Any help with this question please? 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 ≠ 1
B. a = b, c = 0
C. a ≠ b, c = 0
D. a – b = 1, c = 0
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

6.5

step-by-step explanation:

you said it was in the question

~courtney

sine

step-by-step explanation: