# Math Made Easy: Solving Inequalities with Variables Below the Line

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}

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

1. 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. 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

2. What if we have a quadratic function in the denominator do we still follow same method?

1. Thanks for reaching out. yes it will be the same way. Square of the whole quadratic denominators will need to be multiplied for both sides.