Consider this equation: c=ax-bx Joseph claims that if a, b, and c are non-negative integers, then the equation has exactly one solution for x. Select all cases that show Joseph's claim is incorrect. A.a-b=1, c=0B.a=b, c≠0C.a=b, c=0D.a-b=1, c≠1E.a≠b, c=0

The equation given is c = ax - bx. We can factor the right hand side to obtain an equivalent equation which is c = (a-b)x

Let’s explore each answer choice given. We are looking for cases where there is not one solution for the equation.

A
a-b = 1 so the right hand side becomes 1x and we have x=c. Since c is 0 we have one solution that is x=0

B
a=b so a-b =0 and the equation becomes 0=c but the answer choice says c does not equal zero. So in this case there is no solution. This is a correct answer to the problem.

C
This is the same as choice B but since C =0 both sides of the equation equal zero. We get 0=0 but notice that this is true no matter what the value of x is so this equation is called an identity and any value of x will do so there isn’t one solution but rather infinitely many. This is another right answer.

D
Here a-b=1 so we end up with x = c and since c doesn’t equal one any value of x except 1 is a solution so there isn’t one solution but infinitely many. This too is an answer to the question.

E
Since a doesn’t equal b and since c = 0 we have (a-b)x = 0 so. Either a-b is zero but since a and b are different this can’t be or x is zero. This ther is one solution: x=0.

From the above the answer to the question is choices B, C and D