ykaiim wrote:

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.

Answer: A.

Hope it's clear.

Agree about the question. thanks.