Technique

Linearise a system by substituting reciprocals

Algebra • Equations • Simultaneous Equations

Statement

A pair of equations whose every term is

\frac{1}{x}

\frac{1}{y}

or a constant becomes linear under

u = \frac{1}{x}

v = \frac{1}{y}

; an equation such as

p (m x + n y) = q x y

takes that shape after dividing through by

x y

, and the final step is to invert

u

and

v

back.

How to use it

1If every term of a bilinear equation carries $x y$ or a single variable, divide through by $x y$ to expose the reciprocals.
2Solve the resulting linear system by elimination, then report $x = \frac{1}{u}$ and $y = \frac{1}{v}$ .
3Stopping at $u$ and $v$ is the usual slip; the question asks for $x$ and $y$ .

Solve

\frac{2}{x} + \frac{3}{y} = 13

and

\frac{5}{x} - \frac{4}{y} = - 2

Step-by-step

1
$u = \frac{1}{x}, v = \frac{1}{y}$ : $2 u + 3 v = 13$ and $5 u - 4 v = - 2$
Why: Both equations are linear in the reciprocals.
2
$8 u + 12 v = 52$ and $15 u - 12 v = - 6$ add to $23 u = 46$ , so $u = 2$ and $v = 3$
Why: Ordinary elimination.
3
$x = \frac{1}{2}, y = \frac{1}{3}$
Why: Invert to return to the original unknowns.

Answer

x = \frac{1}{2}, y = \frac{1}{3}

Watch out

Dividing by

x y

silently discards

x = 0

y = 0

; when the original equations are satisfied by

x = y = 0

that degenerate solution is still a solution of the system as printed, so check it separately before calling the reciprocal answer the only one.

Builds on

Also known as

reciprocal substitution, equations reducible to linear form, substitute 1/x and 1/y, bilinear system