Question

$$\frac{5x - 3}{(x - 1)(x + 1)} \leq 0$$
type in true or false for each statement.
for test point $x = -2$, the inequality is true.
for test point $x = 0$, the inequality is false.
for test point $x = \frac{4}{5}$, the inequality is true.
for test point $x = 2$, the inequality is false.
identify the graph of the solution set.
(options with number lines from -2 to 2, with different markings)

Explanation:

Step1: Find critical points

Set numerator and denominator equal to zero.
For numerator \(5x - 3=0\), we get \(x=\frac{3}{5}\).
For denominator \((x - 1)(x + 1)=0\), we get \(x = 1\) and \(x=-1\).

Step2: Test intervals

• Interval \((-\infty,-1)\): Let \(x=-2\), \(\frac{5(-2)-3}{(-2 - 1)(-2+1)}=\frac{-10 - 3}{(-3)(-1)}=\frac{-13}{3}\leq0\) (True).
• Interval \((-1,\frac{3}{5})\): Let \(x = 0\), \(\frac{5(0)-3}{(0 - 1)(0+1)}=\frac{-3}{-1}=3>0\) (False).
• Interval \((\frac{3}{5},1)\): Let \(x=\frac{4}{5}\), \(\frac{5(\frac{4}{5})-3}{(\frac{4}{5}-1)(\frac{4}{5}+1)}=\frac{4 - 3}{(-\frac{1}{5})(\frac{9}{5})}=\frac{1}{-\frac{9}{25}}\leq0\) (True).
• Interval \((1,\infty)\): Let \(x = 2\), \(\frac{5(2)-3}{(2 - 1)(2+1)}=\frac{10 - 3}{(1)(3)}=\frac{7}{3}>0\) (False).

Also, \(x=-1\) and \(x = 1\) make the denominator zero (excluded), \(x=\frac{3}{5}\) is included.

Answer:

The graph with a closed - circle at \(x=\frac{3}{5}\), open - circles at \(x=-1\) and \(x = 1\), and the intervals \((-\infty,-1)\) and \((\frac{3}{5},1)\) shaded.