Equations
| English | Chinese | Pinyin |
|---|---|---|
| linear equations | 一次方程 | yī cì fāng chéng |
| simultaneous equations | 联立方程 | lián lì fāng chéng |
| eliminate | 消元 | xiāo yuán |
| quadratic | 二次 | èr cì |
| subject | 主项 | zhǔ xiàng |
The ancient art of balancing
- The word "equation" comes from "equal" — both sides must be the same, like a balance scale.
- Babylonians solved linear equations 一次方程 4000 years ago using clay tablets. Today we use the same principle: do the same to both sides.
Linear equations
- Solve by doing the same operation to both sides until the unknown is alone.
- $3x + 7 = 22 \Rightarrow 3x = 15 \Rightarrow x = 5$.
Watch the signs. $-2x + 5 = 11 \Rightarrow -2x = 6 \Rightarrow x = -3$. Dividing by a negative flips the sign of the answer.

Solve by doing the same to both sides
Solving an equation
y = ax² + bx + c
Solving means finding the roots — where the curve crosses the x-axis.
Solve 3x + 7 = 22.
3x = 22 − 7 = 15, so x = 5.
Simultaneous equations 联立方程
- Two equations, two unknowns. Eliminate 消元 one variable by adding or subtracting.
- $2x + y = 7$ and $3x - y = 8$: adding gives $5x = 15$, so $x = 3$, then $y = 1$.
Check your answer. Substitute back: $2(3) + 1 = 7$ ✓ and $3(3) - 1 = 8$ ✓.
Solve 2x + y = 7 and 3x − y = 8. What is x?
Adding the equations: 5x = 15, so x = 3 (then y = 1).
From 2x + y = 7 and 3x − y = 8, what is y?
x = 3 from adding. Substituting: 2(3) + y = 7 → y = 1.
Quadratic 二次 equations (Extended)
- A quadratic $ax^2 + bx + c = 0$ can have two, one, or zero real solutions.
- By factorising: $x^2 + 5x + 6 = 0 \Rightarrow (x+2)(x+3) = 0 \Rightarrow x = -2$ or $x = -3$.
- By the formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ (given in the exam).

The roots of a quadratic are where the parabola crosses the $x$-axis — the solutions of $y = 0$.
Solve x² + 5x + 6 = 0 by factorising.
(x + 2)(x + 3) = 0, so x = −2 or x = −3.
The quadratic formula is x = (−b ± √(b² − 4ac)) / (2a).
This formula solves any quadratic ax² + bx + c = 0.
Rearranging formulae
- Make a letter the subject 主项 by reversing the operations applied to it.
- Make $r$ the subject of $V = \dfrac{4}{3}\pi r^3$: $\;r^3 = \dfrac{3V}{4\pi}$, so $r = \sqrt[3]{\dfrac{3V}{4\pi}}$.
Make r the subject of V = (4/3)πr³. r = ∛(____).
Multiply by 3/(4π): r³ = 3V/(4π), then take the cube root.
You've got it
- linear: same operation to both sides; simultaneous: add/subtract to eliminate
- quadratics: factorise, complete the square, or use the formula
- rearrange: undo the operations in reverse order to make a letter the subject