# 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, $\displaystyle \frac{x^2-4}{3-x} \le 0$.
Please feel free to let us know if you need any help on this 🙂

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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.

3. You cannot solve Rational Inequalities with a variable in the denominator like this.
The problem is that you don’t know whether expression in the denominator is neg or pos.
If you use your method of solving the inequality and it happens that the denominator expression is pos, you might get the correct answer in some cases, BUT that is misleading as in some cases the solution lies in the denominator expression being a neg and your solution will be incorrect.
Your calculation method is correct, but the mathematical principal behind it is wrong.
An important first principle is to never, never, never just multiply the equation out because we are dealing with an inequality which is a bit more tricky than a normal equation!
If you don’t know whether the expression in the denominator is neg or pos, you must check out both and apply your mind to which present the correct solution.

1. Thanks for reaching out. The method that used in this article, is to multiply squared of the denominators, such as $(x-2)^2, x^2, (x+2)^2$ and so to ensure the solution works always regardless of its sign either positive or negative as squared is always positive. You have pointed that you don’t know whether the denominators is negative or positive. However, it is true, that is why the squared of the denominators is begin multiplied, which gives positive signs allways.

Hope this helps.