# Inequality with Variables in the Denominator

Solving Inequality with Variables in the Denominator requires special cares due to the direction of the inequalities. Let’s have a look at the following key points.

### Key Point 1

\begin{aligned} \displaystyle \require{color} \frac{1}{x} &\ge 2 \\ \frac{1}{x} \times x &\ge 2 \times x &\color{green} \text{Many of you may think this is TRUE.} \\ &&\color{green} \text{This is TRUE only if } x \gt 0. \\ \frac{1}{x} \times x &\le 2 \times x &\color{green} \text{But the inequality changes if } x \lt 0. \\ \frac{1}{x} \times x^2 &\ge 2 \times x^2 &\color{green} \text{This is always TRUE as } x^2 \ge 0. \\ \end{aligned} \\
Make sure to multiply the square of the denominator.

### Key Point 2

\begin{aligned} \displaystyle \require{color} \frac{1}{x} &\ge 2 &\color{green} x \ne 0 \ \because \text{denominator} \ne 0 \cdots (1) \\ \frac{1}{x} \times x^2 &\ge 2 \times x^2 \\ x &\ge 2x^2 \\ 2x^2 -x &\le 0 \\ x(2x-1) &\le 0 \\ 0 &\le x \le \frac{1}{2} \\ \therefore 0 &\lt x \le \frac{1}{2} &\color{green} \text{by } (1) \\ \end{aligned} \\
It is important to exclude some values to make the fraction undefined.

### Key Point 3

\begin{aligned} \displaystyle \require{color} \frac{2}{x-3} &\gt 1 \\ \end{aligned} \\

##### Method 1

\begin{aligned} \displaystyle \require{color} \frac{2}{x-3} \times (x-3)^2 &\gt 1 \times (x-3)^2 \\ 2(x-3) &\gt (x-3)^2 \\ 2x-6 &\gt x^2 – 6x + 9 &\color{red} \text{Not a great idea, but still working} \\ x^2 – 6x + 9 – 2x + 6 &\lt 0 \\ x^2 – 8x + 15 &\lt 0 \\ (x-3)(x-5) &\lt 0 \\ \therefore 3 \lt x &\lt 5 \\ \end{aligned} \\

##### Method 2

\begin{aligned} \displaystyle \require{color} \frac{2}{x-3} \times (x-3)^2 &\gt 1 \times (x-3)^2 \\ 2(x-3) &\gt (x-3)^2 \\ (x-3)^2 – 2(x-3) &\lt 0 &\color{green} \text{Good idea to keep them simple} \\ (x-3)\big[(x-3) – 2\big] &\lt 0 \\ (x-3)(x-5) &\lt 0 \\ \therefore 3 \lt x &\lt 5 \\ \end{aligned} \\
It is recommended not expanding the brackets, as these may cause unexpected mistakes. I found lot of students made silly mistakes while expanding and factorising them.

Now, let’s practice with the following worked examples by ensuring these 3 key points. Enjoy!

### Question 1

Solve $\displaystyle \frac{2x+1}{x-2} \gt 1$.

\begin{aligned} \displaystyle \require{color} \frac{2x+1}{x-2} \times (x-2)^2 &\gt 1 \times (x-2)^2 \\ (2x+1)(x-2) &\gt (x-2)^2 \\ (2x+1)(x-2) – (x-2)^2 &\gt 0 \\ (x-2)\big[(2x+1) – (x-2)\big] &\gt 0 \\ (x-2)(2x+1-x+2) &\gt 0 \\ (x-2)(x+3) &\gt 0 \\ \therefore x \lt -3 \text{ or } x &\gt 2 \\ \end{aligned} \\

### Question 2

Solve $\displaystyle x-4 \le \frac{5}{x}$.

\begin{aligned} \displaystyle \require{color} (x-4) \times x^2 &\lt \frac{5}{x} \times x^2 \\ (x-4)x^2 &\lt 5x \\ x^3 – 4x^2 – 5x &\lt 0 \\ x(x^2 – 4x – 5) &\lt 0 \\ x(x-3)(x+2) &\lt 0 \\ \therefore x \lt -1 \text{ or } 0 &\lt x \lt 5 \\ \end{aligned} \\

### Question 3

Solve $\displaystyle \frac{x(x-3)}{x-2} \gt 2$.

\begin{aligned} \displaystyle \require{color} \frac{x(x-3)}{x-2} \times (x-2)^2 &\gt 2 \times (x-2)^2 \\ x(x-3)(x-2) &\gt 2(x-2)^2 \\ x(x-3)(x-2) – 2(x-2)^2 &\gt 0 \\ (x-2)[x(x-3)-2(x-2)] &\gt 0 \\ (x-2)(x^2-3x-2x+4) &\gt 0 \\ (x-2)(x^2 -4x+4) &\gt 0 \\ (x-2)(x-1)(x-4) &\gt 0 \\ \therefore 1 \lt x \lt 2 \text{ or } x &\lt 4 \\ \end{aligned} \\

### Question 4

Solve $\displaystyle \frac{x^2-4x+3}{x+2} \le 0$.

\begin{aligned} \displaystyle \require{color} \frac{x^2-4x+3}{x+2} \times (x+2)^2 &\le 0 \times (x+2)^2 \\ (x^2-4x+3)(x+2) &\le 0 \\ (x-1)(x-3)(x+2) &\le 0 \\ x \le -2 \text{ or } 1 &\le x \le 3 \\ \therefore x \lt -2 \text{ or } 1 &\le x \le 3 &\color{green} x \ne -2 \\ \end{aligned} \\

### Question 5

Solve $\displaystyle \frac{1}{3x} \ge \frac{1}{x+2}$.

\begin{aligned} \displaystyle \require{color} \frac{1}{3x} \times (3x)^2 (x+2)^2 &\ge \frac{1}{x+2} \times (3x)^2 (x+2)^2 &\color{red} 3x \le x+2 \text{ is incorrect} \\ 3x (x+2)^2 &\ge (3x)^2 (x+2) \\ 3x (x+2)^2 – (3x)^2 (x+2) &\ge 0 &\color{green} \text{Do not expand, but factorise} \\ 3x(x+2)(x+2-3x) &\ge 0 \\ 3x(x+2)(2-2x) &\ge 0 \\ x(x+2)(x-1) &\le 0 &\color{green} \text{multiply both sides by } -\frac{1}{6} \\ x \le -2 \text{ or } 0 &\le x \le 1 \\ \therefore x \lt -2 \text{ or } 0 &\lt x \le 1 &\color{green} x \ne -2,0 \\ \end{aligned} \\

Try to solve this question, $\displaystyle \frac{x^2-4}{3-x} \le 0$.
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