Thread: Solving a system of equations?

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

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

Originally Posted by tonydeanxo

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

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

where ln(x) is log to the base e of x. Is this right? Or are they log base 10?

In either case, what have you been able to do with them?

-Dan

Have you considered transforming to get rid of the logs? Hint:

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.

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

Originally Posted by tonydeanxo

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

That's correct. Where would you go from there?

-Dan

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?

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?

Originally Posted by tonydeanxo

so i get. (9y/7)^2 + y^2 =130
(81y^2/49) + y^2 = 130
mulitply all by 49
81y^2 + y^2 = 6370

81y^2 + 49y^2 = 6730

You will get y = (+/-) 7. What is x? (Then you have to figure out which +/- signs actually work in the original equation.)

-Dan

yay. i ended up graphing it and getting 9,7. -9,7 doesn't work thought wuth the logs.