Welcome to Algebraic Fractions!
Welcome to one of the core building blocks of CCEA GCSE Further Mathematics (Unit 1: Pure Mathematics). If you can handle regular numerical fractions, you already know the basic rules! An algebraic fraction is simply a fraction that contains letters (unknown variables like \(x\), \(y\), or \(n\)) in the numerator, the denominator, or both.
Why are these important? Algebraic fractions show up everywhere in pure mathematics, mechanics, calculus, and engineering. Mastering them gives you the confidence to manipulate equations smoothly without breaking a sweat. Don't worry if this seems a bit daunting at first — we will break every process down into clear, easy-to-follow steps!
Did you know? The word "fraction" comes from the Latin word fractio, which means "to break". Think of algebraic fractions as breaking complex algebraic relationships down into manageable bite-sized pieces!
1. Simplifying Algebraic Fractions
Just like simplifying \(\frac{6}{8}\) down to \(\frac{3}{4}\) by dividing out the common factor of \(2\), we simplify algebraic fractions by dividing the numerator and the denominator by their common factors.
The Golden Rule of Simplifying
Always factorise first! You can only cancel common factors (things that are multiplied together), never individual terms (things separated by \(+\) or \(-\)).
Common Mistake to Avoid:
Students often try to cancel terms incorrectly. For example, in \(\frac{x + 5}{x + 2}\), you cannot cross out the \(x\)'s! The \(x\) is tied to the numbers by addition. You can only cancel an entire bracket that appears in both the numerator and the denominator.
Quick Review: Factorisation Tools in Your Toolkit
• Single Bracket (Common Factor): \(3x^2 + 6x = 3x(x + 2)\)
• Difference of Two Squares (DOTS): \(x^2 - 9 = (x - 3)(x + 3)\) or \(4x^2 - 25 = (2x - 5)(2x + 5)\)
• Quadratic Trinomials: \(x^2 + 5x + 6 = (x + 2)(x + 3)\)
• Harder Quadratics (\(ax^2 + bx + c\)): \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)
Worked Example: Simplifying
Problem: Simplify fully: \(\frac{x^2 - 4}{2x^2 + 3x - 14}\)
Step 1: Factorise the numerator.
\(x^2 - 4\) is a difference of two squares: \((x - 2)(x + 2)\)
Step 2: Factorise the denominator.
\(2x^2 + 3x - 14 = (2x + 7)(x - 2)\)
Step 3: Cancel identical factors.
\(\frac{(x - 2)(x + 2)}{(2x + 7)(x - 2)} = \frac{x + 2}{2x + 7}\)
Key Takeaway: Factorise the top, factorise the bottom, and cancel out identical matching brackets!
2. Multiplying and Dividing Algebraic Fractions
Multiplying Algebraic Fractions
Multiplying algebraic fractions is straightforward: multiply the numerators together and multiply the denominators together. However, to save yourself a lot of messy algebra, factorise everything first and cancel common factors before multiplying.
Formula: \(\frac{A}{B} \times \frac{C}{D} = \frac{A \times C}{B \times D}\)
Dividing Algebraic Fractions
Remember the simple rule from arithmetic: Keep, Change, Flip (KCF)!
• Keep the first fraction as it is.
• Change the division sign \(\div\) to a multiplication sign \(\times\).
• Flip the second fraction upside down (this is called finding the reciprocal).
Worked Example: Division
Problem: Simplify \(\frac{x^2 - 9}{x + 4} \div \frac{2x - 6}{x^2 + 4x}\)
Step 1: Apply Keep, Change, Flip (KCF).
\(\frac{x^2 - 9}{x + 4} \times \frac{x^2 + 4x}{2x - 6}\)
Step 2: Factorise every single part fully.
• Top left: \(x^2 - 9 = (x - 3)(x + 3)\)
• Bottom left: \(x + 4\) (cannot be factorised further)
• Top right: \(x^2 + 4x = x(x + 4)\)
• Bottom right: \(2x - 6 = 2(x - 3)\)
Our expression becomes:
\(\frac{(x - 3)(x + 3)}{x + 4} \times \frac{x(x + 4)}{2(x - 3)}\)
Step 3: Cancel common factors across numerators and denominators.
• Cancel \((x - 3)\) from top and bottom.
• Cancel \((x + 4)\) from top and bottom.
