CDM Question

[Div2388]

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Studying for the QR section with CDM and need some help please.

X + Y = 3

(X^2) + (Y^2) = 8

XY = ??



I start out with Y = 3 - X and plug that into the second equation, which gives a binomial:

2(x^2) - 6X + 1 = 0

Factoring out 2 gives:

(x^2) - 3x + (1/2) = 0

This is about as far as I get. Normally I am excellent with algebra, but I just can't seem to get anywhere with this one. An explanation would be greatly appreciated. Thanks all!

 

[Zakl]

you foo, im trying to study for my finals, but i can't figure this problem out either 🙁 sorry, damn i totally would remember how to do this back when i had to do SAT II math

 

[UCB05]

X + Y = 3
(X^2) + (Y^2) = 8

if you square the first equation, you get
x^2 + 2xy + y^2 = 9
since you're given that x^2 + y^2 = 8, you can substitute to get
2xy + 8 = 9
xy = 1/2 🙂

Studying for the QR section with CDM and need some help please.

X + Y = 3

(X^2) + (Y^2) = 8

XY = ??



I start out with Y = 3 - X and plug that into the second equation, which gives a binomial:

2(x^2) - 6X + 1 = 0

Factoring out 2 gives:

(x^2) - 3x + (1/2) = 0

This is about as far as I get. Normally I am excellent with algebra, but I just can't seem to get anywhere with this one. An explanation would be greatly appreciated. Thanks all!

 

[Zakl]

X + Y = 3
(X^2) + (Y^2) = 8

if you square the first equation, you get
x^2 + 2xy + y^2 = 9
since you're given that x^2 + y^2 = 8, you can substitute to get
2xy + 8 = 9
xy = 1/2 🙂

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[Div2388]

X + Y = 3
(X^2) + (Y^2) = 8

if you square the first equation, you get
x^2 + 2xy + y^2 = 9
since you're given that x^2 + y^2 = 8, you can substitute to get
2xy + 8 = 9
xy = 1/2 🙂

Ahhh didn't even think of going about it that way. Thank you for the swift response!