Algebra · Paper 1
Fractions
Higher Level · Tap NEXT to begin
Algebraic fractions look intimidating — they're not. The work is the same as with numbers, just done with letters: simplify by cancelling factors top and bottom; add/subtract by getting a common denominator; multiply/divide by combining. The whole chapter rests on one rule learners break constantly.
Read this before anything else
Cancel whole brackets, never parts of brackets.
Why? $3x + 3$ represents one number. You can't cancel the $3$ inside it with a $3$ elsewhere unless it's the only thing there. Just like $\dfrac{2 + 4}{2} = \dfrac{6}{2} = 3$, not $5$. The top is one number $(6)$, not "$2 + 4$ with cancellable parts."
Section 1 of 5
Simplifying Fractions
Strip the fraction to its lowest form. With algebra: factor the top and the bottom completely first, then cancel any whole bracket that appears in both.
Question 1 · Numeric warm-up — same logic, just numbers. Simplify $\dfrac{2}{4}$ and $\dfrac{25}{30}$.
$\dfrac{2}{4} = \dfrac{2(1)}{2(2)} = \dfrac{1}{2}$ — the whole factor $2$ cancels.
$\dfrac{25}{30} = \dfrac{5(5)}{5(6)} = \dfrac{5}{6}$
$\dfrac{1}{2} \qquad \dfrac{5}{6}$
Same logic with letters: factor, then cancel whole factors.
Question 2 · The cancellation rule, properly explained. Simplify $\dfrac{x^{2} + x}{3x + 3}$.
Factor each piece fully: top $x^{2} + x = x(x + 1)$, bottom $3x + 3 = 3(x + 1)$
Cancel the matching bracket: $\dfrac{x(x + 1)}{3(x + 1)} = \dfrac{x}{3}$
$\dfrac{x}{3}$
⚠ The mistake. Writing $\dfrac{2 + 4}{2} =$ "cancel the $2$'s" $= 1 + 4 = 5$ is wrong. The top is $6$, one single number. Cancellation is between whole factors, never pieces of a sum.
Question 3 · Cubes over squares — both share the $(x - y)$ bracket. Simplify $\dfrac{x^{3} - y^{3}}{x^{2} - y^{2}}$.
Top is difference of cubes: $x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2})$
Bottom is DOTS: $x^{2} - y^{2} = (x - y)(x + y)$
Cancel the common $(x - y)$: $\dfrac{(x - y)(x^{2} + xy + y^{2})}{(x - y)(x + y)} = \dfrac{x^{2} + xy + y^{2}}{x + y}$
$\dfrac{x^{2} + xy + y^{2}}{x + y}$
⚠ Common mistake. You can't cancel further — $(x^{2} + xy + y^{2})$ is one expression, you can't peel an $x$ off it to pair with the $x$ below. Fully simplified.
Section 2 of 5
Adding & Subtracting Fractions
Same recipe as primary school: get a common denominator, write everything over it, then add the numerators. With algebra, factor the denominators first to find the smallest common denominator.
Question 4 · Numeric warm-up — set the rhythm. Simplify $\dfrac{1}{2} + \dfrac{1}{3}$.
Common denominator $2 \times 3 = 6$: $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3}{6} + \dfrac{2}{6}$
Add over the common denominator: $= \dfrac{5}{6}$
$\dfrac{5}{6}$
Question 5 · Two unrelated denominators — multiply them together. Simplify $\dfrac{3}{x + 1} + \dfrac{5}{2x - 3}$.
No common factor, so the common denominator is the product $(x + 1)(2x - 3)$.
Each numerator times the other denominator: $\dfrac{3(2x - 3) + 5(x + 1)}{(x + 1)(2x - 3)}$
Expand the top: $\dfrac{6x - 9 + 5x + 5}{(x + 1)(2x - 3)} = \dfrac{11x - 4}{(x + 1)(2x - 3)}$
$\dfrac{11x - 4}{(x + 1)(2x - 3)}$
Leave the bottom factored — never expand $(x + 1)(2x - 3)$ unless you have a specific reason.
Question 6 · Factor first — the LCD is smaller than you'd think. Simplify $\dfrac{x}{x^{2} - 1} + \dfrac{1}{x - 1}$.
