Use algebraic techniques to expand, factor and simplify expressions and algebraic fractions, and to solve linear and simultaneous equations

What this dot point is asking

This is the groundwork the whole Advanced course is built on. NESA expects you to manipulate algebra fluently: expand brackets, factor by every standard method, simplify and combine algebraic fractions, and solve linear, literal and simultaneous equations. None of it is hard in isolation, but every later topic, differentiating a quotient, finding a domain, completing the square, solving a trig equation, leans on doing this quickly and without slips. The aim here is not just to know the methods but to make them automatic, and to understand why each one works so you can recover it under pressure.

The answer

Expanding brackets and the three special expansions

Expanding means using the distributive law $a(x + y) = ax + ay$, then collecting like terms. For a product of two brackets, multiply every term in the first by every term in the second:

Three expansions occur so often that you should recognise them instantly, in both directions:

• Square of a sum: $(A + B)^2 = A^2 + 2AB + B^2$.
• Square of a difference: $(A - B)^2 = A^2 - 2AB + B^2$.
• Difference of squares: $(A + B)(A - B) = A^2 - B^2$.

The middle term $2AB$ in the first two is the one students drop. $(A + B)^2$ is not $A^2 + B^2$; there is a cross term, twice. The difference of squares is the most useful of the three because it runs backwards as a factoring tool: any expression of the form "a square minus a square" factors on sight.

The four methods of factoring

Factoring reverses expanding, and it is the skill the rest of the course depends on most. There are four methods, and the order matters: always take out the highest common factor first, then pick the method by counting the terms.

• Highest common factor (HCF). Always try this first. Pull out the largest factor common to every term: $4x^3 + 3x^2 = x^2(4x + 3)$.
• Difference of squares (two terms). If you have a square minus a square, use $A^2 - B^2 = (A + B)(A - B)$. Sometimes take out a common factor first: $80x^2 - 5y^2 = 5(16x^2 - y^2) = 5(4x - y)(4x + y)$.
• Quadratic trinomials (three terms). Find two numbers that combine correctly; see below for monic and non-monic cases.
• Grouping (four or more terms). Split into groups, factor each, then take out the shared bracket: $12xy - 9x - 16y + 12 = 3x(4y - 3) - 4(4y - 3) = (3x - 4)(4y - 3)$.

Keep going until every factor is irreducible, meaning it cannot be broken down any further. A common slip is to stop after one step, for example writing $5(16x^2 - y^2)$ and not noticing the bracket is itself a difference of squares.

Monic quadratics have a leading coefficient of $1$. To factor $x^2 + bx + c$, find two numbers whose sum is $b$ and whose product is $c$; those two numbers are the constants in the brackets. For $x^2 - 13x + 36$, the numbers are $-9$ and $-4$ (sum $-13$, product $36$), so $x^2 - 13x + 36 = (x - 9)(x - 4)$.

Non-monic quadratics have a leading coefficient other than $1$, and they trip up more students than anything else on this page. The reliable method is to split the middle term. To factor $ax^2 + bx + c$, find two numbers whose sum is $b$ and whose product is $a \times c$ (not just $c$). Split $bx$ into those two terms, then factor by grouping. The worked example below shows this in full for $6x^2 + 11x - 35$.

Algebraic fractions

An algebraic fraction is just a fraction with pronumerals, and the arithmetic rules carry over unchanged, with factoring doing most of the work.

• Adding or subtracting. Factor each denominator, find the lowest common denominator, then combine. For example $\dfrac{1}{x - 4} - \dfrac{1}{x} = \dfrac{x - (x - 4)}{x(x - 4)} = \dfrac{4}{x(x - 4)}$.
• Cancelling. Factor the top and bottom completely first, then cancel common factors. You can only cancel a factor of the whole numerator against a factor of the whole denominator, never a single term: $\dfrac{x^2 - x}{x^2 - 1} = \dfrac{x(x - 1)}{(x + 1)(x - 1)} = \dfrac{x}{x + 1}$.
• Multiplying and dividing. Factor everything, cancel, then multiply across. To divide, multiply by the reciprocal.
• Compound fractions (a fraction inside a fraction). The fastest route is to multiply the top and the bottom of the big fraction by one common multiple of every small denominator; that clears all the little fractions in one move.

