This is fundamental to physics and other sciences. Algebra equations can be rearranged so that one variable can equal the equation. This is very important as it allows us to get a certain formula in the form we want it and then plug the numbers in to solve it. This is another fundamental tool for algebra and maths in general.
Firstly, let’s remind ourselves that a letter can stand for a number. For example we are given this equation:
$$ 2x -3y, x = 4, y = 2 $$
We can solve it really easily:
$$ 2(4) - 3(2) = 8 - 6 = 2 $$
Now we need to be able to rearrange equations with just symbols. Remember that:we must balance the force. Let’s start off with a simple equation to rearrange, “Rearrange the following to make x the subject” (subject means making one side contain that and the other side contain the rest):
$$ \dfrac{2x - 4}{y} \color{red}{\times y} = y \color{red}{\times y} $$ $$ 2x - 4 \color{red}{+ 4} = y^2 \color{red}{+ 4} $$ $$ \dfrac{2x}{\color{red}{2}} = \dfrac{y^2 + 4}{\color{red}{2}} $$ $$ x = \dfrac{y^2 + 4}{2} $$
Notice how by doing the same thing to both sides we can rearrange the equation and also keep both sides of the equation equal? Another important concept to note is that each time we want to get rid of something we do the opposite (or reverse) mathematical operation. If we want to get rid of a division we multiply both sides, if we want to get rid of an addition we subtract from both sides, if we want to get rid of square roots we square both sides… This is a very powerful property in mathematics known as inverse operations.
Let’s try a harder example; making l the subject of:
$$ T = 2\pi \sqrt{\dfrac{l}{g}} $$ $$ (T)\color{red}{^2} = (2\pi \sqrt{\dfrac{l}{g}})\color{red}{^2} $$ $$ T^2 = 2\color{red}{^2} \pi\color{red}{^2} \times \dfrac{l}{g} = 4 \pi^2 \times \dfrac{l}{g} $$ $$ T^2 \color{red}{\div 4\pi^2} = \dfrac{4\pi^2}{\color{red}{4 \pi^2}} \times \dfrac{l}{g}$$ $$ \dfrac{T^2}{4 \pi^2} \color{red}{\times g} = \dfrac{l}{g} \color{red}{\times g} $$ $$ \dfrac{T^2 g}{4 \pi^2} = l $$
The key to this is just practicing, let’s try making a the subject:
$$ 3a + 5 = \dfrac{4 - a}{r} $$ $$ \color{red}{r}(3a + 5) = \dfrac{4 - a}{r} \color{red}{\times r} $$ $$ 3ar + 5r \color{red}{-5r + a} = 4 - a \color{red}{-5r + a} $$ $$ 3ar + a = 4 - 5r $$ $$ \dfrac{\color{red}{a}(3r + 1)}{\color{red}{3r + 1}} = \dfrac{4 - 5r}{\color{red}{3r + 1}} $$ $$ a = \dfrac{4 - 5r}{3r + 1} $$
It’s an important trick to remember that we can factorise to get the letter we want then divide the bracket to remove it to make a the subject.
More coming soon