# Simultaneous Equation With Two Unknowns

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## Theorem

A pair of simultaneous linear equations of the form:

\(\ds a x + b y\) | \(=\) | \(\ds c\) | ||||||||||||

\(\ds d x + e y\) | \(=\) | \(\ds f\) |

where $a e \ne b d$, has as its only solution:

\(\ds x\) | \(=\) | \(\ds \frac {c e - b f} {a e - b d}\) | ||||||||||||

\(\ds y\) | \(=\) | \(\ds \frac {a f - c d} {a e - b d}\) |

## Proof 1

\(\ds a x + b y\) | \(=\) | \(\ds c\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {c - b y} a\) | rearranging | ||||||||||

\(\ds d x + e y\) | \(=\) | \(\ds f\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds d \paren {\frac {c - b y} a } + e y\) | \(=\) | \(\ds f\) | substituting $x = \dfrac {c - b y} a$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \frac {c d - b d y} a + e y\) | \(=\) | \(\ds f\) | Multiplying out brackets | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds c d - b d y + a e y\) | \(=\) | \(\ds a f\) | multiplying by $a$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a e y - b d y\) | \(=\) | \(\ds a f - c d\) | subtracting $cd$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds y \paren {a e - b d}\) | \(=\) | \(\ds a f - c d\) | factorising | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac {a f - c d} {a e - b d}\) | dividing by $a e - b d$ |

The solution for $x$ can be found similarly.

When $a e = b d$ we have that $a e - b d = 0$ and hence no solution exists.

$\blacksquare$

## Proof 2

This is an example of Solution to Simultaneous Linear Equations and can be solved using the technique of matrices.

$\blacksquare$