Componendo and Dividendo

When participating in math competitions, there are a handful of useful tricks that can be used for solving rational equations. ^[src]

Assume that `\displaystyle \frac ab=\frac cd` for nonzero `b` and `d`, and `k` is such that no denominators are zero.

By performing some algebraic manipulation, one can find the following identity:
`\displaystyle \frac ab=\frac cd\implies\frac ab+k=\frac cd+k\implies\frac ab+\frac {kb}b=\frac cd+\frac {kd}d\implies\boxed{\frac{a+kb}b=\frac{c+kd}d}\pod{\text{Componendo}}`

Similarly:
`\displaystyle \frac ab=\frac cd\implies\frac ab-k=\frac cd-k\implies\frac ab-\frac {kb}b=\frac cd-\frac {kd}d\implies\boxed{\frac{a-kb}b=\frac{c-kd}d}\pod{\text{Dividendo}}`

Next, divide the above two results:

`\displaystyle \frac{\frac{a+kb}b}{\frac{a-kb}b}=\frac{\frac{c+kd}d}{\frac{c-kd}d} = \boxed{\frac{a+kb}{a-kb}=\frac{c+kd}{c-kd}}\pod{\text{Componendo et Dividendo}}`

The following are some more basic, but helpful identities: ^[src]

`\displaystyle \frac ab = \frac{a(b+kd)}{b(b+kd)} \implies \frac ab = \frac{ab+kad}{b(b+kd)} \implies \frac ab = \frac{a+\frac abkd}{b+kd} \implies \frac ab = \frac{a+\frac cdkd}{b+kd} \implies\boxed{\frac ab=\frac{a+kc}{b+kd}}`

`\displaystyle \frac ab=\frac cd\implies\boxed{\frac ac=\frac bd}\pod{\text{Alternendo}}`

`\displaystyle \frac ab=\frac cd\implies\boxed{\frac ba=\frac dc}\pod{\text{Invertendo}}`