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Extremely hard algebra! 
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#1
Nov805, 07:51 AM

P: 67

hi this is a hard algebra qestion
i have these 2 formulas 12x^2 2xy = 0 12y^2  2xy = 0 for some reason x and y are both 2 how do you get this i tried and got y= 6/x 0.5x tried to sub into second one but gets messy any way to do this on a calculator by any chance i have a Ti83 but i want to know how to do it algebraically thanks! 


#2
Nov805, 08:53 AM

HW Helper
P: 1,422

[tex]\left\{ \begin{array}{l} x ^ 2  2xy + 12 = 0 \ \ (1) & y ^ 2  2xy + 12 = 0 \ \ (2) \end{array} \right.[/tex]
You should note that there are 12, and 2xy in both (1), and (2), so it's common to subtract both sides of (1) from both sides of (2), or to subtract both sides of (2) from both sides of (1). So (1)  (2) gives: y^{2}  x^{2} = 0 [tex]\Leftrightarrow \left[ \begin{array}{l} x = y & x = y \end{array} \right.[/tex] So you have 2 cases: x = y and x = y. Case 1: x = y, substitute that back to either (1) or (2), to solve for y, then for x (in this case x = y). Case 2: x = y, substitute that back to either (1) or (2), to solve for y, then for x (in this case x = y). Can you go from here? 


#3
Nov805, 09:58 AM

HW Helper
P: 1,422




#4
Nov805, 10:03 AM

HW Helper
P: 1,025

Extremely hard algebra!
Line #9!!! My bad.



#5
Dec409, 08:19 PM

P: 1

Take a look at this page
http://www.wolframalpha.com/input/?i=12x^2+2xy+%3D+0 it should give you the awensers 


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