Question about linear Diophantine equations

Question

Helo guys!! I have a two questions about linear Diophantine equations ($ax+by=c$) that I am not able to solve. The equations are: a)$9x-10y=-4$ and b)$-2x+3y=5$,

I'm solving them like this:

1. I found the gcd (d = 1 for $a$ and $b$) . This indicates that the two equations can be solved. 2)I used Bezóut's identity to find particular solutions: a)$x_0$=-1 and $y_0$=1 and b)$x_0$=1 and $y_0$=-1
2. I used the theorem to find x and y using the particular solutions ( $x=x_0 + \frac{bk}{d}$ and $y=y_0 - \frac{ak}{d}$): a) $x=1-10k$ and $y=1-9k$ and b)$x=1+3k$ and $y=2k-1$

The problem appears next, when I will find the interval that satisfies the equations: 4)I assume $x \geq 0$ and $y \geq 0$, and use these x and y to find the range of my solutions, but I just find: a) $k \leq 0$ and b) $k \geq 0$.

Can anyone tell me where I'm going wrong? Thanks.

Answer 1

Equation a) is solved by $$x=10 k+4,y=9 k+4$$ If we want both positive we must assume $$10k+4\ge 0,9k+4\ge 0\to k\ge 0$$

Equation b) has solutions $$x=3 h+2,y=2 h+3$$ setting the positivity condition $$3h+2\ge 0,2h+3\ge 0\to h\ge 0$$

Answer 2

I don't see how Bezóut's identity or finding a GCD has anything to do with this problem. Do you mean LCM? Any "interval" you seek can be idefified by the exact solutions to these equations.

This is often covered in Algebra but sometimes is delayed until Lindear Equations. By adding or subtracting polynomials, we can cancel one variable at a time. Finding the Least Common Multiple (LCM) of $y$ and multiplying through we get

$$9x-10y=-4\implies 27x-30y=-12$$ $$-2x+3y=5\implies -20x+30y=50$$

\begin{align*} (27x-30y=-12)&\\ +(-20x+30y=50)&\\ –––––––––––––&\\ 7x=38\implies x=\frac{38}{7} \end{align*}

$$-2x+3y=5\implies -2\cdot\frac{38}{7}+3y=5\implies y=\frac{37}{7}$$

Another option given $n$ equations in $n$ unknowns is to use Cramer's Rule shown here