First off, you can rewrite your equation like so: c=(a-b)x. You can then plug in your given constraints for your choices. If a-b = 1 and c = 0 leaving: 0=1(x), x must equal 0 and only 0 as any constant multiplied by 0 equals 0. So that choice is eliminated. Now let's consider when a=b and c != 0. Since we are given a-b and a=b and c != 0, we have: c = 0x. This contradicts our claim we made about our constraints. C cannot equal zero but we have a-b=0. Therefore, this claim makes no sense as any value for x will not satisfy the equation. This choice is valid. When a=b and c=0, we have: 0 = 0x. Here, x can be any value and still return 0 as an answer. This choice is valid. If a-b=1 and c != 1, we have: c = 1x. Our only rule here is that c cannot equal 1. This means that x can be any value other than 1 so this choice can be marked down. If a != b and c=0, this gives: 0 = (a-b)x. Given that a-b can be any value, x must be equal to only 0 to satisfy this equation so this choice can't be correct. So the right answers are: option 2, option 3, option 4 and option 5.