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=0
b. a=b, c≠0
c. a=b, c=0
d. a-b=1,c≠1
e. a≠b, c=0

B.a=b, c≠0

C.a=b, c=0

D.a-b=1, c≠1

Step-by-step explanation:

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 no 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 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 there is one solution: x=0.

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

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