How do solve (x+3)/(x-1)>=0 and write the answer as a inequality and interval notation?

Jun 13, 2017

Solution for $\frac{x + 3}{x - 1} \ge 0$ is the region $\left(- \infty , - 3\right] \cup \left[1. \infty\right)$ i.e. {x:x<=-3 or x>=1}

Explanation:

The sign of $\frac{x + 3}{x - 1}$ depends on the values $\left(x + 3\right)$ and $\left(x - 1\right)$ take at any point. It is apparent that change of sign will appear around $x = - 3$. Observe before that $\left(x + 3\right)$ is negative and after that it is positive. Similarly, before $x = 1$, $\left(x - 1\right)$ is negative and after that it is positive.

Now, leave the equality sign temporarily and let us divide the real number line in three regions. One $x < - 3$, then second region where we have $- 3 < x < 1$ and finally the third region $x > 1$.

In first region, we have $\left(x + 3\right)$ and $\left(x - 1\right)$ both negative and hence $\frac{x + 3}{x - 1}$ is positive and as we are looking at $\frac{x + 3}{x - 1} \ge 0$, this is a solution.

In second region, we have $\left(x + 3\right)$ as positive but $\left(x - 1\right)$ is negative and hence $\frac{x + 3}{x - 1}$ is negative and this region is not a solution.

In the third region, we have both $\left(x + 3\right)$ and $\left(x - 1\right)$ positive and hence $\frac{x + 3}{x - 1}$ is positive and so, this is a solution .

Hence solution for $\frac{x + 3}{x - 1} \ge 0$ is the region $\left(- \infty , - 3\right] \cup \left[1. \infty\right)$ i.e. {x:x<=-3 or x>=1}

Note : This method is known as sign chart and can be used to solve such sums involving both algebraic fractions (ratio of two one degree i.e. linear binomials as in the case above) as well as polynomials whose factors are one degree i.e. linear binomials.