Question

solve the rational inequality. write the solution set in interval notation.\\(\frac{2}{3x - 1} > \frac{-5}{x}\\)\\(\\)\\(\\)the solution set is \\(\square\\).\\((\text{use integers or fractions for any numbers in the expression. type your answer in interval notation.})\\)

Explanation:

Step1: Move all terms to left side

To solve \(\frac{2}{3x - 1}>\frac{-5}{x}\), we first move \(\frac{-5}{x}\) to the left side: \(\frac{2}{3x - 1}+\frac{5}{x}>0\).

Step2: Find a common denominator

The common denominator of \(3x - 1\) and \(x\) is \(x(3x - 1)\). So we rewrite the left side: \(\frac{2x+5(3x - 1)}{x(3x - 1)}>0\).

Step3: Simplify the numerator

Simplify the numerator \(2x + 5(3x-1)=2x+15x - 5=17x - 5\). So the inequality becomes \(\frac{17x - 5}{x(3x - 1)}>0\).

Step4: Find critical points

Set the numerator and denominator equal to zero:

• Numerator: \(17x-5 = 0\Rightarrow x=\frac{5}{17}\)
• Denominator: \(x = 0\) or \(3x - 1=0\Rightarrow x=\frac{1}{3}\)

These critical points divide the number line into intervals: \((-\infty,0)\), \((0,\frac{5}{17})\), \((\frac{5}{17},\frac{1}{3})\), \((\frac{1}{3},\infty)\)

Step5: Test intervals

• For \(x\in(-\infty,0)\), let \(x=-1\). Then \(\frac{17(-1)-5}{(-1)(3(-1)-1)}=\frac{-22}{(-1)(-4)}=\frac{-22}{4}<0\), not satisfying the inequality.
• For \(x\in(0,\frac{5}{17})\), let \(x=\frac{1}{17}\). Then \(\frac{17(\frac{1}{17})-5}{(\frac{1}{17})(3(\frac{1}{17})-1)}=\frac{1 - 5}{(\frac{1}{17})(\frac{3 - 17}{17})}=\frac{-4}{(\frac{1}{17})(\frac{-14}{17})}=\frac{-4\times17\times17}{-14}>0\), satisfying the inequality.
• For \(x\in(\frac{5}{17},\frac{1}{3})\), let \(x=\frac{1}{5}\). Then \(\frac{17(\frac{1}{5})-5}{(\frac{1}{5})(3(\frac{1}{5})-1)}=\frac{\frac{17 - 25}{5}}{(\frac{1}{5})(\frac{3 - 5}{5})}=\frac{\frac{-8}{5}}{(\frac{1}{5})(\frac{-2}{5})}=\frac{-8\times5\times5}{-2}<0\), not satisfying the inequality.
• For \(x\in(\frac{1}{3},\infty)\), let \(x = 1\). Then \(\frac{17(1)-5}{(1)(3(1)-1)}=\frac{12}{2}=6>0\), satisfying the inequality.

Answer:

\((0,\frac{5}{17})\cup(\frac{1}{3},\infty)\)