I tried the solution using other ways like:
$12x+5y=7$
$12x=7-5y$
$x=7-5y/12$
putting the value of $x$
$12(7-5y/12)+5y=7$
both $12$ will cancel out
$7-5y+5y=7$ here comes the problem please help me ?
I tried the solution using other ways like:
$12x+5y=7$
$12x=7-5y$
$x=7-5y/12$
putting the value of $x$
$12(7-5y/12)+5y=7$
both $12$ will cancel out
$7-5y+5y=7$ here comes the problem please help me ?
You can use the theorem of diophantine equations that if a equation
$ax+by=c$
is given and $a\neq0$, $b\neq0$, then let $g=gcd(a,b)$. If $g \nmid c$ then no solutions. If $g \mid c$, then infinitely many solutions. If the pair $(x_1,y_1)$ is a solution, then all other solutions are $x=x_1+kb/g$, $y=y_1-ka/g$ where $k\in\mathbb{Z}$.
So here $a=12$, $b=5$, $g=1$, $c=7$. $g \mid c$ and $(1,-1)$ is one solution.
Using above theorem
$x=1+5k$
$y=-1-12k$, $k\in\mathbb{Z}$
Observe that $7 = 12-5\implies 12(-x+1) = 5(y+1)$, and since $\text{gcd}(5,12) = 1$, we have: $-x+1 = 5k, y+1 = 12k\implies x = -5k+1, y = 12k-1, k \in \mathbb{Z}$