Inequality with Variables in Denominator

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 3 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} \\

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 🙂

1. Nathan Robinson

There are some mistakes in your working out:
– Q2: factorisation of the quadratic is wrong, however solution is correct.
-Q3: Solution is wrong as x>4 would be solution rather than x<4

1. iitutor Post author

Hi Nathan.

Thank you so much for pointing out these mistakes and we really appreciated that. While we discovered more minor mistakes as well, all the bugs are now amended.
Thanks again, Nathan

PS It would be greatly appreciated if you could get back to us for further mistakes and/or feedback!

Cheers iitutor team