Factor the first denominator (DOTS): $x^{2} - 1 = (x - 1)(x + 1)$
The second denominator $(x - 1)$ is "missing" an $(x + 1)$. Multiply the second fraction by $\dfrac{x + 1}{x + 1}$ to match.
Common denominator $(x - 1)(x + 1)$: $\dfrac{x}{(x - 1)(x + 1)} + \dfrac{x + 1}{(x - 1)(x + 1)}$
Add & simplify: $\dfrac{x + (x + 1)}{(x - 1)(x + 1)} = \dfrac{2x + 1}{(x - 1)(x + 1)}$
$\dfrac{2x + 1}{(x - 1)(x + 1)}$
Question 7 · Two quadratic denominators — one bracket overlaps. Simplify $\dfrac{3}{x^{2} - 5x + 6} + \dfrac{7}{x^{2} - 9}$ (for $x$ ≠ $2, 3, -3$).
Factor bottom 1: $\text{GN} = +6$, sum $-5$, pair $-2, -3$: $x^{2} - 5x + 6 = (x - 2)(x - 3)$
Factor bottom 2 (DOTS): $x^{2} - 9 = (x - 3)(x + 3)$
The shared bracket is $(x - 3)$ — count it once. LCD $= (x - 2)(x - 3)(x + 3)$: three brackets, not four.
Bottom 1 needs an extra $(x + 3)$, bottom 2 needs an extra $(x - 2)$: $\dfrac{3(x + 3) + 7(x - 2)}{(x - 2)(x - 3)(x + 3)}$
Expand & tidy: $\dfrac{3x + 9 + 7x - 14}{(x - 2)(x - 3)(x + 3)} = \dfrac{10x - 5}{(x - 2)(x - 3)(x + 3)}$
$\dfrac{10x - 5}{(x - 2)(x - 3)(x + 3)}$
You could pull a $5$ out the top: $5(2x - 1)$. Either form is acceptable.
Section 3 of 5
The Sign-Flip Trick
A whole class of "show this simplifies to a constant" questions hangs on one identity. Anytime two denominators look like reverses of each other, factor a $-1$ out of one — suddenly they match, and everything cancels.
The reversed-bracket identity
$(b - a) = -(a - b)$
So $(7 - x) = -(x - 7)$, $(3 - x) = -(x - 3)$, $(y - x) = -(x - y)$. Two reversed-bracket denominators are off by a single $-1$.
Question 8 · Show this simplifies to a constant. Show that $\dfrac{x}{x - 7} + \dfrac{7}{7 - x}$ simplifies to a constant ($x \in \mathbb{R}$, $x$ ≠ $7$).
Sign-flip the second denominator, since $7 - x = -(x - 7)$: $\dfrac{x}{x - 7} + \dfrac{7}{-(x - 7)} = \dfrac{x}{x - 7} - \dfrac{7}{x - 7}$
Same denominator — combine: $\dfrac{x - 7}{x - 7}$
Cancel the matching brackets: $= 1$
$1$ — a constant, for every $x$ ≠ $7$.
Question 9 · Same trick, different numbers. Simplify $\dfrac{x}{x - 3} + \dfrac{3}{3 - x}$ ($x \in \mathbb{R}$, $x$ ≠ $3$).
Use $3 - x = -(x - 3)$: $\dfrac{x}{x - 3} - \dfrac{3}{x - 3}$
Combine and cancel: $\dfrac{x - 3}{x - 3} = 1$
$1$
Section 4 of 5
Complex Fractions
A complex fraction has fractions inside the top, the bottom, or both. The technique: combine the top into one fraction, combine the bottom into one fraction, then divide — flip and multiply.
Question 10 · Numeric warm-up — builds the rhythm. Simplify $\dfrac{1 + \frac{1}{2}}{2 + \frac{1}{3}}$.
Top into one fraction: $1 + \dfrac{1}{2} = \dfrac{3}{2}$
Bottom into one fraction: $2 + \dfrac{1}{3} = \dfrac{7}{3}$
Flip the bottom, multiply: $\dfrac{3}{2} \div \dfrac{7}{3} = \dfrac{3}{2} \times \dfrac{3}{7} = \dfrac{9}{14}$
$\dfrac{9}{14}$
Question 11 · Same recipe with letters. Simplify $\dfrac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$.
