Calculate the value of M

[maxmin]

 

[Deleted member 4993]

View attachment 8118

What are 'x' & 'y'? Where are 'x' & 'y'?

$\displaystyle \displaystyle{Calculate \ \ expressions \ \ value \ \ M \ \ = \ \ a^2 \ \ + \ \ b^2 \ \ know \ \ x \ \ and \ \ y \ \ satisfy:}$

$\displaystyle \displaystyle{\frac{3b^2}{a^2} + \frac{2}{a^3} \ = 1}$ and

$\displaystyle \displaystyle{\frac{3a^2}{b^2} + \frac{1}{b^3} \ = 1}$

 

[maxmin]

What are 'x' & 'y'? Where are 'x' & 'y'?

sr, x=a,y=b.hiii

 

[tkhunny]

Perhaps you could share the ENTIRE problem and show ANY of your own work? That might give us a clue.

 

[mmm4444bot]

x=a, y=b

In this case, your exercise involves solving a system of two equations.

$\displaystyle \dfrac{3y^2}{x^2} + \dfrac{2}{x^3} = 1$

$\displaystyle \dfrac{3x^2}{y^2} + \dfrac{1}{y^3} = 1$

Both of these are cubic equations. Solving this system by hand seems too complex, for me.

I used software to graph the functions; there are three Real solutions. Wolfram-Alpha reports five additional solutions, each containing an imaginary part.

 

[Deleted member 4993]

In this case, your exercise involves solving a system of two equations.

$\displaystyle \dfrac{3y^2}{x^2} + \dfrac{2}{x^3} = 1$

$\displaystyle \dfrac{3x^2}{y^2} + \dfrac{1}{y^3} = 1$

Both of these are cubic equations. Solving this system by hand seems too complex, for me.

I used software to graph the functions; there are three Real solutions. Wolfram-Alpha reports five additional solutions, each containing an imaginary part.

Assuming $\displaystyle x \ne 0$ & $\displaystyle y \ne 0$

3y²x + 2 = x³

3x²y + 1 = y³

Subtracting:

(x-y)(x² +xy + y²) + 3xy(x-y) = 1


(x-y)(x² + 4xy + y²) = 1

Adding:

(x+y)(x² - xy + y²) - 3xy(x+y) = 3

(x+y)(x² - 4xy + y²) = 3

x² + y² = 1/2 * [1/(x-y) + 3/(x+y)]

I have not gone any further....