Question

Solve$\frac{x+2}{x-3}<2$

Short answer

The solution set of inequality is$\left(-\infty ,3\right)\cup \left(8,\infty \right)$

Step-by-step solution

Step 1. Given Information

The given inequality is$\frac{x+2}{x-3}<2$

Step 2. Explanation

Solve the given inequality,

$$\begin{gathered} \frac{x+2}{x-3}-2<0 \\ \frac{x+2}{x-3}-2\left(\frac{x-3}{x-3}\right)<0 \\ \frac{8-x}{x-3}<0 \end{gathered}$$

To determine the real zero of function, we will find the value of x where $f\left(x\right)=0$

$$\begin{gathered} \frac{8-x}{x-3}=0 \\ 8-x=0 \\ x=8 \end{gathered}$$

Thus, real zero of function is 8.

Also function is not defined at$x=3$

Step 3. Calculation

Construct the table by dividing the real number line into 3 intervals.

Thus, the required solution set is$\left(-\infty ,3\right)\cup \left(8,\infty \right)$