http://i52.tinypic.com/2ug14c0.jpg

Here is a picture of the problem. It gives me a headache haha

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- May 14th 2011, 03:20 PMtonydeanxoSolving a system of equations?
http://i52.tinypic.com/2ug14c0.jpg

Here is a picture of the problem. It gives me a headache haha - May 14th 2011, 03:37 PMtopsquark
- May 14th 2011, 03:38 PMTKHunny
Have you considered transforming to get rid of the logs? Hint: $\displaystyle 10^{log(8)} = 8$

Before transforming, you should worry about Domain issues. Don't lose track of these:

x + y > 0

x - y > 0

Note: I assumed Base 10 logs. - May 14th 2011, 04:27 PMtonydeanxo
Yes the log bases is 10!

When i approached the problem. I moved log(13) to the other side and condensed them.

so it would be log(x^2 +y^2)/13) = 1

so.. (x^2 +y^2/13) = 10 right?

so x^2 + y^2 = 130?

and for the second one

log(x+y)= log(8(x+y))

so x+y=8x+8y?

hmm - May 14th 2011, 04:34 PMtopsquark
- May 14th 2011, 04:40 PMtonydeanxo
Opps on that last comment it is supposed to be 8x-8y :)

so then I would solve for an x or y on the second equation.

so x +y = 8x-8y

so 7x= 9y

so either x= 9y/7

or y= 7x/9

and plug one of those into the previous equation? - May 14th 2011, 04:44 PMtonydeanxo
so i get. (9y/7)^2 + y^2 =130

(81y^2/49) + y^2 = 130

mulitply all by 49

81y^2 + y^2 = 6370

divide by 82. y^2 = 77.68

y= + or - 8.81? - May 14th 2011, 05:59 PMtopsquark
- May 14th 2011, 07:37 PMtonydeanxo
yay. i ended up graphing it and getting 9,7. :) -9,7 doesn't work thought wuth the logs.