My favorite way is through continued fractions.

But since this seems like the beginning of number theory course I will not mention it.

100y+x-68=2(100x+y)

100y+x-68=200x+2y

98y-200x=68

Let x'=-x,

200x'+98y=68

Divide,

100x'+49y=34

This is a linear diophantine equation.

gcd(100,49)=1, solutions therefore exist in Z.

Using Euclidean Algorithm we find that:

100=2(49)+2

49= 24(2)+1

2=2(1)+0

Thus, the gcd(49,100)=1 as expected.

Working backwards we find that,

49-24(2)=1

49-24(100-2*49)=1

49-24*100+48*49=1

49(49)+100(-24)=1

100(-24)+49(49)=1

Multiply by 34,

100(-816)+49(1666)=34

Thus,

x'=-816+49t

y=1666-100t

Thus,

x=816-49t

y=1666-100t