# Simultaneous Equation With Two Unknowns

## 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.

$\blacksquare$