Solving linear equations

The governing principle is simple: do the same thing to both sides to keep the equation balanced. You may add or subtract any number from both sides, and multiply or divide both sides by any non-zero number. Work towards isolating the unknown.

When the equation contains algebraic fractions, multiply through by the lowest common denominator first to clear them, then solve the resulting linear equation.

Changing the subject of a formula (literal equations)

A literal equation is one where you rearrange to make a chosen pronumeral the subject, treating the other letters as constants. The same balancing moves apply. Where the target pronumeral appears in more than one term, gather those terms on one side and factor it out, then divide. For example, to make $x$ the subject of $y = \dfrac{x + 1}{x + 2}$: multiply out to $xy + 2y = x + 1$, gather $xy - x = 1 - 2y$, factor $x(y - 1) = 1 - 2y$, and divide to get $x = \dfrac{1 - 2y}{y - 1}$.

Solving simultaneous equations

Two linear equations in two unknowns usually have one solution: the single pair $(x, y)$ that satisfies both. There are two algebraic methods, and you choose whichever is tidier.

• Elimination. Add or subtract multiples of the equations so one unknown cancels, leaving one equation in the other unknown. Best when the coefficients line up easily.
• Substitution. Rearrange one equation for a single unknown and substitute into the other. Best when one equation already has $x$ or $y$ by itself.

Geometrically, each linear equation is a straight line, and the solution is the point where the lines cross. That is why a simultaneous pair normally has exactly one solution: two non-parallel lines meet at exactly one point. (Parallel lines never meet, giving no solution; identical lines meet everywhere, giving infinitely many. Recognising these two special cases is worth a mark.)

Read the solution off the graph, stage by stage

The cleanest way to see what "solving simultaneously" means is to graph both lines and watch them meet. Below, the pair $3x + 2y = 12$ and $5x - 2y = 4$ is built up one line at a time; the meeting point is the algebraic solution $x = 2$, $y = 3$. (This is the same worked example solved by elimination below.)

Stage 1, draw the first line. A straight line needs only two points. For $3x + 2y = 12$, the intercepts are quickest: set $x = 0$ to get $2y = 12$, so $y = 6$, giving $(0, 6)$; set $y = 0$ to get $3x = 12$, so $x = 4$, giving $(4, 0)$. Rule the line through them.

Stage 2, draw the second line. For $5x - 2y = 4$, the intercepts are $(0, -2)$ (set $x = 0$: $-2y = 4$, so $y = -2$) and $(4, 8)$ (set $x = 4$: $20 - 2y = 4$, so $y = 8$). Rule this line on the same axes. The two lines are not parallel, so they cross exactly once.

Stage 3, find where they cross. The point that lies on both lines is the only $(x, y)$ that satisfies both equations at once. Drop a dashed line down to the $x$-axis and across to the $y$-axis from the intersection: it sits at $x = 2$ and $y = 3$.

Stage 4, read off and confirm the solution. The intersection $(2, 3)$ is the solution of the simultaneous pair. Confirm it satisfies both equations: $3(2) + 2(3) = 6 + 6 = 12$ and $5(2) - 2(3) = 10 - 6 = 4$. Both check, so $x = 2$, $y = 3$ is correct. The graph and the algebra agree, which is exactly what "solving simultaneously" guarantees.

How exam questions ask about algebraic techniques

These skills rarely appear as a standalone question; they are the working inside almost every other question. Recognise the cue words that tell you which technique is wanted:

• "Factor" or "factorise" means take it apart into a product. Count the terms first: two terms suggests a common factor or difference of squares, three suggests a quadratic trinomial, four suggests grouping. Always take out the HCF before anything else.
• "Expand" or "expand and simplify" means multiply out and collect like terms. If you see a perfect square, do not forget the cross term.
• "Simplify" on a fraction means factor everything and cancel. State the simplified form, and watch for an excluded value where a cancelled factor was zero.
• "Solve" an equation means find the value(s) of the unknown. For a linear equation, isolate the unknown; if there are fractions, clear them first.
• "Make $[\,\cdot\,]$ the subject" or "rearrange" is a literal equation: treat the other letters as constants, and if the target appears in two terms, factor it out before dividing.
• "Solve simultaneously" means two equations, two unknowns: use elimination or substitution, then check both equations. Words like "the point of intersection of the lines $\dots$" are the same task in disguise.
• "Hence" tells you to reuse a result you just found (often a factorisation) rather than starting again, so do not throw away your previous line.