Componendo and dividendo

Everything here starts from one statement: two ratios are equal, $\dfrac{a}{b} = \dfrac{c}{d}$. This single fact can be rearranged in several standard ways, each with its own name.

• Invertendo flips both sides: if $\dfrac{a}{b} = \dfrac{c}{d}$, then $\dfrac{b}{a} = \dfrac{d}{c}$.
• Alternendo swaps the middle terms (the means): if $\dfrac{a}{b} = \dfrac{c}{d}$, then $\dfrac{a}{c} = \dfrac{b}{d}$.
• Componendo adds the denominator to the numerator on both sides: if $\dfrac{a}{b} = \dfrac{c}{d}$, then $\dfrac{a+b}{b} = \dfrac{c+d}{d}$.
• Dividendo subtracts instead: if $\dfrac{a}{b} = \dfrac{c}{d}$, then $\dfrac{a-b}{b} = \dfrac{c-d}{d}$.

Dividing the componendo result by the dividendo result cancels the common denominators $b$ and $d$ and produces the rule you will use in almost every exam question — the combined componendo-dividendo property:

where: $a, c$ are the numerators of the original proportion, $b, d$ are the denominators, and the rule requires $a \neq b$ and $c \neq d$ so that neither new denominator vanishes. The figure below shows all four moves starting from the same box $\dfrac{a}{b}=\dfrac{c}{d}$, so you can see at a glance which one changes what.

The four properties all rest on the same trick: introduce a common value $k$ so that $\dfrac{a}{b}=\dfrac{c}{d}=k$, which forces $a=bk$ and $c=dk$. Substituting this back into each rearranged expression is exactly how the derivation below proves the combined rule. It helps to notice what each property preserves and what it changes. Invertendo and alternendo only rearrange the four existing terms $a,b,c,d$ — no new quantity like $a+b$ appears, so these two never risk creating a zero denominator on their own (beyond the obvious $a,c\neq 0$ already needed for the original ratio to make sense). Componendo and dividendo, by contrast, introduce brand-new quantities, the sum $a+b$ and the difference $a-b$, which is exactly why they are the pair worth combining: sums and differences are precisely the shapes that show up naturally in algebraic and surd expressions in exam questions, far more often than a bare swap of terms. That is also why the combined rule, not the four individual properties in isolation, is the one worth having ready to use on sight.

Question: If $\dfrac{x}{y} = \dfrac{3}{5}$, use componendo-dividendo to find $\dfrac{x+y}{x-y}$.

Step 1. Match the given proportion to the rule's pattern, with $a=x, b=y, c=3, d=5$.

Step 2. Apply componendo-dividendo directly: $\dfrac{x+y}{x-y} = \dfrac{c+d}{c-d} = \dfrac{3+5}{3-5} = \dfrac{8}{-2}$.

Step 3. Simplify. $\dfrac{8}{-2} = -4$, so $\dfrac{x+y}{x-y} = \boldsymbol{-4}$.

Check: take $x=3, y=5$ (any multiple works, since only the ratio matters). Then $x+y=8$ and $x-y=-2$, giving $\dfrac{8}{-2}=-4$, matching the shortcut answer directly.

Competitive stretch

Question: Solve for $x$: $\dfrac{\sqrt{x+4}+\sqrt{x-4}}{\sqrt{x+4}-\sqrt{x-4}} = 5$.

Step 1. Let $a=\sqrt{x+4}$ and $b=\sqrt{x-4}$, so the equation reads $\dfrac{a+b}{a-b} = \dfrac{5}{1}$ — already in componendo-dividendo output form, with $c=5, d=1$.

Step 2. Run the rule backward to recover the simple ratio: $\dfrac{a}{b} = \dfrac{c+d}{c-d} = \dfrac{5+1}{5-1} = \dfrac{6}{4} = \dfrac{3}{2}$, so $\dfrac{\sqrt{x+4}}{\sqrt{x-4}} = \dfrac{3}{2}$.

Step 3. Square both sides to clear the square roots: $\dfrac{x+4}{x-4} = \dfrac{9}{4}$. Cross-multiplying, $4(x+4) = 9(x-4)$, so $4x+16 = 9x-36$, giving $52 = 5x$ and $x = \dfrac{52}{5} = \boldsymbol{10.4}$.

Check: $x-4 = 6.4 > 0$, so both square roots are real, and the solution is valid.

• Confusing plain componendo ($\dfrac{a+b}{b}$) with the combined componendo-dividendo rule ($\dfrac{a+b}{a-b}$) — read the question carefully to see which form is actually wanted.
• Forgetting the rule needs $a \neq b$ (and $c\neq d$), since $a-b$ sits in a denominator; a proportion where $a=b$ makes dividendo undefined.
• Dropping or mishandling the negative sign when $c-d$ comes out negative, as in the worked example above.
• Reaching for cross-multiplication on a surd expression instead of noticing it already fits the $\dfrac{a+b}{a-b}$ pattern, which usually makes the algebra far shorter.
• Mixing up alternendo ($\dfrac{a}{c}=\dfrac{b}{d}$) with invertendo ($\dfrac{b}{a}=\dfrac{d}{c}$) — they look similar but rearrange the proportion differently.
• Applying componendo-dividendo when the proportion is not actually given in that exact $\dfrac{a}{b}=\dfrac{c}{d}$ form yet — if the question hands you an equation like $\dfrac{p+3}{q} = \dfrac{5}{2}$, first identify what plays the role of $a,b,c,d$ correctly before substituting into the rule.

Ask the tutor: not sure whether your problem needs plain componendo, dividendo, or the combined rule — ask the AI tutor to spot the pattern and walk through the steps with you.