The equation can be divided throughout by a as long as a is not zero. Otherwise a=0 would be a solution. So when we divide by a we get 10bc+7cd=3d. There isn't enough info to find values for b, c and d, but we can factorise c(10b+7d)=3d or 10bc=d(3-7c). d=10bc/(3-7c) or c=3d/(10b+7d). Can we find whole numbers to fit the equation? We'll try. 10b+7d=3d/c, so c must divide into 3d. If c=3, then 10b+7d=d, so 10b=-6d, or 5b=-3d. So b/d=-3/5, and b=-3 and d=5 would fit, as would b=3 and d=-5. The complete answer would be: a=any number (but a=0 means we can't find the other variables), b=-3, c=3 and d=5. There are many solutions, but this is an example.