Question

Solve the following inequation and graph the solution set on the number line: $\frac{x+3}{x-2}\le 2,x>2$.

Answer 1

Step 1: Given inequations: $\frac{x+3}{x-2}\le 2,x>2$.

Consider the inequality: $\frac{x+3}{x-2}\le 2$.

Subtract $2$ to both sides of the inequality.

$$\begin{gathered} \frac{x+3}{x-2}-2\le 2-2 \\ \frac{x+3-2x+4}{x-2}\le 0 \\ \frac{-x+7}{x-2}\le 0 \\ -\frac{x-7}{x-2}\le 0 \end{gathered}$$

Step 2: Multiply $\left(-1\right)$ to both sides of the inequality and change the sign of inequality.

$$\begin{gathered} -\frac{x-7}{x-2}\times \left(-1\right)\le 0\times \left(-1\right) \\ \frac{x-7}{x-2}\ge 0 \end{gathered}$$

Step 3: Solve the inequality $\frac{x-7}{x-2}\ge 0$, we need to consider that 2 different cases.

Case-1:

$x-7\ge 0$ and $x-2>0$

or, $x\ge 7$ and $x>2$

or, $x\ge 7$

Case-2:

$x-7\le 0$ and $x-2<0$

or, $x\le 7$ and $x<2$

or, $x<2$

Thus, the solution set of the given inequality, $\left(-\infty ,2\right)\cup [7,\infty )$

Step 4: Represent the solution set on the number line

Therefore, the solution set of $\left(-\infty ,2\right)\cup [7,\infty )$ and $x>2$ can be given by $[7,\infty )$.

The solution set $[7,\infty )$ can be plotted on the number line as,