Topic 2: Algebra

Algebra is the language of GCSE Mathematics. Instead of working only with numbers, you use letters to stand for unknown or varying quantities, which lets you describe patterns, model real situations and solve problems that arithmetic alone cannot handle. Topic 2 is the largest part of the AQA 8300 specification and it underpins almost everything else, from ratio problems to graph interpretation. This guide covers the core skills in order of difficulty: first tidying up and rewriting expressions, then expanding and factorising, then solving equations and rearranging formulae. After that we move into graphs, including straight lines through $y=mx+c$ and the geometry of parallel and perpendicular lines, then on to quadratic graphs and the three main ways of solving quadratics. We finish with simultaneous equations, inequalities, sequences, functions and iteration. Throughout, the key idea is balance: whatever you do to one side of an equation you must do to the other, and whatever you do to an expression must leave its value unchanged.

Simplifying expressions and the laws of indices

An algebraic expression is built from terms separated by plus or minus signs. Like terms have exactly the same letters raised to the same powers, so $3x$ and $5x$ are like terms but $3x$ and $3x^{2}$ are not. To simplify, you collect like terms by adding their coefficients: $4a+7b-2a+3b=2a+10b$. When multiplying terms you multiply the numbers and combine the letters, for example $3x \\times 4y=12xy$ and $2a \\times 5a=10a^{2}$. The laws of indices let you handle powers efficiently. When multiplying powers of the same base you add the indices: $x^{m} \\times x^{n}=x^{m+n}$, so $x^{3} \\times x^{4}=x^{7}$. When dividing you subtract: $x^{m} \\div x^{n}=x^{m-n}$, so $x^{6} \\div x^{2}=x^{4}$. A power raised to a power multiplies the indices: $(x^{m})^{n}=x^{mn}$, so $(x^{2})^{5}=x^{10}$. Any non-zero base to the power zero equals one, $x^{0}=1$, a negative index means a reciprocal, $x^{-n}=\\frac{1}{x^{n}}$, and a fractional index means a root, $x^{\\frac{1}{2}}=\\sqrt{x}$. A common exam item is something like $\\frac{12x^{5}y^{3}}{4x^{2}y}$, which simplifies to $3x^{3}y^{2}$ by dividing the coefficients and subtracting the indices for each letter.

Expanding brackets

Expanding (or multiplying out) brackets means writing an expression without brackets while keeping its value. For a single bracket you multiply every term inside by the term outside: $3(2x+5)=6x+15$ and $-2(4-x)=-8+2x$. Be careful with the signs when the multiplier is negative. To expand two brackets you multiply each term in the first by each term in the second, a process many students remember as FOIL (First, Outer, Inner, Last). For example $(x+3)(x+5)=x^{2}+5x+3x+15=x^{2}+8x+15$. With a negative sign, $(x-4)(x+2)=x^{2}+2x-4x-8=x^{2}-2x-8$. A perfect square such as $(x+6)^{2}$ must be written as $(x+6)(x+6)$ first, giving $x^{2}+12x+36$; a frequent mistake is to write only $x^{2}+36$. At higher tier you may expand three brackets by multiplying two of them first and then multiplying the result by the third, collecting like terms at the end.

Factorising, difference of two squares and quadratics

Factorising is the reverse of expanding: you write an expression as a product of factors. The first step is always to take out the highest common factor. For example $6x+9=3(2x+3)$ and $4x^{2}-10x=2x(2x-5)$. A quadratic of the form $x^{2}+bx+c$ factorises into two brackets when you can find two numbers that multiply to give $c$ and add to give $b$. To factorise $x^{2}-5x+6$ you need two numbers multiplying to $+6$ and adding to $-5$, which are $-2$ and $-3$, so the answer is $(x-2)(x-3)$. When there is a coefficient in front of $x^{2}$, such as $2x^{2}+7x+3$, you look for factors of $2 \\times 3=6$ that add to $7$, namely $6$ and $1$, split the middle term, then factorise in pairs to get $(2x+1)(x+3)$. The difference of two squares is a special pattern worth memorising: $a^{2}-b^{2}=(a+b)(a-b)$. So $x^{2}-49=(x+7)(x-7)$ and $9x^{2}-25=(3x+5)(3x-5)$. It only applies to a subtraction of two square terms, never a sum.

Solving linear equations and rearranging formulae

A linear equation contains an unknown to the first power only. You solve it by doing the same operation to both sides until the unknown is alone. To solve $3x+4=19$, subtract $4$ from both sides to get $3x=15$, then divide by $3$ to get $x=5$. When the unknown appears on both sides, gather the variable terms on one side first: from $5x-2=2x+10$ you get $3x=12$, so $x=4$. If the equation contains brackets or fractions, expand or multiply through to clear them before collecting terms. Rearranging a formula (changing the subject) uses exactly the same balancing rules, but you treat all the other letters as if they were numbers. To make $r$ the subject of $A=\\pi r^{2}$, divide by $\\pi$ to get $\\frac{A}{\\pi}=r^{2}$, then square root to get $r=\\sqrt{\\frac{A}{\\pi}}$. When the new subject appears more than once, collect those terms together and factorise it out. For example, to make $x$ the subject of $y=\\frac{x+a}{x-1}$ you multiply up, expand, gather the $x$ terms and factorise before dividing.

