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.
$\blacksquare$