The Algebra Toolkit

Starting point: Years 9 and 10

Like terms and substitution

Like terms have exactly the same letters and powers. You can add them: $3x+5x=8x$, but $3x+5x^2$ cannot be joined.

To substitute, swap each letter for its number, using brackets. If $x=-2$, then $3x^2=3(-2{)}^2=3\times 4=12$.

Expanding brackets

Multiply everything inside by the term outside.

$$3(2x-5)=6x-15$$

For two brackets, multiply every term in the first by every term in the second.

$$(x+3)(x-5)=x^2-5x+3x-15=x^2-2x-15$$

Factorising

Factorising is expanding backwards. Always try these in order:

1. Common factor: $6x^2+9x=3x(2x+3)$
2. Difference of two squares: $x^2-49=(x+7)(x-7)$
3. Quadratic trinomial: find two numbers that multiply to the constant and add to the middle number. For $x^2+7x+12$, the numbers 3 and 4 work, so it is $(x+3)(x+4)$.

Index laws

$$a^m\times a^n=a^{m+n}\frac{a^m}{a^n}=a^{m-n}(a^m{)}^n=a^{mn}$$

$$a^0=1a^{-n}=\frac{1}{a^n}{a}^{\frac{1}{n}}=\sqrt[n]{a}$$

So $8^{\frac{2}{3}}=(\sqrt[3]{8}{)}^2=2^2=4$.

Surds

A surd is a root that is not a whole number, like $\sqrt{2}$. Simplify by finding a square factor:

$$\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$$

To rationalise a denominator, multiply top and bottom by the surd:

$$\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$$

Solving equations

Do the same thing to both sides until the letter is alone.

$$\frac{2x-1}{3}=5\Rightarrow 2x-1=15\Rightarrow 2x=16\Rightarrow x=8$$

Simultaneous equations have two unknowns. Use elimination or substitution.

Quick summary

• Expand: multiply every term inside by every term outside.
• Factorise: common factor, then difference of squares, then trinomial.
• Index laws: add powers when multiplying, subtract when dividing, multiply for a power of a power.
• $a^{-n}=\frac{1}{a^n}$ and $a^{\frac{1}{n}}=\sqrt[n]{a}$.
• Simplify surds with square factors. Rationalise to clear roots from the bottom.
• Solve by doing the same thing to both sides. Check by substituting back.

Quiz