Question

In the following exercises, solve each rational inequality and write the solution in interval notation.

$\frac{4}{(\mathrm{x}-2)}\ge \frac{3}{(x+1)}$

Short answer

The solution of rational inequality is $\mathrm{x}\in [-10,-1)\cup (2,\infty )$.

Step-by-step solution

Step 1. Given Information

The given rational inequality is$\frac{4}{(\mathrm{x}-2)}\ge \frac{3}{(x+1)}.$

Step 2. Simplification

On taking the terms on left hand side,

$$\begin{gathered} \frac{4}{(\mathrm{x}-2)}-\frac{3}{(\mathrm{x}+1)}\ge \frac{3}{(\mathrm{x}+1)}-\frac{3}{(\mathrm{x}+1)} \\ \frac{4}{(\mathrm{x}-2)}-\frac{3}{(\mathrm{x}+1)}\ge 0 \\ \mathrm{LCD}\mathrm{of}\left[\right(\mathrm{x}-2),(\mathrm{x}+1\left)\right]=(\mathrm{x}+1)(\mathrm{x}-2). \end{gathered}$$

Making the denominator,

localid="1662724842006" $$\begin{gathered} \frac{4}{(\mathrm{x}-2)}\times \frac{(\mathrm{x}+1)}{(\mathrm{x}+1)}-\frac{3}{(\mathrm{x}+1)}\times \frac{(\mathrm{x}-2)}{(\mathrm{x}-2)}\ge 0 \\ \frac{\mathrm{x}+10}{(\mathrm{x}+1)(\mathrm{x}-2)}\ge 0 \end{gathered}$$

Step 3. Critical points

To find the critical points,

$$\begin{gathered} \mathrm{x}+10=0 \\ \mathrm{x}=-10 \\ (\mathrm{x}+1)(\mathrm{x}-2)=0 \\ \mathrm{x}=-2,1 \end{gathered}$$

These three critical points cannot be in solution set and undefined inequality. On testing the points,

Step 4. Testing of critical points

To test points between critical points,

$$\begin{gathered} \mathrm{At}\mathrm{x}=-15,\frac{-15+10}{(-15+1)(-15-2)}\ge 0 \\ \frac{-5}{238}\ge 0\mathrm{False} \\ \mathrm{At}\mathrm{x}=-5,\frac{-5+10}{(-5+1)(-5-2)}\ge 0 \\ \frac{5}{28}\ge 0\mathrm{True} \\ \mathrm{At}\mathrm{x}=1,\frac{1+10}{(1+1)(1-2)}\ge 0 \\ \frac{-11}{2}\ge 0False \\ \mathrm{At}\mathrm{x}=5,\frac{5+10}{(5+1)(5-2)}\ge 0 \\ \frac{5}{6}\ge 0\mathrm{True} \end{gathered}$$

Thus, the solution set is$\mathbf{x}\mathbf{\in }\mathbf{[}\mathbf{-}\mathbf{10}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{\cup }\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{\infty }\mathbf{)}\mathbf{.}$