Find all natural solutions of $14x+21y=91$

Question

Find all natural solutions of $14x+21y=91$ where $x,y\in \mathbb N$

My attempt:

Divided by $7$:

$$2x+3y=13$$

And $\gcd(2,3)=1\mid13$

I found a private solution $x_0=\color{blue}5,\quad y_0=\color{blue}1$

$$2\cdot \color{blue}5+3\cdot \color{blue}1=13$$

So the general solution:

$$\begin{cases}x=5+2t\\ y=1-3t\end{cases}$$

Where $t\in \mathbb Z$

But WA's answer is diffrent, why?

Answer 1

WA says $$ 2 + 3k\\ 3 - 2k $$ You say $$ 1 - 3t\\ 5 + 2t $$ You're close, but wrong, because pugging in $t = 1$, you get $$ x = 7\\ y = -2 $$ and $$ 2 \cdot 7 + 3 \cdot -2 \ne 13. $$

So that's where you went wrong. The difficulty is that it should probably be $$ x = 5 - 3t \\ y = 1 + 2t $$ instead.

Answer 2

There are an infinite number of ways to parametrize equations so two sets of equations could both generate the same set. Your parametrization of the equation is just a different from that WA's. For your purpose both answers are correct.

Many problems that ask you to parametrize something will give you constraints. It is often related to the $df/dt$ but these are unique to each area of study.