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How many roots does this equation have? A*x + A = B

1) A does not equal 0 2) B does not equal 0

OA is A and OE is: Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is (B-A)/A.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is: For S1, x = (B-A)/A

If B=0 then x= -1 If A=B then x=0 If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

How many roots does this equation have? A*x + A = B

1) A does not equal 0 2) B does not equal 0

OA is A and OE is: Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is (B-A)/A.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is: For S1, x = (B-A)/A

If B=0 then x= -1 If A=B then x=0 If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.

Perhaps we need to take a closer look at the problem statement. The question is : How many roots does the equation have?

With Statement 1, you can definitely answer, 3 roots. With Statement 2, we need to know something about A to answer the question .

Thats correct. But, my question is untill we know both A and B we are not sure how many roots can be possible. If the question asked - What maximum number of roots this equation can have, then we could answer that it can be 3. _________________

How many roots does this equation have? A*x + A = B

1) A does not equal 0 2) B does not equal 0

OA is A and OE is: Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is (B-A)/A.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is: For S1, x = (B-A)/A

If B=0 then x= -1 If A=B then x=0 If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.

Not really the question GMAT would ask, but anyway:

I think the question asks about the # of roots for ax+a=b, where a and b are some particular real numbers and x is the variable.

3 cases: 1. if a=0 and b\neq{0}, equation will have no root (0 roots), as no value of x can satisfy the equation 0*x+0=b, when b\neq{0};

2. if a\neq{0} and b=any \ number, equation will have have one root: x=\frac{b-a}{a}. This root can take any value depending on a and b (not only 3 values);

3. if a=0 and b=0, equation will have have infinite # of roots, as any value of x will satisfy the equation 0*x+0=0.

(1) a\neq{0} --> we have the second case, hence equation has one root. Sufficient. (2) b\neq{0} --> we can have the first case, for which equation has no root OR the second case, for which equation has one root. Not sufficient.

How many roots does this equation have? A*x + A = B

1) A does not equal 0 2) B does not equal 0

OA is A and OE is: Statement (1) by itself is sufficient. If A is not 0, the equation has one root which is (B-A)/A.

Statement (2) by itself is insufficient. If B is not 0, two variants are possible. If A is 0, the equation has no roots, if A is not 0, the equation has one root.

My question is: For S1, x = (B-A)/A

If B=0 then x= -1 If A=B then x=0 If B!=0 and A!=B then x !=0

So, we can have different solutions for S1. So, how A can be the OA?

I dont know where I am missing.

Not really the question GMAT would ask, but anyway:

I think the question asks about the # of roots for ax+a=b, where a and b are some particular real numbers and x is the variable.

3 cases: 1. if a=0 and b\neq{0}, equation will have no root (0 roots), as no value of x can satisfy the equation 0*x+0=b, when b\neq{0};

2. if a\neq{0} and b=any \ number, equation will have have one root: x=\frac{b-a}{a}. This root can take any value depending on a and b (not only 3 values);

3. if a=0 and b=0, equation will have have infinite # of roots, as any value of x will satisfy the equation 0*x+0=0.

(1) a\neq{0} --> we have the second case, hence equation has one root. Sufficient. (2) b\neq{0} --> we can have the first case, for which equation has no root OR the second case, for which equation has one root. Not sufficient.