Equations

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.
• $3x + 7 = 22 \Rightarrow 3x = 15 \Rightarrow x = 5$.

Solve by doing the same to both sides

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 $y = 1$.

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 $y = 0$.

Powers in rearranging formulae (Extended)

• 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 $r = \sqrt[3]{\dfrac{3V}{4\pi}}$.

Build an equation, then choose a method

• A ticket problem gives $2a+c=31$ and $a+2c=26$. Double the second and subtract the first: $3c=21$, so $c=7$, then $a=12$. Check both original totals.
• Core rearrangement: $C=5t+8$ gives $C-8=5t$, then $t=(C-8)/5$. The subject appears once and has no power or root.

Completing the square and the formula (Extended)

• $x^2+6x+5=0$ becomes $(x+3)^2-4=0$, so $x+3=\pm2$, giving $x=-1,-5$.
• For $2x^2+x-3=0$, choose $a=2,b=1,c=-3$ in $x=(-b\pm\sqrt{b^2-4ac})/(2a)$. Thus $x=(-1\pm5)/4=1,-1.5$. Substitution checks each root.

Fractional equations (Extended)

• For $x/(x+2)=3/5$, exclude $x=-2$. Multiply by $5(x+2)$: $5x=3(x+2)$, so $2x=6$ and $x=3$. Check $3/5=3/5$.
• For $2/(x+1)+1/(x-1)=1$, exclude $x=-1,1$. Multiply through by $(x+1)(x-1)$: $2(x-1)+(x+1)=x^2-1$. This gives $x^2-3x=0$, so $x=0$ or 3; both are allowed.

One line and one curve (Extended)

• Solve $y=x+2$ and $y=x^2$ by substituting: $x^2=x+2$, so $(x-2)(x+1)=0$. Thus $x=2$ or $-1$.
• Find the matching $y$ for each: the pairs are $(2,4)$ and $(-1,1)$. Do not mix one root with the other root's $y$ value.

A subject appearing twice (Extended)

• Rearrange $y=(3x+2)/(x-1)$ for $x$: $y(x-1)=3x+2$, then $yx-3x=y+2$.
• Factor the subject: $x(y-3)=y+2$, giving $x=(y+2)/(y-3)$, $y\ne3$. The original also requires $x\ne1$.

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