rightarrow x\le -3\] or \[x>1\]
\[\Rightarrow x\in \left( -\infty ,-3 \right]\cup \left( 1,\infty \right)\]
Hence, the above two relations are the answer in the inequality and interval notation respectively.

Note: One may note that there can be two more cases for the above inequality. In these cases we will consider the signs of the numerator and denominator opposite to each other. But such a case will not give any solution set for the value of x because if numerator and denominator will be of opposite signs then the inequality will become less than, which is not our question. You may remember the direct formula for the solution of inequality \[\dfrac{x-a}{x-b}\ge 0\] given as \[x\le a\] or \[x>b\], if \[b>a\] and \[x b\].