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Algebra (LC Ordinary Level) — Key Notes & Examples

Concise notes aligned with recent LC OL papers (2019–2025), with one representative example per skill.

Expanding and Simplifying
  • Use the distributive law: multiply each term in the linear factor by each term in the quadratic, then collect like terms.

Example. Expand and simplify \((x+3)(2x^2 - x + 4)\).

Area model (2 × 3): rows correspond to \(x\) and \(+3\); columns correspond to \(2x^2,\ -x,\ +4\).

\(\,2x^2\,\)
\(-x\)
\(+4\)
\(x\)
\(\;x\cdot 2x^2=2x^3\)
\(\;x\cdot(-x)=-x^2\)
\(\;x\cdot 4=4x\)
\(+3\)
\(\;3\cdot 2x^2=6x^2\)
\(\;3\cdot(-x)=-3x\)
\(\;3\cdot 4=12\)

Combine like terms: \(2x^3 + (-x^2+6x^2) + (4x-3x) + 12 = 2x^3 + 5x^2 + x + 12\).

Answer: \(\boxed{2x^3 + 5x^2 + x + 12}\).

Linear Equations (including fractional)
  • Simplify carefully; collect like terms; isolate \(x\).
  • For fractional equations, clear denominators using the LCM.
  • Check domain restrictions if denominators depend on \(x\).

Example. Solve \( \dfrac{3x+1}{5}+\dfrac{x-2}{2}=\dfrac{47}{10} \) for \(x\in\mathbb{R}\).

LCM \(10\): \(2(3x+1)+5(x-2)=47 \Rightarrow 6x+2+5x-10=47 \Rightarrow 11x=55 \Rightarrow x=5.\)
Quadratic Equations
  • Standard form \(ax^2+bx+c=0\) \((a\neq0)\).
  • Quadratic formula: \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\).
  • Factor where convenient; otherwise use the formula and round as requested.

Example. Solve \(3x^2-5x+1=0\) (to 2 d.p.).

\(a=3,\ b=-5,\ c=1.\quad x=\dfrac{5\pm\sqrt{13}}{6}\approx\dfrac{5\pm3.606}{6}\Rightarrow x\approx1.43\ \text{or}\ 0.23.\)
Inequalities
  • Maintain the inequality direction; if you multiply/divide by a negative, reverse it.

Example. Solve \(2x+4\ge 6x-8\) for \(x\in\mathbb{R}\).

\(2x+4\ge 6x-8 \Rightarrow -4x\ge -12 \Rightarrow \boxed{x\le 3}.\)
Simultaneous Equations — Two Linear Equations
  • Use substitution or elimination; present a clear chain of working.

Example. Solve \(\begin{cases}3x+2y=11\\ x-4y=-1\end{cases}\).

From \(x-4y=-1\): \(x=4y-1\). Substitute into \(3x+2y=11\): \(3(4y-1)+2y=11\Rightarrow 12y-3+2y=11\Rightarrow14y=14\Rightarrow y=1\). Then \(x=3\). \(\boxed{(x,y)=(3,1)}\).
Simultaneous Equations — Linear with Circle
  • Substitute the linear relation into \(x^2+y^2=r^2\); solve the resulting quadratic.

Example. Solve \(\begin{cases}2x+y=5\\ x^2+y^2=25\end{cases}\).

\(y=5-2x\Rightarrow x^2+(5-2x)^2=25\Rightarrow 5x^2-20x=0\Rightarrow 5x(x-4)=0\Rightarrow x=0\ \text{or}\ x=4.\) Points: \((0,5)\) and \((4,-3)\).
Algebraic Division (Quadratic ÷ Linear)
  • Find the quotient so that divisor × quotient = dividend (plus remainder, if any). The 2×2 area model shows the products that reconstruct the dividend.

Example. Simplify \( \dfrac{6x^2 - 23x + 20}{\,2x - 5\,} \).

Idea: Seek a quotient \(ax+b\) so that \((2x-5)(ax+b)=6x^2-23x+20\).

\(\,3x\,\)
\(-4\)
\(2x\)
\((2x)(3x)=6x^2\)
\((2x)(-4)=-8x\)
\(-5\)
\((-5)(3x)=-15x\)
\((-5)(-4)=20\)

Check the middle term: \(-8x + (-15x) = -23x\). Hence \((2x-5)(3x-4)=6x^2-23x+20\).

\(\displaystyle \frac{6x^2 - 23x + 20}{2x - 5} = \boxed{3x - 4},\quad x \ne \tfrac{5}{2}.\)

Manipulating Formulae (literal equations)
  • Treat letters as numbers; isolate the required variable step by step.
  • State domain restrictions if division by a variable expression occurs.

Example. Write \(q\) in terms of \(p,t\) given \(3(2p+q)=t\).

\(6p+3q=t\Rightarrow 3q=t-6p\Rightarrow \boxed{q=\tfrac{t}{3}-2p}.\)
Substitution & Functions
  • Substitute accurately; simplify to a single rational number where asked.
  • Expand and collect like terms to rewrite \(f(x)\) in \(ax^2+bx+c\).

Example. Evaluate \( \dfrac{3x+5}{10}-\dfrac{1}{x+3}\) at \(x=2\) and express as \(\tfrac{a}{b}\) in lowest terms.

\(\dfrac{11}{10}-\dfrac{1}{5}=\dfrac{11}{10}-\dfrac{2}{10}=\dfrac{9}{10}.\)
Indices & Surds
  • Exponent laws: \(a^m a^n=a^{m+n}\), \((a^m)^n=a^{mn}\), \(a^{-n}=1/a^n\).
  • Convert surds to fractional indices and simplify.

Example. Find \(x\in\mathbb{Q}\) such that \(2^x=\sqrt{32}\).

\(\sqrt{32}=2^{5/2}\Rightarrow x=\tfrac{5}{2}.\)
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