Linear graphs, y = mx + c, parallel and perpendicular lines

Every straight-line graph can be written as $y=mx+c$, where $m$ is the gradient (steepness) and $c$ is the y-intercept (where the line crosses the y-axis). The gradient is the change in $y$ divided by the change in $x$, often written as $m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$. A positive gradient slopes uphill from left to right and a negative gradient slopes downhill. To find the equation of a line through two points, calculate the gradient first, then substitute one point to find $c$. Two lines are parallel when they have the same gradient. So a line parallel to $y=3x+2$ has gradient $3$, and if it must pass through a given point you use that to find its own intercept. Two lines are perpendicular when the product of their gradients is $-1$; equivalently, the perpendicular gradient is the negative reciprocal. If a line has gradient $2$, a perpendicular line has gradient $-\\frac{1}{2}$, because $2 \\times -\\frac{1}{2}=-1$. You can also draw a straight line quickly from its equation by plotting the intercept and then using the gradient as a step of so many up for one across.

Quadratic graphs, the formula and completing the square

A quadratic graph $y=ax^{2}+bx+c$ is a smooth U-shaped curve called a parabola. It opens upward when $a$ is positive and downward when $a$ is negative. The points where the curve crosses the x-axis are the solutions (roots) of $ax^{2}+bx+c=0$, and the lowest or highest point is the turning point (vertex). There are three ways to solve a quadratic equation. First, factorising: $x^{2}-5x+6=0$ becomes $(x-2)(x-3)=0$, so $x=2$ or $x=3$. Second, the quadratic formula, which always works: $x=\\frac{-b\\pm\\sqrt{b^{2}-4ac}}{2a}$. You read off $a$, $b$ and $c$ and substitute carefully, watching the signs. The quantity $b^{2}-4ac$ is the discriminant; if it is positive there are two solutions, if zero there is one repeated solution, and if negative there are no real solutions. Third, completing the square rewrites $x^{2}+bx+c$ as $(x+\\frac{b}{2})^{2}-(\\frac{b}{2})^{2}+c$. For example $x^{2}+6x+1=(x+3)^{2}-8$. This form is useful because it both solves the equation and reveals the turning point, which for $(x+3)^{2}-8$ is at $(-3,-8)$.

Simultaneous equations and inequalities

Simultaneous equations are two equations that must be true at the same time, and solving them finds the pair of values that satisfies both. For two linear equations you can use elimination: make the coefficients of one variable match, then add or subtract the equations to remove it. From $3x+2y=12$ and $x-2y=4$, adding gives $4x=16$, so $x=4$, and substituting back gives $y=0$. Alternatively, substitution rearranges one equation for a variable and inserts it into the other; this is the method needed when one equation is quadratic, such as a line meeting a curve. Inequalities use the symbols $\\lt$, $\\gt$, $\\leq$ and $\\geq$ and are solved much like equations, with one crucial rule: when you multiply or divide both sides by a negative number you must reverse the inequality sign. So solving $-2x \\gt 6$ gives $x \\lt -3$. Solutions can be shown on a number line, using an open circle for a strict inequality and a filled circle for $\\leq$ or $\\geq$. You may also be asked to list integer values that satisfy a double inequality such as $-2 \\leq x \\lt 3$, or to identify a region on a graph satisfied by several inequalities at once.

Sequences, functions and iteration

A sequence is an ordered list of numbers following a rule. In an arithmetic sequence the difference between consecutive terms is constant; its nth term is $an+b$, where $a$ is the common difference. For $5, 8, 11, 14$ the difference is $3$ and the nth term is $3n+2$. In a geometric sequence each term is multiplied by a constant ratio, as in $2, 6, 18, 54$ with ratio $3$. A quadratic sequence has a constant second difference; its nth term has the form $an^{2}+bn+c$, and the coefficient $a$ equals half the second difference. Other named sequences include the triangular numbers and the Fibonacci sequence, where each term is the sum of the previous two. A function is a rule that turns an input into a single output, written for example as $f(x)=2x+1$, so $f(3)=7$. Composite functions apply one then another, written $fg(x)$, meaning do $g$ first, and the inverse function $f^{-1}(x)$ reverses the rule. You should recognise the shapes of standard graphs: linear, quadratic, cubic such as $y=x^{3}$, reciprocal such as $y=\\frac{1}{x}$, and exponential such as $y=2^{x}$. Iteration is a method for finding approximate solutions to equations that cannot be solved exactly. You rearrange the equation into the form $x=g(x)$, then use an iterative formula $x_{n+1}=g(x_{n})$, feeding each answer back in as the next input. Starting from a sensible first value $x_{0}$ and repeating, the results often settle towards a root, which you then state to a required degree of accuracy.