Thread: how can beh 7 divides y

1. how can beh 7 divides y

show: if 7 | 2x - 3y and 7 | 28x+35y then 7|y
(2x - 3y = 28 + 35 y) / 7

only 28 and 35 is divisible by 7 which is 4 and 5 quotient respectively... how come 7 divides y?

2. Re: how can beh 7 divides y

7 | 28x + 35y = 7(4x + 5y) doesn't provide us with any useful information...this is true for all integral values of x and y.

7 | 2x - 3y let x = 5 and y = 1...7 does not divide 1.

3. Re: how can beh 7 divides y

show: if $7 | 2x - 3y$ and $7 | 4x + 5y$ then $7|y$

4. Re: how can beh 7 divides y

Thanks Idea. That makes a doable problem.

$Given\ x,\ y \in \mathbb Z,\ 7\ |\ 2x - 3y,\ and\ 7\ |\ 4x + 5y.$

Spoiler:
$\therefore \exists\ m,\ n \in \mathbb Z\ such\ that\ 7m = 2x - 3y\ and\ 7n = 4x + 5y.$

$7m = 2x - 3y \implies x = 0.5(7m + 3y).$

$\therefore 7n = 4x + 5y \implies 7n = 14m + 11y \implies y = \dfrac{7(n - 2m)}{11} = \dfrac{7}{11} * (n - 2m).$

$But\ y,\ (n - 2m) \in \mathbb Z \implies \exists\ k \in \mathbb Z\ such\ that\ 11k = (n - 2m) \implies$

$y = \dfrac{7}{11} * 11k = 7k$

$THUS\ 7\ |\ y.$

5. Re: how can beh 7 divides y

Note that if $7|a$ then $7|2a$. If also $7|b$, then $7|(2a-b)$.

6. Re: how can beh 7 divides y

Thank you sir Jeff