$1234x+567y=89$

Question

I need to solve this with congruence. What I already done is :
$$1234x+567y=89$$ $$1234x\equiv 89\bmod 567$$ $$1234\cdot17x\equiv 89\cdot17\bmod 567$$ $$20978x\equiv 1513\bmod 567$$ $$x\equiv -1513\bmod 567$$
$x_0=-1513$, so $x=-1513+567t$
Now I'm stuck here. I don't know how to search for $y_0$

Edit :
I found out I can change the equation into $567y\equiv 89\bmod 1234$
$$567y\equiv 89\bmod 1234$$ $$567\cdot 37y\equiv 89\cdot 37\bmod 1234$$ $$20979y\equiv 3293\bmod 1234$$ $$y\equiv 3293\bmod 1234$$ So $y_0=3293$ and $y=3293+1234t$

Did I get something wrong? Do I need to do more than this?

Answer 1

It seems like you have already solved the equation. Simply plug in $y$ and $x$ into your original equation and then solve for $t$ in the equation below:

$$1234x+567y=1234(-1513+567t)+567(3293+1234t)=89$$

It seems that $t=0$ is a solution. Which means $x=-1513$ and $y=3293$ are solutions to this equation.

Answer 2

You have $100x\equiv89\pmod{567}$ and some calculation gives $x=188+567t$ Taking $t=0$ we get $y=\dfrac{89-1234\cdot188}{567}=-409$ and the general solution can be given as $$x=188+567t\\y=-409-1234t$$ which comes from $$1234x+567y=89\\1234\cdot188+567\cdot(-409)=89$$ making a difference giving $1234(x-x_0)-567(y-y_0)=0$ thus....