elementary number theory - GCD (342,2240) such that 342x +2240y=g

I was given the following question to answer. But I am currently stuck on the last part.

The question is as follows: a) Find $gcd(342, 2240)$ call it $g$

b) find $integers$ $x,y$ such that $342x+2240y=g$

c) Find another pair of $integers$, call the $a$ and $b$, such that $342a+2240b=g$

For part A:

2 ways I solved. First I used prime factorization of each number

$342$=$2*3^2*19$ and $2240=2^6*5*7$ and found the $gcd$, $g$ was $2$.

Another way was the Euclidean algorithm:

$2240=342*6+188$

$342=188*1+154$

$188=154*1+34$

$154=34*4+18$

$34=18*1+16$

$18=16*1+2$

$16=2*8+0$ leaving me with the $gcd$, $g$ was $2$

Part B:

Working from the algorithm backwards I can find it or do division and work backwards.

So:

$2=18-1*16$

$=18-1(34-1*18)$

$=2*18-1*34$

$=2(154*-4*34)-1*34$

$=2*154-9*34$

$=2*154-9(188-1*154)$

$=11*154-9*188$

$=11(342-1*188)-9*188$

$=11*342-20*188

$=11*342-20(2240-6*342)$

$=131*342-20*2240$

Which leaves me with $342(131)+2240(-20)=g$ were $g=2$
Did I leave this statement correctly?

And how exactly do I find 2 additional number $a,b$ if you can assist with this problem, direct me to one or do a smaller but similar question with it I would appreciate it.

Answer

Just solve the homogeneous equation $342x+2240y=0$. For example $x=2240$, $y=-342$ would be a solution. Then $x+x_0$ and $y+y_0$ is a new solution of your original equation, where $x_0, y_0$ is the solution of non-homogeneous equation $342x+2240y=g$