cartera wrote:What is the value of X?

1. x^2+y^2=25
2. xy=12

OA E


Where Am I wrong?

x=12/y -------> (12/y)^2+y^2=25

144/y^2+y^2=25 --------> (y^4+144)/y^2=25 ------> y^4+144-25y^2=0

Fine till here


But you can't do the following

cartera wrote: Sqrt [y^4+144-25y^2=0] --------> y^2-5y+12=0

Then the solution should be C

If you want to solve the equation y^4+144-25y^2=0

substitute y^2 = z

you'll get z^2 -25z +144 = 0

solve for z

plug z into the equation y^2 = z

and then solve for x by using the equation xy=12

OR

when you had xy=12
you could've substituted for y i.e. y = 12/x and then the whole equation would've been in terms of x and not y

hope this helps..

Stmt I

x^2=9 y^2=16
x^2=16 y^2=9

X COULD BE 3 OR 4 INSUFF

Stymt II

xy=12
x=1 y=12
x=3 y=4

INSUFF

Combined

still x Could be 3 or 4

INSUFF
E)

STM1: x^2+y^2 = 25: Insuff
STM2: xy=12: Insuff

Together: (x+y)^2 = x^2+y^2 + 2xy
= 25+2*12
=49
Therefore, (x+y) could be (+/-)7...insuff
Answer: E