Find two rational numbers with denominators 11 and 13, respectively, and a sum of 7/143.
I got x/11 + y/13=7/143.
Two unknowns and one equation. Kind of stuck.
Don't forget $\displaystyle x$ and $\displaystyle y$ are integers. The equation they satisfy is $\displaystyle 13x+11y=7$.
You should first find one particular solution $\displaystyle (x_0,y_0)$ (you may start by finding $\displaystyle x,y\in\mathbb{Z}$ such that $\displaystyle 13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $\displaystyle 13x+11y=13x_0+11y_0$, hence $\displaystyle 13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $\displaystyle x=x_0+11k$ and $\displaystyle y=y_0-13k$ for some integer $\displaystyle k$.
Laurent.
Ok, so solve for x and y in 13x+11y=1, using the Euclidean algorithm (http://en.wikipedia.org/wiki/Euclidean_algorithm). Actually, do the Euclidian algorithm over 13 and 11, and observe...
Then, multiply by 7 :
13*(7x)+11*(7y)=7
Uh.
Use the Euclidean Algorithm to get the gcd of 11 and 13 (yes, you know this is 1, but the calculations you did in the algorithm help you find out what values of a and b give you 11a + 13b = 1.
You now want two numbers x and y such that 11x + 13y = 7.
Well, you just got 11a + 13b = 1, so multiply everything by 7:
$\displaystyle 11 \times 7a + 13 \times 7b = 7$
So 7a and 7b are the numbers you want for x and y. Job done.
You do not need to use Euclidean algorithm.
It is quite obvious that $\displaystyle 13(6) + 11(-7) = 1$.
Thus, $\displaystyle 13(42) + 11(-49) = 7$.
This means the solutions to $\displaystyle 13x+11y=7$ are: $\displaystyle x=42 - 11t \mbox{ and }y=-49 + 42t$.