Inequalities with fractions
Task number: 2758
Solve the following inequalities over the real domain:
Variant 1
\( \frac{x+1}{x-2}\ge 0 \)
Resolution
The expresssion is undefined for \(x=2\).
- \(x<2: \longrightarrow x+1\le 0 \longrightarrow x\le-1 \longrightarrow x\in (-\infty,-1\rangle\)
- \(x>2: \longrightarrow x+1\ge 0 \longrightarrow x\ge-1 \longrightarrow x\in (2,\infty)\)
Result
The solution is \(x\in (-\infty,-1\rangle \cup (2,\infty)\).
Variant 2
\( \frac{x-2}{2x-8}\ge 1 \)
Resolution
The expression is undefined for \(2x-8=0\) or \(x=4\).
- \(x< 4: x-2\le 2x-8 \longrightarrow x\ge 6 \longrightarrow \) has no solution
- \(x> 4: x-2\ge 2x-8 \longrightarrow x\le 6 \longrightarrow x\in (4{,}6\rangle\)
Result
The solution is \(x\in (4{,}6\rangle\).
Variant 3
\( \frac{x+3}{x-1}\ge \frac{x+1}{x-5} \)
Solution
The expression is undefined for \(x=1\) or \(x=5\). Multiply the inequality by both denominators.
- for \(x\lt 1\) are both negative, the inequality will not turn: \((x+3)(x-5)\ge x^2-1 \Rightarrow x^2-2x-15\ge x^2-1 \Rightarrow -2x\ge 14 \Rightarrow x\le -7 \Rightarrow x\in(-\infty,-7\rangle\)
- for \(1\lt x\lt 5\) is \(x-1\) positive and \(x-5\) is negative, the inequality will turn: \((x+3)(x-5)\le x^2-1 \Rightarrow x\ge -7 \Rightarrow x\in(1{,}5)\)
- for \(x\gt 5\) are both positive \((x+3)(x-5)\ge x^2-1 \Rightarrow x\le -7 \Rightarrow \) no solution
Result
The solution is \(x\in(-\infty,-7\rangle \cup (1{,}5)\).
Variant 4
\( \frac1{x+2}<\frac{x}{x-1} \)
Hint
The expression is undefined for \(x=-2\) or \(x=1\).
- \(x<-2: x-1< x(x+2) \longrightarrow x^2+x+1 > 0 \longrightarrow x_{1{,}2}=\frac{-1\pm \sqrt{1-4}}{2} \longrightarrow x\in(-\infty,-2)\)
- \(-2< x<1: x-1> x(x+2) \longrightarrow x^2+x+1 < 0 \longrightarrow\) has no solution
- \(x>1: x-1< x(x+2) \longrightarrow x^2+x+1 > 0 \longrightarrow x\in (1,\infty)\)
Resolution
The solution is \(x\in(-\infty,-2)\cup(1,\infty)\).
