First, find restriction(s):

[latex]x-3 \ne 0[/latex]

[latex]x \ne 3[/latex]

Second, let’s rewrite the inequality as an equation and then solve it:

[latex]\frac{5x}{x-3}=2[/latex]

[latex]\frac{5x}{x-3} = \frac{2}{1}[/latex]

[latex]5x=2(x-3)[/latex]

[latex]5x=2x-6[/latex]

[latex]3x=-6[/latex]

[latex]x=-2[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an open circle. Since the inequality has “[latex]\leq[/latex]” (less than or equal to), which has “[latex]=[/latex]” (equal sign), the solution should be a closed circle:

Note that now there are three intervals on the number line: [latex](-\infty, -2], [-2, 3),[/latex] and [latex](3, \infty)[/latex].

Fourth, choose your choice of test value from each interval:

I chose [latex]-3, 0,[/latex] and [latex]4[/latex] from those three intervals.

fifth, test each interval using the test value:

When [latex]x=-3[/latex], [latex]\frac{5(-3)}{(-3)-3} \leq 2 \rightarrow \frac{-15}{-6} \leq 2 \rightarrow 2.5 \leq 2[/latex] (False)  [latex]\therefore[/latex] not a solution

When [latex]x=0[/latex], [latex]\frac{5(0)}{(0)-3} \leq 2 \rightarrow \frac{0}{-3} \leq 2 \rightarrow 0 \leq 2[/latex] (True)  [latex]\therefore[/latex] solution

When [latex]x=4[/latex], [latex]\frac{5(4)}{(4)-3} \leq 2 \rightarrow \frac{20}{1} \leq 2 \rightarrow 20 \leq 2[/latex] (False)  [latex]\therefore[/latex] not a solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{5x}{x-3}-2[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{5x}{x-3} \leq 2[/latex] is [latex][-2, 3)[/latex].

First, find restriction(s):

[latex]x+1 \ne 0[/latex]

[latex]x \ne -1[/latex]

Second, let’s rewrite the inequality as an equation and then solve it:

[latex]\frac{4x-5}{x+1}=0[/latex]

[latex]\frac{4x-5}{x+1} = \frac{0}{1}[/latex]

[latex]4x-5=0[/latex]

[latex]4x=5[/latex]

[latex]x=\frac{5}{4}[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an open circle. Since the inequality has “[latex]\geq[/latex]” (greater than or equal to), which has “[latex]=[/latex]” (equal sign), the solution should be a closed circle:

Note that now there are three intervals on the number line: [latex](-\infty, -1), (-1, \frac{5}{4}],[/latex] and [latex][\frac{5}{4}, \infty)[/latex].

Fourth, choose your choice of test value from each interval:

I chose [latex]-2, 0,[/latex] and [latex]2[/latex] from those three intervals.

fifth, test each interval using the test value:

When [latex]x=-2[/latex], [latex]\frac{4(-2)-5}{(-2)+1} \geq 0 \rightarrow \frac{-13}{-1} \geq 0 \rightarrow 13 \geq 0[/latex] (True)  [latex]\therefore[/latex] solution

When [latex]x=0[/latex], [latex]\frac{4(0)-5}{(0)+1} \geq 0 \rightarrow \frac{-5}{1} \geq 0 \rightarrow -5 \geq 0[/latex] (False)  [latex]\therefore[/latex] not a solution

When [latex]x=2[/latex], [latex]\frac{4(2)-5}{(2)+1} \geq 0 \rightarrow \frac{3}{3} \geq 0 \rightarrow 1 \geq 0[/latex] (True)  [latex]\therefore[/latex] solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{4x-5}{x+1}[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{4x-5}{x+1} \geq 0[/latex] is [latex](-\infty, -1) \cup [\frac{5}{4}, \infty)[/latex].