Expressions and Formulae

This links in very well with the science equations you need to know (especially physics).

Formulas are written rules that describe the relationship between something. For example:

$A = \pi r^2$ which describes the relationship of the area of a circle to it's radius and pi.

Expressions are just groups of numbers (no inequalities) like:

$4x + 5y$

The triangle method is bullshit when the proper way to rearrange formulas is easy.

Rearranging Algebraic Equations

Remember: There is balance in the force

When rearranging you will usually be asked to make a certain letter "the subject". This just means to get the letter on one side and everything else on the other side of the equals sign.

Follow this order in rearranging and you will be fine:

1. Remove roots by exponentiating both sides (square roots and squaring...)
2. Get rid of fractions
3. Multiply out brackets
4. Collect all subject terms on one side and rest on other side.
5. Simplify the form

Rearranging involves using the inverse operation. For example:

if you want to get rid of multiplication, divide from both sides if you want to get rid of addition, subtract from both sides if you want to get rid of roots, exponentiate from both sides

If we want to rearrange this to make $b$ the subject we can do this:

$\dfrac{a \color{red}{-a}}{b} = t \color{red}{-a}$ Why is this wrong? Because you haven't done the same to the bottom of the fraction. Here is the correct solution:

$\begin{aligned} \dfrac{a}{b} \color{red}{* b} &= t \color{red}{*b} \\[7mu] \dfrac{a}{b \color{blue}{\div b}} \color{red}{* \dfrac{b \color{blue}{\div b}}{1}} &= a = tb \end{aligned}$

Another way to work it out would be to:

$\begin{aligned} \dfrac{a}{b} \color{red}{* \dfrac{b}{1}} &= t \color{red}{*b} \\[7mu] \dfrac{ab}{b} &= \dfrac{b}{b} * a = a = tb \end{aligned}$

Here is a slightly harder one, make $h$ the subject:

$\begin{aligned} d &= \sqrt{\dfrac{3h}{2}} \\ d\color{red}{^2} &= (\sqrt{\dfrac{3h}{2}})\color{red}{^2} \\ d^2\color{red}{* 2} &= \dfrac{3h}{2} \color{red}{*\dfrac{2}{1}} \\ \dfrac{2d^2}{\color{red}{3}} &= \dfrac{3h}{\color{red}{3}} \\ \dfrac{2d^2}{3} &= h \end{aligned}$