I'd go with C
1. y = x + 1
Xy = x(x+1)
Xy = x^2+x Insuff.

2. y = x^2+1
Xy = x(x^2+1)
Xy = X^3+x Insuff.

1+2

X+1 = x^2+1...............or x = x^2
0 = X^2 - x
0 = X(x-1)

Case 1: x = 0
Then XY must equal 0

Case 2: x-1=0.........................or x = -1
Plug this back into first equation y = x+1
Y = 0
Then XY must equal 0.

Sufficient.

naval,

quick q on this part -

1+2

X+1 = x^2+1...............or x = x^2

at this point, i test x = 1 or x = 0 as both meet that rule.

so to determine xy, if x = 0, then xy = 0. if x = 1, then y is 2.

since we get 2 values for xy, i chose E. your method looks to be in line though so could you tell me where i missed this?

Actually you might be correct. I never plugged -1 into the second equation. I would have gotten 2 as well.

OA please.

Answer should be E

(1) insuff as above
(2) insuff as above

combined you can try solving for x

x+1 = x^2+1
x=x^2

can be either 1 or 0

x, y, xy
1, 2, 2
0, 1, 0

so insuff

Hope that helps

Xbond wrote:Hi there,

Could you explain in the simplest this concept and how to resolve it


What is the value of xy ?
(1) y = x + 1
(2) y = x (square) + 1

Will be E. xy can either be 0 or 2.

navalpike wrote:I'd go with C
1. y = x + 1
Xy = x(x+1)
Xy = x^2+x Insuff.

2. y = x^2+1
Xy = x(x^2+1)
Xy = X^3+x Insuff.

1+2

X+1 = x^2+1...............or x = x^2
0 = X^2 - x
0 = X(x-1)

Case 1: x = 0
Then XY must equal 0

Case 2: x-1=0.........................or x = -1
Plug this back into first equation y = x+1
Y = 0
Then XY must equal 0.

Sufficient.

How do you get x=-1.It should be 1

OA is E