A LevelFurther MathsEdexcelFurther Pure 1Revision NotesInequalitiesInequalitiesSolving Inequalities involving Algebraic Fractions

Solving Inequalities involving Algebraic Fractions (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 8 January 2026

Solving inequalities involving algebraic fractions

What are inequalities involving algebraic fractions?

Here are some examples of inequalities involving algebraic fractions
- $\frac{1}{x - 3} < 1$
- $\frac{1}{x + 2} \geq 6$
- $\frac{1}{x - 1} < \frac{1}{x - 2}$
- $\frac{1}{x - 1} > \frac{x + 4}{x - 5}$
- $\frac{1}{x^{2} - 4} \geq \frac{2}{x} + 1$

Why can't I just multiply both sides by the lowest common denominator?

The lowest common denominator may be
- positive for some values of $x$
- but negative for other values of $x$
I.e. multiplying both sides of an inequality by the lowest common denominator
- may keep the same inequality sign
- but may flip the inequality sign
  - when it's negative
  - meaning the original inequality is no longer true
For example
- $\frac{1}{x - 3} < 1$ is true for $x < 3$ or $x > 4$
  - e.g. substitute in $x = 0$ and $x = 5$ to see this
- but if you multiply both sides by $(x - 3)$
  - $1 < (x - 3)$ is only true for $x > 4$

How do I solve inequalities involving algebraic fractions using algebra?

To solve inequalities involving algebraic fractions:
STEP 1
Multiply both sides by the squares of any algebraic denominators
- as these are always positive
- so the inequality sign is never flipped
STEP 2
Bring the terms to one side
- i.e. zero is on the other side
- Do not expand the brackets!
STEP 3
Factorise out of the brackets and solve the polynomial inequality
- e.g.
  - find the critical values
  - sketch to see the ranges of $x$
- The solutions to the polynomial inequality
  - are the solutions to the original inequality
STEP 4
Decide whether to include or exclude the end points of the solution
- Do not include them if they make the original denominators zero
  - as these values are undefined
- This problem does not come up if the original inequality was strict
  - I.e. if the original inequality was < or > (rather than $\leq$ or $\geq$ )

Examiner Tips and Tricks

To solve the polynomial inequality, you are allowed to draw a quick sketch, despite the exam question saying "use algebra to solve the inequality"!

Worked Example

Use algebra to determine the values of $x$ for which

$\frac{4}{x - 2} \geq \frac{x}{x - 1}$

Answer:

Multiply both sides by the squares of the denominators

$\frac{4}{x - 2} \times {(x - 2)}^{2} {(x - 1)}^{2} \geq \frac{x}{x - 1} \times {(x - 2)}^{2} {(x - 1)}^{2}$

Cancel the denominators with their respective numerators

$4 (x - 2) {(x - 1)}^{2} \geq x {(x - 2)}^{2} (x - 1)$

Do not expand, but instead bring all terms to one side

$4 (x - 2) {(x - 1)}^{2} - x {(x - 2)}^{2} (x - 1) \geq 0$

Factorise out as many brackets as possible

$(x - 2) (x - 1) [4 (x - 1) - x (x - 2)] \geq 0$

Simplify inside the remaining bracket

$\begin{array}{rcl} (x - 2) (x - 1) [4 x - 4 - x^{2} + 2 x] & \geq & 0 \\ (x - 2) (x - 1) (- x^{2} + 6 x - 4) & \geq & 0 \end{array}$

It helps to multiply both sides by $- 1$ to make $- x^{2}$ a positive $x^{2}$

but this also flips the inequality sign

$\begin{array}{rcl} (x - 2) (x - 1) (x^{2} - 6 x + 4) & \leq & 0 \end{array}$

Two critical values are 2 and 1

Find the others, solve $x^{2} - 6 x + 4 = 0$ , e.g. by completing the square

$\begin{array}{rcl} {(x - 3)}^{2} - 9 + 4 & = & 0 \\ {(x - 3)}^{2} & = & 5 \\ x - 3 & = & \pm \sqrt{5} \\ x & = & 3 \pm \sqrt{5} \end{array}$

Hence write out all the critical values

$x = 3 - \sqrt{5}, 1, 2 or 3 + \sqrt{5}$

Do a rough sketch of the quartic graph $y = (x - 2) (x - 1) (x^{2} - 6 x + 4)$

It is a positive quartic shape
It has $x$ -intercepts of the four critical points

Graph of the function y = (x-2)(x-1)(x²-6x+4) showing a curve with roots at x = 1, 2, 3 ± √5 on a cartesian plane.

To solve the polynomial inequality $\begin{array}{rcl} (x - 2) (x - 1) (x^{2} - 6 x + 4) & \leq & 0 \end{array}$

find the ranges of $x$ for which the graph is below the $x$ -axis

Graph of a polynomial inequality, showing a curve intersecting the x-axis at points 3-sqrt(5), 1, 2 and 3+sqrt(5), with shaded regions below the axis. The inequality (x-2)(x-1)(x^2-6x+4)<=0 is written.

