Linear Diophantine Equation/Examples/2x + 3y = 4

Example of Linear Diophantine Equation

The linear diophantine equation:

$2 x + 3 y = 4$

has the general solution:

$x = -4 + 3 t, y = 4 - 2 t$

Graphical Presentation

Proof

We have that:

$\gcd \set {2, 3} = 1$

which is (trivially) a divisor of $4$.

So, from Solution of Linear Diophantine Equation, a solution exists.

First we find a single solution to $2 x + 3 y = 4$:

\(\ds 1\)

\(=\)

\(\ds 1 \times 3 - 1 \times 2\)

\(\ds \leadsto \ \ \)

\(\ds 4\)

\(=\)

\(\ds 4 \times 3 - 4 \times 2\)

and so:

\(\ds x_0\)

\(=\)

\(\ds -4\)

\(\ds y_0\)

\(=\)

\(\ds 4\)

is a solution.

From Solution of Linear Diophantine Equation, the general solution is:

$\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$

where $d = \gcd \set {a, b}$.

In this case:

\(\ds x_0\)

\(=\)

\(\ds -4\)

\(\ds y_0\)

\(=\)

\(\ds 4\)

\(\ds a\)

\(=\)

\(\ds 2\)

\(\ds b\)

\(=\)

\(\ds 3\)

\(\ds d\)

\(=\)

\(\ds 1\)

giving:

\(\ds x\)

\(=\)

\(\ds -4 + 3 t\)

\(\ds y\)

\(=\)

\(\ds 4 - 2 t\)

$\blacksquare$

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-3}$ The Linear Diophantine Equation: Exercise $1 \ \text {(a)}$