Solving Inequality with Variables in the Denominator requires special care 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 to expand the brackets, as these may cause unexpected mistakes. I found a lot of students made silly mistakes while expanding and factorising them.

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

Question 1

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

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

Question 2

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

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

Question 3

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

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

Question 4

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

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

Question 5

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

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

It’s your turn now!
Try to solve this question, \( \displaystyle \frac{x^2-4}{3-x} \le 0 \).
Please feel free to let us know if you need any help on this 🙂

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