This leaves us with:
\(\frac{x(x + 3)}{2}\) or \(\frac{x^2 + 3x}{2}\)
Key Takeaway: For division, flip the second fraction and multiply. Always factorise everything before you multiply across!
3. Adding and Subtracting Algebraic Fractions
Just like basic fractions, you cannot add or subtract algebraic fractions until they share a Common Denominator.
Step-by-Step Method:
1. Find a common denominator: Usually, this is the product of the denominators.
2. Rewrite each fraction: Multiply the numerator and denominator of each fraction by whatever was needed to make the common denominator.
3. Combine into a single fraction: Write the combined numerators over the common denominator.
4. Expand and simplify the numerator: Leave the denominator in factorised bracket form.
Danger Zone with Negative Signs:
When subtracting, be extremely careful! Put brackets around the second numerator to ensure the minus sign applies to every term inside it.
For example: \(-\,(2x - 3) = -2x + 3\).
Worked Example: Addition and Subtraction
Problem: Express \(\frac{3}{x + 2} - \frac{2}{x - 1}\) as a single fraction in its simplest form.
Step 1: Determine the common denominator.
The common denominator is \((x + 2)(x - 1)\).
Step 2: Adjust both fractions.
\(\frac{3(x - 1)}{(x + 2)(x - 1)} - \frac{2(x + 2)}{(x + 2)(x - 1)}\)
Step 3: Combine over the common denominator.
\(\frac{3(x - 1) - 2(x + 2)}{(x + 2)(x - 1)}\)
Step 4: Expand the numerator carefully.
Numerator \(= 3x - 3 - 2x - 4 = x - 7\)
Final Answer:
\(\frac{x - 7}{(x + 2)(x - 1)}\)
Key Takeaway: Find a common denominator, cross-multiply numerators, and always watch out for minus signs distributing across brackets!
4. Solving Equations Involving Algebraic Fractions
When an algebraic fraction is inside an equation with an equals sign (\(=\)), your main goal is to eliminate the fractions entirely in the first step by multiplying every term by the common denominator.
Step-by-Step Strategy:
1. Identify the lowest common multiple (LCM) of all denominators.
2. Multiply every single term on both sides of the equation by this LCM to clear the fractions.
3. Expand brackets and simplify.
4. Rearrange into standard form (e.g., \(ax^2 + bx + c = 0\) if quadratic, or \(ax = b\) if linear).
5. Solve using factorisation, the quadratic formula, or basic linear rearrangement.
Worked Example: Solving a Quadratic Resulting Equation
Problem: Solve the equation: \(\frac{4}{x + 1} + \frac{3}{x - 2} = 2\)
Step 1: Multiply every term by \((x + 1)(x - 2)\).
\(4(x - 2) + 3(x + 1) = 2(x + 1)(x - 2)\)
Step 2: Expand all brackets.
Left side: \(4x - 8 + 3x + 3 = 7x - 5\)
Right side: \(2(x^2 - x - 2) = 2x^2 - 2x - 4\)
Equation now reads:
\(7x - 5 = 2x^2 - 2x - 4\)
Step 3: Rearrange to equal zero (\(ax^2 + bx + c = 0\)).
\(0 = 2x^2 - 2x - 7x - 4 + 5\)
\(2x^2 - 9x + 1 = 0\)
Step 4: Solve using the Quadratic Formula.
Here, \(a = 2\), \(b = -9\), \(c = 1\):
\(x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(1)}}{2(2)}\)
\(x = \frac{9 \pm \sqrt{81 - 8}}{4}\)
\(x = \frac{9 \pm \sqrt{73}}{4}\)
Evaluating to 2 decimal places:
\(x \approx 4.39\) or \(x \approx 0.11\)
Key Takeaway: Multiply the entire equation by the common denominator to clear the fractions immediately, then solve the resulting linear or quadratic equation!
Quick Review: Top Tips for CCEA Exam Success
• Always leave denominators factorised unless specifically told to expand them.
• Watch the signs: When subtracting fractions, write the second numerator in brackets after the minus sign.
• Look out for Difference of Two Squares: Spotting forms like \(x^2 - 16 = (x - 4)(x + 4)\) or \(9 - x^2 = (3 - x)(3 + x)\) will save you valuable time in your exam.
• Check for zero denominators: Values that make any denominator equal to zero are undefined and cannot be solutions!