Question

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

$\frac{5}{(\mathrm{x}-1)}\le \frac{4}{(x+2)}$

Short answer

Solution of rational inequality is$\mathrm{x}\in (-\infty ,-14]\cup (-2,1).$

Step-by-step solution

Step 1. Given Information

Given that the rational inequality is$\frac{5}{(\mathrm{x}-1)}\le \frac{4}{(\mathrm{x}+2)}.$

Step 2. Critical points

On taking terms on left hand side,

$$\begin{gathered} \frac{5}{(\mathrm{x}-1)}\le \frac{4}{(\mathrm{x}+2)} \\ \frac{5}{(\mathrm{x}-1)}-\frac{4}{(\mathrm{x}+2)}\le 0 \\ \frac{5}{(\mathrm{x}-1)}\times \frac{(\mathrm{x}+2)}{(\mathrm{x}+2)}-\frac{4}{(\mathrm{x}+2)}\times \frac{(\mathrm{x}-1)}{(\mathrm{x}-1)}\le 0 \\ \frac{5(\mathrm{x}+2)}{(\mathrm{x}-1)(\mathrm{x}+2)}-\frac{4(\mathrm{x}-1)}{(\mathrm{x}-1)(\mathrm{x}+2)}\le 0 \\ \frac{\mathrm{x}+14}{(\mathrm{x}-1)(\mathrm{x}+2)}\le 0 \\ (\mathrm{x}+14)=0 \\ \mathrm{x}=-14 \\ (\mathrm{x}-1)(\mathrm{x}+2)=0 \\ \mathrm{x}=1,-2 \end{gathered}$$

At this critical point, the inequality is undefined. Thus, it cannot be a solution set.

Step 3. Testing of points

To test the points between critical points,

Step 4. Testing

It can be expressed as,

$$\begin{gathered} \mathrm{At}\mathrm{x}=-15,\frac{-15+14}{(-15-1)(-15+2)}\le 0 \\ \frac{-1}{208}\le 0\mathrm{True} \\ \mathrm{At}\mathrm{x}=-5,\frac{-5+14}{(-5-1)(-5+2)}\le 0 \\ \frac{9}{18}\le 0\mathrm{False} \\ \mathrm{At}\mathrm{x}=-1,\frac{-1+14}{(-1-1)(-1+2)}\le 0 \\ \frac{-13}{2}\le 0\mathrm{True} \\ \mathrm{At}\mathrm{x}=2,\frac{2+14}{(2-1)(2+2)}\le 0 \\ 4\le 0\mathrm{False} \end{gathered}$$

Thus, the solution set is$\mathbf{x}\mathbf{\in }\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{-}\mathbf{14}\mathbf{]}\mathbf{\cup }\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{.}$