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16 November, 13:20

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=0

B. a=b, c≠0

C. a=b, c=0

D. a-b=1, c≠1

E. a≠b, c=0

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Answers (1)
  1. 16 November, 16:52
    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
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