adkikani wrote:
niks18Thanks for prompt support.
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I assume that highlighted section does not refer to the question per se but is just an illustration.
Correct, we are on same page.
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There is a difference between your illustration and this question. This question can be factorized to \(y*(x-3)^2=0\). Now when you get this equation, then there could be only two solutions - either y=0 or x=3. Then from the statements you know the boundary of x which will eventually lead to y=0.
In your illustration \(x^2+y-1=0\), we know from equation 1 that \(x^2=1\), hence you can substitute the value of \(x^2\) in equation (2) to get \(1+y-1=0 => y=0\)
Apologize to have not mentioned earlier about possibility of (2) being solved independently.
I could, in no way, solve (2) independently was my point.
My two basic queries are :
1. In a typical DS question as in this, where in we are asked to find product of x and y,
I need unique values for both x and y. (Since this is a value Q)
As per
VeritasPrepKarishma we DO GET unique values of both x and y.
She did not mention MULTIPLE values of x as in below theoretical example.
\(x^2\) - 1 = 0
x = 1, -1
So, why do I care if I am loosing solution of another variable?
2. Why is factorization so crucial in simplifying in original q stem?
6xy=\(x^2\)y + 9y
We know numerical values CAN NOT be equal to zero.
So Q stem reduces to :
Is xy = \(x^2\)y +yThe above equation is NOT solvable on its own, not because I am loosing a solution,
but because I can get multiple ways to solve this equality since we are NOT given that
either x or y are integers.
Let me know if my approach is correct.
Hi
adkikaniThe highlighted portion is incorrect.
\(6xy=x^2y+9y\) cannot be simply reduced to \(6xy=x^2y+9y\). The co-efficient 6 & 9 are an integral part of the equation. In equality you can divide both sides of the equation by same constant to simplify the equation.
also note that in the equation \(6xy=x^2y+9y\), you have two variables, x & y, so you need to solve for both and that can be done only by factorizing the equation. Here you cannot simply divide both sides of the equation by y to get \(6x=x^2+9\) because if y=0, then division by 0 is simply not possible. Hence you should never cancel out common "Variables". if you have common "constants/numbers", then you can cancel out them.
As in this question you are required to find the product xy, so you need values of both x & y. Suppose if you only know that x=1 or -1, then xy=y or -y. this will not give you any solution. if y=0, then irrespective of the value of x, xy=0. if you do not care about the variable y, then you will get option E or C as an answer, which in this case will be incorrect.
So, DS question should be looked in TOTALITY and not on the basis of any one of the equation or statement.