Sorry but what is that symbol in the middle?

I can't see any relationship between any of the statements. I would say E

Giorgio wrote:Sorry but what is that symbol in the middle?

i.e square x^2 - y = w

maihuna wrote:

Giorgio wrote:Sorry but what is that symbol in the middle?

i.e square x^2 - y = w

mihuna,

is 2-y all to the power of x or only 2 ?

IMO C.


From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)

From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)

combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.

heshamelaziry wrote:IMO C.


From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)

From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)

combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.

But When x=-2, (-2-1)(-2+1) = (-3)*(-1) = 3 = w so condition is still satisfied isn't

maihuna wrote:

heshamelaziry wrote:IMO C.


From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)

From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)

combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.

But When x=-2, (-2-1)(-2+1) = (-3)*(-1) = 3 = w so condition is still satisfied isn't

Yu are correct ! still the answer should be C because we need to combine the statements to get the value of w. I am not sure why -2 works here ? it could be that the question is not well designed or so; this is not unusual with some DS questions. MAybe experts can help ?

You can not solve 2nd equation like that! (x-1)(x+1)=3 does not give you X=2 and x=4 ! 3 is not 0!

You just get X=+2 or -2 , since equation is X^2-1=3

So answer is E !

Giorgio wrote:You can not solve 2nd equation like that! (x-1)(x+1)=3 does not give you X=2 and x=4 ! 3 is not 0!

You just get X=+2 or -2 , since equation is X^2-1=3

So answer is E !

Algebraically, why can't I do x-1=3 and x+1 = 3 ???????? if instead of 3 there were a zero, we do this. What is the difference ?

heshamelaziry wrote:

Giorgio wrote:You can not solve 2nd equation like that! (x-1)(x+1)=3 does not give you X=2 and x=4 ! 3 is not 0!

You just get X=+2 or -2 , since equation is X^2-1=3

So answer is E !

Algebraically, why can't I do x-1=3 and x+1 = 3 ???????? if instead of 3 there were a zero, we do this. What is the difference ?

The difference is the special properties of the number 0.

The ONLY way to get a product of 0 is to multiply by 0. There are infinite ways to get a product of 3.

Hesham,

Let's say we have (x-5)(x+7) = 0

See that this is just saying:

(one number)(another number) = 0

Or

x*y = 0

The only way for x*y = 0 to be true is if at least one of x or y equals 0.

This is why (x-5)(x+7) = 0 necessitates that either x = 5 or x = -7.

But when you have a non-zero number there will be a whole bunch of numbers whose product will be that number; in fact, if you consider non-integers, there are an infinite pairs of numbers whose product is 3.