It’s worth pointing out that even without a change of variables you can solve the original equation by noticing that it is homogeneous in x and y. This means that if (x,y,z,w) is a solution then so is (dx,dy,z,w) for any d. In particular, if you divide through by the gcd(x,y) you may assume that gcd(x,y) = 1.

But then notice that in the equation 10x + y - yz - 10yw = 0 you have that y divides 10x. But if gcd(x,y) = 1 then y divides 10 and so y = 1,2,5 or 10. You can easily solve the original equation by plugging in the various values of y and then solving for z.

For example if y = 5 then

10x + 5 - 5z - 50w = 0 yields z = 2x - 10w + 1. So you are free to use any x,w and then you have the solutions (dx, 5d, 2x-10w+1, w) where d,x,w are arbitrary positive integers and gcd(5,x) = 1.

You can do this for each value of y to parametrize all solutions to the original equation.