Bunuel wrote:
kashishh wrote:
If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?
(1) y-x = 2
(2) x^3 <0
what I did : 4xy = x^2y+4y
4x = x^2+4
x^2-4x+4 = 0
(x-2)^2=0
x = 2
is this wrong?
You cannot divide 4xy=x^2y+4y by y and write 4x=x^2+4, because y can be zero and division be zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So when you divide by y you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution.
If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?\(4xy=x^2*y+4y\) --> \(x^2*y-4xy+4y=0\) --> \(y(x^2-4x+4)=0\) --> \(y(x-2)^2=0\) --> either \(x=2\) or \(y=0\) (or both).
(1) y-x = 2 --> if \(x=2\) then \(y=4\) and \(xy=8\) but if \(y=0\) then \(x=-2\) and \(xy = 0\). Not sufficient.
(2) x^3<0 --> \(x<0\), so \(x\neq{2}\) which means that \(y\) must equal to zero --> \(xy=0\). Sufficient.
Answer: B.
Hope it's clear.
instead of taking LHS to RHS , I brought RHS to LHS
and got 4xy-x^2y - 4y = 0
now y (4x-x^2- 4 )= 0
y (-x^2 +4x - 4)= 0
y (x+2)^2= 0 either y = 0 or x = -2
now obviously x=-2 does not satisfy 4xy = x^2*y+4y, now I was expecting the same answer both ways.
Now am I doing something wrong,or is it just one of those cases where the equation works a certain way and not the other way
by the way factorization of
-x^2 +4x-4 =0 is (x+2)^2 right? if x^2 has a coefficient we have to multiply the constant (-4) with the coefficient of x^2 do we not.
You could expand \((x+2)^2\) to check your reasoning: \((x+2)^2=x^2+2x+4\), while \(-x^2 +4x-4=-(x^2-4x+4)=-(x-2)^2\). Thus, \(y(4x-x^2- 4 )=0\) can be written as \(-y(x-2)^2=0\) --> \(x=2\) or \(y=0\).