Question 11 (nov. 14th)

Substitute y =

I. 1/(2x)
II. – 2/x
III. 3/x

in the x^2y^2 - xy equation.

Only II and II will equal 6


I) 1/2x

x^2 * (1/2x)^2 - x*1/2x is not equal to 6


II) – 2/x

X^2 ( -2/X)^2 - X (-2/X)

= 4+2
=6

Could be true

III) 3/x

X^2 (3/X)^2 - X * 3/X
= 9 - 3
=6

COULD BE TRUE

Hence E)

x^2 y^2 - xy = 6
xy(xy - 1) = 6

xy=3 or xy=-2 because these are only two possible solutions to the equation

so y=3/x and y=-2/x

E

My technique for this problem, wasn't the best, but it lead me to the correct answer:

x^2y^2-xy = 6

you can factor out xy, therefore:

xy(xy-1) = 6

xy has to be an integer, so the product of a certain integer and that integer minus 1 = 6, so I figured the intgers could be 3 and 2

so

X = 3; y=1
or
x=1; y=3

from here I just plugged in X into the options:

I. 1/(2x) = 1/6 or 1/3; no good- eliminate A, C, & D

** only B & E remain, II must work

III. 3/x = 1 or 3, this works- choose E

If my post causes more harm than, good I apologize. Once I realized that I couldn't simply isolate y, I realized this problem had a solution that wasn't intuitive to me, so I tried the best I could and it worked (in under two minutes). I think that's the moral of my story.

Naturally Cramya has an great solution- use hers for this type of problem Thanks Cramya!

dmateer25 wrote:x^2 y^2 - xy = 6
xy(xy - 1) = 6

xy=3 or xy=-2 because these are only two possible solutions to the equation

so y=3/x and y=-2/x

E

Perfect ! I just wanted to do this...

I am sure you got xy=3 and xy=-2 from options right

let xy = a
a^2 - a - 6 = 0
(a-3)(a+2)=0

a=xy=3 .......so y=3/x

a=xy=-2.......so y=-2/x

E