Problem solving equation system

[Taturana]

Homework Statement

Solve this equation system for x, y and lambda.

$$\left\{\begin{matrix}<br /> 2x = \lambda (2x-6y)\\ <br /> 2y = \lambda(-6x-14y)\\ <br /> x^2-6xy-7y^2+80=0<br /> \end{matrix}\right.$$

The Attempt at a Solution

I really tried A LOT of things, but I can't solve it. I think it is not helpful to post here all the arithmetic ways I tried.

Thank you for the help!

 

[rcgldr]

Can you just explain what you've tried so far?

 

[Taturana]

Can you just explain what you've tried so far?

Thank you for your response, rcgldr.

I have tried simple substitution. Tried to come up with one equation of one variable from substituting, summing and manipulating the system's equations. But I could not get an equation of one variable, at least with the substitutions I tried.

I don't know if this helps, but the real problem is to find the distance from the curve to the origin. The curve is the last equation in the system. So I'm using Lagrange where the distance equation if the f(x,y) and the curve equation is the g(x,y). This system came up from gradient(x^2 + y^2) = Lambda * gradient(x^2 - 6xy -7y^2 + 80).

 

[ehild]

Isolate y from the first equation and substitute into the second. What do you get?

ehild

 

[Ray Vickson]

Homework Statement

Solve this equation system for x, y and lambda.

$$\left\{\begin{matrix}<br /> 2x = \lambda (2x-6y)\\ <br /> 2y = \lambda(-6x-14y)\\ <br /> x^2-6xy-7y^2+80=0<br /> \end{matrix}\right.$$

The Attempt at a Solution

I really tried A LOT of things, but I can't solve it. I think it is not helpful to post here all the arithmetic ways I tried.

Thank you for the help!

You can solve the first equation for x in terms of y and λ (although not for some, special values of λ---they would need separate treatment). Substituting that into the second equation gives you an equation of the form y*A(λ) = 0, so either y = 0 or A(λ) = 0.

RGV