Simultaneous Equation With Two Unknowns

Theorem

A pair of simultaneous linear equations of the form:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a x + b y\)	\(=\)	\(\displaystyle c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle d x + e y\)	\(=\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {c e - b f} {a e - b d}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle \frac {a f - c d} {a e - b d}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Proof 1

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle ax+by\)	\(=\)	\(\displaystyle c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac{c-by}{a}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Rearranging
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle dx+ey\)	\(=\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle d \left({\frac{c-by} a }\right) + e y\)	\(=\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Substituting $x=\dfrac{c-by} a$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac{cd-bdy}{a}+ey\)	\(=\)	\(\displaystyle f\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Multiplying out brackets
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle cd-bdy+aey\)	\(=\)	\(\displaystyle af\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Multiplying by $a$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle aey-bdy\)	\(=\)	\(\displaystyle af-cd\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subtracting $cd$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y(ae-bd)\)	\(=\)	\(\displaystyle af-cd\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Factorising
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle \frac{af-cd}{ae-bd}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Dividing by $ae-bd$

The solution for $x$ can be found similarly.

$\blacksquare$

Note that when $a e = b d$ the solution is invalid.

Proof 2

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

$\blacksquare$