Combine the top over $x$: $1 + \dfrac{1}{x} = \dfrac{x + 1}{x}$
Combine the bottom: $1 - \dfrac{1}{x} = \dfrac{x - 1}{x}$
Flip, multiply, cancel $x$: $\dfrac{x + 1}{x} \times \dfrac{x}{x - 1} = \dfrac{x + 1}{x - 1}$
$\dfrac{x + 1}{x - 1}$
Question 12 · Constants mixed with algebraic fractions. Simplify $\dfrac{2 + \frac{5}{x - 3}}{\frac{7}{x - 3} - 8}$.
Top — write $2$ over $(x - 3)$: $\dfrac{2(x - 3) + 5}{x - 3} = \dfrac{2x - 1}{x - 3}$
Bottom — write $8$ over $(x - 3)$: $\dfrac{7 - 8(x - 3)}{x - 3} = \dfrac{31 - 8x}{x - 3}$
Flip & multiply — the $(x - 3)$ cancels: $\dfrac{2x - 1}{x - 3} \times \dfrac{x - 3}{31 - 8x} = \dfrac{2x - 1}{31 - 8x}$
$\dfrac{2x - 1}{-8x + 31}$
Carl writes it with the $x$ term first: $\dfrac{2x - 1}{-8x + 31}$. Same fraction.
Question 13 · The big one — same recipe. Simplify $\dfrac{x - \frac{1}{x - 3}}{\frac{5}{x - 3} + 7}$.
Top — write $x$ over $(x - 3)$: $\dfrac{x(x - 3) - 1}{x - 3} = \dfrac{x^{2} - 3x - 1}{x - 3}$
Bottom — write $7$ over $(x - 3)$: $\dfrac{5 + 7(x - 3)}{x - 3} = \dfrac{7x - 16}{x - 3}$
Flip & multiply — $(x - 3)$ cancels: $\dfrac{x^{2} - 3x - 1}{x - 3} \times \dfrac{x - 3}{7x - 16} = \dfrac{x^{2} - 3x - 1}{7x - 16}$
$\dfrac{x^{2} - 3x - 1}{7x - 16}$
Section 5 of 5
Bracket-Shape Tricks
Some products of brackets look like nothing simplifies — until you spot the pattern. $(a + b + c)(a + b - c)$ is secretly DOTS, with $(a + b)$ as the "first" and $c$ as the "second."
Question 14 · Three multiplications — spot the DOTS in disguise. Simplify (i) $(x + 5)(x - 5)$, (ii) $(a + b + 7)(a + b - 7)$, (iii) $(x - y + 3)(x + y - 3)$.
(i) Straight DOTS: $(x + 5)(x - 5) = x^{2} - 25$
The pattern: $(A + B)(A - B) = A^{2} - B^{2}$. Two brackets matching except one sign flip.
(ii) Treat $(a + b)$ as one block: $(a + b)^{2} - 7^{2} = a^{2} + 2ab + b^{2} - 49$
(iii) The trick: read $(x - y + 3)$ as $x - (y - 3)$ and $(x + y - 3)$ as $x + (y - 3)$. Now it's DOTS with those blocks.
$[x - (y - 3)][x + (y - 3)] = x^{2} - (y - 3)^{2}$
$= x^{2} - (y^{2} - 6y + 9) = x^{2} - y^{2} + 6y - 9$
(i) $x^{2} - 25$ (ii) $a^{2} + 2ab + b^{2} - 49$ (iii) $x^{2} - y^{2} + 6y - 9$
Five techniques. One golden rule.
Simplify: factor everything, cancel whole brackets only. Add/subtract: common denominator first — factor the bottoms before deciding the LCD. Sign-flip: reversed brackets like $(7 - x)$ and $(x - 7)$ are off by a single $-1$. Complex fractions: combine top, combine bottom, flip and multiply. Bracket tricks: mismatched-looking products are often DOTS in disguise.
Stuck on a question, or want one looked at before the next test? Send it over (photo, screenshot, anything) and I'll come back with worked notes inside 24 hours. That's what these notes — and the grinds — are for. — Carl