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

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by GmatMathPro » Thu Oct 27, 2011 9:11 am
For any normal linear equation of this form, you would typically have one solution for x. For example, if A=3 and B=2, 3x+3=2--->x=-1/3. More generally:

AX=B-A
X=(B-A)/A

This is valid for all real values of A and B EXCEPT when A=0, because we cannot divide by zero.

For example, if A=0 and B=2 we have 0x+0=2. The left side is always zero no matter what x is, so there are no solutions. Suppose A=0 and B=0, we have 0x+0=0. In this case the equation is always true regardless of the value of x, so there are infinitely many solutions.

Statement 1: If we know A is not zero, X=(B-A)/A is a valid solution. Whatever values B and A have, (B-A)/A will have exactly one value, regardless of whether B is zero or not. SUFFICIENT.

Statement 2: If we know B is not zero, there may be one, zero, or infinitely many solutions as outlined above, depending on whether or not A is zero. INSUFFICENT.

ANS: A
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by user123321 » Thu Oct 27, 2011 8:53 pm
lenagmat wrote:How many roots does this equation have? A*X+A=B

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

[spoiler]OA is A.

And can this equation be smth like that- 1X+1=Y ?? (Because I don't have any information of B).
[/spoiler]
for a polynomial of nth degree, the coefficient of highest degree term should not be zero.
hence in ax+b = a, a =! 0
hence should be A

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