A Two-Line Trick Solves Equations That Look Too Messy to Cross-Multiply

If $\frac{a}{b}=\frac{c}{d}$, you can rearrange the terms in several ways and the equality still holds. These rearrangements have names — invertendo, alternendo, componendo and dividendo — and together they turn awkward fractions into simple equations.

This part covers the properties, solving equations with componendo and dividendo, equations with surds and cubes, and proofs by the k-method.

**If $\frac{a}{b}=\frac{c}{d}$, invertendo flips both ratios, alternendo swaps the means, componendo adds the denominator to the numerator, dividendo subtracts it, and componendo and dividendo together give $\frac{a+b}{a-b}=\frac{c+d}{c-d}$.

- Invertendo**: $\frac{b}{a}=\frac{d}{c}$
- Alternendo: $\frac{a}{c}=\frac{b}{d}$
- Componendo: $\frac{a+b}{b}=\frac{c+d}{d}$
- Dividendo: $\frac{a-b}{b}=\frac{c-d}{d}$
- Componendo and dividendo: $\frac{a+b}{a-b}=\frac{c+d}{c-d}$

**Numerical check with $\frac{2}{3}=\frac{4}{6}$:**

$$\text{componendo: }\frac{5}{3}=\frac{10}{6}\text{dividendo: }\frac{-1}{3}=\frac{-2}{6}\text{C and D: }\frac{5}{-1}=\frac{10}{-2}=-5$$

Why C and D works. Divide the componendo result by the dividendo result; the $b$ and $d$ cancel.

An everyday example. **If $2$ kg of dal costs ₹240 and $3$ kg costs ₹360**, then $2:240=3:360$, but alternendo gives $2:3=240:360$ — comparing kilograms with kilograms and rupees with rupees.

The boundary case. Alternendo needs all four quantities to be of the same kind; in the dal example it works only because each new ratio compares like with like.

**Apply componendo and dividendo to both sides so the $a$ terms cancel in the numerator and the $x$ terms cancel in the denominator, then solve the simple equation left.

Worked example 1.** Solve $\frac{x+4}{x-4}=3$.

$$\frac{(x+4)+(x-4)}{(x+4)-(x-4)}=\frac{3+1}{3-1}\frac{2x}{8}=2x=8$$

Check: $\frac{12}{4}=3$.

Worked example 2. Solve $\frac{3x+5}{3x-5}=\frac{9}{4}$.

$$\frac{6x}{10}=\frac{13}{5}x=\frac{13}{5}\times \frac{10}{6}=\frac{13}{3}$$

Worked example 3 — leads to a quadratic. Solve $\frac{x^2+x+1}{x^2-x+1}=\frac{7}{3}$.

$$\frac{2x^2+2}{2x}=\frac{10}{4}\frac{x^2+1}{x}=\frac{5}{2}2x^2-5x+2=0x=2\text{ or }\frac{1}{2}$$

An everyday example. **The sum and difference of two brothers' ages are in the ratio $7:3$.** Componendo and dividendo on $\frac{a+b}{a-b}=\frac{7}{3}$ gives $\frac{a}{b}=\frac{10}{4}$, so their ages are in the ratio $5:2$.

The substance. **The method works only when $x\ne a$**, since the original denominator must not be zero.

Apply componendo and dividendo to clear the sum and difference of surds or to reveal perfect cubes, then square or take cube roots and solve.

Worked example 1 — surds. Solve $\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=3$.

$$\frac{2\sqrt{x+1}}{2\sqrt{x-1}}=\frac{3+1}{3-1}=2\frac{x+1}{x-1}=4x+1=4x-4x=\frac{5}{3}$$

Worked example 2 — cubes. Solve $\frac{x^3+3x}{3x^2+1}=\frac{14}{13}$.

Apply componendo and dividendo:

$$\frac{x^3+3x^2+3x+1}{x^3-3x^2+3x-1}=\frac{27}{1}{\left(\frac{x+1}{x-1}\right)}^3=27\frac{x+1}{x-1}=3$$

Apply componendo and dividendo again:

$$\frac{2x}{2}=\frac{4}{2}x=2$$

Check: $\frac{8+6}{12+1}=\frac{14}{13}$.

An everyday example. An experienced shopkeeper spots at a glance that ₹500 is five ₹100 notes; in the same way, spotting $(x+1{)}^3$ turns a messy fraction into a simple one.

The substance. Recognise the expansions $(x\pm 1{)}^3=x^3\pm 3x^2+3x\pm 1$ — they are what make the second example work.

**Set each equal ratio equal to $k$, write every term in terms of $k$, and simplify both sides of the identity until they match.

Worked example 1.** If $\frac{a}{b}=\frac{c}{d}$, prove $\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}$.

Let $a=bk$ and $c=dk$.

$$\text{LHS}=\frac{b^2{k}^2+d^2{k}^2}{b^2+d^2}=k^2\text{RHS}=\frac{bk\cdot dk}{bd}=k^2$$

Worked example 2. If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, prove $\frac{a+c+e}{b+d+f}=\frac{a}{b}$.

$$\frac{bk+dk+fk}{b+d+f}=\frac{k(b+d+f)}{b+d+f}=k=\frac{a}{b}$$

Worked example 3 — continued proportion. If $a,b,c$ are in continued proportion, prove $\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}$.

Let $b=ck$ and $a=bk=c{k}^2$.

$$\frac{c^2{k}^4+c^2{k}^2}{c^2{k}^2+c^2}=\frac{k^2(k^2+1)}{k^2+1}=k^2\frac{a}{c}=\frac{c{k}^2}{c}=k^2$$

An everyday example. **Enlarging a $4\times 6$ photo to $8\times 12$** multiplies both sides by the same $k=2$, so every ratio in the picture is kept.

The substance. The k-method turns any proportion proof into algebra with one letter.