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.

$\blacksquare$