Question

PLEASE HELP!

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

Answer 1

B.a=b, c≠0

C.a=b, c=0

D.a-b=1, c≠1

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

Answer 2

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.