In math we use a lot of variables, which are placeholder names for any number or unknown.

Your friend has some weirdly shaped shooter glasses and you can't quite tell if it is 25[ml] of vodka in there or 50[ml] or somewhere in between. Since you can't say how much booze there is in each shot glass we will say there was $x$[ml] in there. So how much alcohol did you drink over the whole evening? Say you had three shots then you drank $3x$[ml] of vodka. If you want to take it one step further, you can say that you drank $n$ shots then the total amount of alcohol you drank is $nx$[ml].

As you see, variables allow us to talk about quantities without knowing the details. This is abstraction and is very powerful stuff: it allows you to get drunk without knowing how drunk exactly!

There are common naming patterns for variables:

• $x$: general name for the unknown in equations. Also used to denote the input to a function

and the position in physics problems.

• $v$: velocity.
• $\theta,\varphi$: the Greek letters “theta” and “phi” are often used to denote angles.
• $x_i,x_f$: Denote initial and final position in physics problems.
• $X$: A random variable in probability theory.
• $C$: Costs in business along with $P$ profit, and $R$ revenues.

We often need to “change variables” and replace some unknown variable with another. For example, say you don't feel comfortable with square roots. Every time you see a square root, you freak out and you find yourself on an exam trying to solve for $x$ in the following: \[ \frac{6}{5 - \sqrt{x}} = \sqrt{x}. \] Needless to say that you are freaking out big time! Substitution can help with your root phobia. You just write down “Let $u=\sqrt{x}$” and then you are allowed to rewrite the equation in terms of $u$: \[ \frac{6}{5 - u} = u, \] which contains no square roots.

The next step when trying to solve for $u$ is to undo the fraction by multiplying both sides of the equation by $(5-u)$ to obtain: \[ 6 = u(5-u) = 5u - u^2. \] This can be rewritten as a quadratic equation $u^2-5u+6=(u-2)(u-3)=0$ for which $u_1=2$ and $u_2=3$ are the solutions. The last step is to convert our $u$-answers into $x$-answers by using $u=\sqrt{x}$, which is equivalent to $x = u^2$. The final answers are $x=2^2=4$ and $x=3^2=9$. You should try plugging these values of $x$ into the original equation with the square root to verify that they satisfy the equation.

Symbolic manipulation is very powerful, because it allows you to manage complexity. Say you are solving a physics problem in which you are told the mass of an object is $m=140$[kg]. If there are many steps in the calculation, would you rather use the number $140$[kg] in each step, or the shorter variable $m$? It is much better to use the variable $m$ throughout your calculation, and only substitute the value $140$[kg] in the last step when you are computing the final answer.