$3 - \sqrt{5} \leq x \leq 1$ or $2 \leq x \leq 3 + \sqrt{5}$

Finally, exclude any end points that make the denominators in $\frac{4}{x - 2} \geq \frac{x}{x - 1}$ equal to zero

$x \neq 2$ and $x \neq 1$

This gives the correct inequality signs

$3 - \sqrt{5} \leq x < 1$ or $2 < x \leq 3 + \sqrt{5}$

Examiner Tips and Tricks

You do not need to give your solution in set notation unless asked, but if you do, make sure you use $\cup$ and $\cap$ correctly.

How do I solve inequalities involving algebraic fractions by first sketching graphs?

Inequalities can also be solved by sketching both sides first
- then seeing where the graphs satisfy the inequality sign
You are expected to know how to sketch the reciprocal graph $y = \frac{a x + b}{c x + d}$
- It has two L-shaped branches
  - e.g. $y = \frac{1}{x}$
- It may or may not have $x$ and $y$ intercepts
- It has a vertical asymptote at $x = - \frac{d}{c}$
  - found by setting the denominator equal to zero
- It has a horizontal asymptote at $y = \frac{a}{c}$
  - found by dividing top and bottom by $x$
  - so $y = \frac{a + \frac{b}{x}}{c + \frac{d}{x}} \to \frac{a + 0}{c + 0}$ as $x \to \infty$
It often helps to write $y = \frac{a x + b}{c x + d}$ in the form $y = e + \frac{f}{c x + d}$
- e.g. $y = \frac{2 x - 1}{x - 1} \equiv \frac{2 (x - 1) + 1}{x - 1} \equiv \frac{2 (x - 1)}{x - 1} + \frac{1}{x - 1} \equiv 2 + \frac{1}{x - 1}$
  - or use polynomial division to find this
- The new form shows that:
  - the graph is a translation of $y = \frac{1}{x}$ by the vector $(\begin{matrix} 1 \\ 2 \end{matrix})$
  - the horizontal asymptote is $y = 2$ (because $x \to \infty$ means $y \to 2 + 0$ )

Worked Example

By first sketching the graphs of $y = \frac{4}{x - 2}$ and $y = \frac{x}{x - 1}$ on the same axes, determine the values of $x$ for which

$\frac{4}{x - 2} \geq \frac{x}{x - 1}$

Answer:

The graph of $y = \frac{4}{x - 2}$ is a vertical stretch of $y = \frac{1}{x}$ by scale factor 4 and a translation by $(\begin{matrix} 2 \\ 0 \end{matrix})$

The graph of y=4/(x-2) is drawn as two L-shaped reciprocal branches about a vertical asymptote of x=2. The x-axis is a horizontal asymptote.

The graph of $y = \frac{x}{x - 1}$ can be sketched as follows

Find its vertical asymptote by setting the denominator equal to zero

$\begin{array}{rcl} x - 1 & = & 0 \\ x & = & 1 \end{array}$

Find its horizontal asymptote by dividing top and bottom by $x$ then letting $x \to \infty$

$y = \frac{1}{1 - \frac{1}{x}} \to \frac{1}{1 - 0} = 1$ as $x \to \infty$

Find any points of intersection with the axes

$(0, 0)$

The graph of y=x/(x-1) is drawn as two L-shaped reciprocal branches about a vertical asymptote of x=1 and a horizontal asymptote of y=1. The lower branch passes through the origin.

Sketch the two graphs on the same axes

The graph of y=4/(x-2) is drawn as two L-shaped reciprocal branches about a vertical asymptote of x=2. The graph of y=x/(x-1) is drawn as two L-shaped reciprocal branches about a vertical asymptote of x=1 and a horizontal asymptote of y=1. The two graphs intersect twice.

Determine the points of intersection using algebra

i.e. solve the equations simultaneously

$\begin{array}{rcl} \frac{4}{x - 2} & = & \frac{x}{x - 1} \\ 4 (x - 1) & = & x (x - 2) \\ 4 x - 4 & = & x^{2} - 2 x \\ 0 & = & x^{2} - 6 x + 4 \end{array}$

Solve the quadratic to find the $x$ -coordinates of intersection

e.g. by completing the square

$\begin{array}{rcl} {(x - 3)}^{2} - 9 + 4 & = & 0 \\ {(x - 3)}^{2} & = & 5 \\ x - 3 & = & \pm \sqrt{5} \\ x & = & 3 \pm \sqrt{5} \end{array}$

Hence find the ranges where $\frac{4}{x - 2} \geq \frac{x}{x - 1}$

i.e. where $y = \frac{4}{x - 2}$ is vertically higher than $y = \frac{x}{x - 1}$

If an end point is at an asymptote of either graph, it must be excluded

$x \neq 1$ and $x \neq 2$

This gives the solution

$3 - \sqrt{5} \leq x < 1$ or $2 < x \leq 3 + \sqrt{5}$

Examiner Tips and Tricks

Sometimes finding the points of intersection algebraically first helps to sketch the two graphs in hindsight!

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Solving Inequalities involving Algebraic Fractions (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Solving inequalities involving algebraic fractions

What are inequalities involving algebraic fractions?

Why can't I just multiply both sides by the lowest common denominator?

How do I solve inequalities involving algebraic fractions using algebra?

How do I solve inequalities involving algebraic fractions by first sketching graphs?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis