Algebra is used by basically everyone in any applied sciences career. Not only does it define the basis for 90% of all math someone will do in their lifetime, but it also can be some of the hardest math in many cases. Often the difficulty comes from having to evaluate or manipulate a massive expression. Like anything in life, there is a nice shortcut to help reduce the pain that you have to tolerate. This method was taught to me by a professor some time ago and has become one of my most useful math revelations. Simply called “Bucket Algebra”, it is a method of simply grouping terms with respect to variables to reduce the complexity of a given problem. For example, lets take this rather nasty equation, which is a solution to a rather simple circuit analysis.

$-R_2(u-y)+R_1y = (R_4+R_5)\left[(u-y)-\left(\frac{-R_2(u-y)+R_1y}{R_3}\right)\right]$

Now in the above example, we are trying to relate ‘u’ to ‘y’. Thus we want to
solve for y. This, by normal methods of algebra, is a tall task and is *highly*
error prone. We can to better! Instead, we will use a different approach to
simplify the problem.

$y[\hat{B}_y] = u[\hat{B}_u]$

Where $\hat{B}_y$ is a “bucket” of terms that have ‘y’ as a factor. This bucket will contain all the terms which we can factor a ‘y’ out of. We work down each side of the equation adding each term to the ‘bucket’ if it can fit. Generally, you want to pick buckets for variables that appear the least overall and/or are the variables you’re trying to solve for. In this case, those variables would be ‘y’ and ‘u’.

To begin my example, lets create the bucket for ‘y’. Starting with the left hand side, we can see the term $-R_2(u-y)$. This contains both ‘-R2’ and ‘-y’, the negatives cancel and we get the term ‘R2’. This term is then placed into our bucket. Next is $-R_1y$ and I shouldn’t have to explain myself, we throw that in the bucket. We continue until we have something that looks like this for the ‘y’ side.

$y\left[R_2 + R_1 + (R_4 + R_5) + (R_4 + R_5)\left(\frac{R_2}{R_3}\right) + (R_4+R_5)\left(\frac{R_1}{R_3}\right)\right]$

We then repeat the process for ‘u’, taking care to be wary of signs.

$u \left[R_2 + (R_4 + R_5) + (R_4 + R_5)\left(\frac{R_2}{R_3}\right)\right]$

Now stick both together and we get this result. While it is *extremely* long, it
is much easier to solve for our target variables ‘u’ and ‘y’. We also have a
much easier equation to deal with if we have substitution values for the other
variables.

$y\left[R_2 + R_1 + (R_4 + R_5) + (R_4 + R_5)\left(\frac{R_2}{R_3}\right) + (R_4+R_5)\left(\frac{R_1}{R_3}\right)\right] = u \left[R_2 + (R_4 + R_5) + (R_4 + R_5)\left(\frac{R_2}{R_3}\right)\right]$

We can begin evaluating using this ‘bucketed’ expression. For the sake of simplicity, lets say all R values equal one. We can find ‘y’ by dividing the right hand side by the ‘y’ coefficient, and the reverse if we want to find ‘u’. Regardless, this is what that looks like:

$y[1+1+2+2+2] = u[1+2+2]$ $y=\frac{5}{8}u$

This method of dealing with algebra vastly reduces the amount of effort you need
to put in and reduces the chance for subtle errors. By grouping terms based on
variables you want to solve for, you can reduce the complexity of a problem by
making the overall problem simpler. Even if you don’t have all the values for R,
the process of simplifying is *monumentally* easier than with the original
problem.