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Equation question - how??

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by Baldini » Tue Mar 10, 2009 1:58 am
Please could someone explain how to get the answer (OA is B).

Thanks[/spoiler]
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by cubicle_bound_misfit » Tue Mar 10, 2009 2:38 am
given x+y != 0

the expression is a(x+y)/(x+y) hence knowing value of a will suffice.
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by Baldini » Wed Mar 11, 2009 1:49 am
Hi Cubicle, thanks for the response. Could you please however explain why knowing that (x+y does not = 0) allows you to get the value of the equation knowing only that a=6?

Thanks again
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by DanaJ » Wed Mar 11, 2009 5:15 am
Baldini: x + y cannot be zero since you'd have a case of dividing smth to zero, since x + y is the numerator in this fraction.
The fraction presented is (ax + ay)/(x + y) = [a(x + y)]/(x + y) - as you can see, x + y can be eliminated and you finally get that the whole thing is actually a. This is why knowing the value of a is enough to solve the equation.
However, just knowing x + y is not sufficient, since what you're really looking for is a.
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by marcusking » Wed Mar 11, 2009 6:16 am
cubicle_bound_misfit wrote:given x+y != 0

the expression is a(x+y)/(x+y) hence knowing value of a will suffice.
ALWAYS, ALWAYS, ALWAYS factor out whatever information is given in the question stem.

[If its x^2+4x+4 make it (x+2)^2 if its (x-3)(x-3) make it x^2-6x+9]

That being said look at this....

(ax+ay)/x+y => a(x+y)/(x+y) the term (x+y) immediately cancel each other out and all we are left with is a = ?

statement 1 gives us information about x & y but we don't have x & y in our equation anymore throw it out.

statement 2 is plain and clear it says a = 6 this indisputably solves our equation.

Answer must be B.
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