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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

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

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sr, x=a,y=b.hiiiWhat are 'x' & 'y'? Where are 'x' & 'y'?

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In this case, your exercise involves solving a system of two equations.x=a, y=b

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

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

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.

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Assuming \(\displaystyle x \ne 0\) & \(\displaystyle y \ne 0\)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.

3y

3x

Subtracting:

(x-y)(x

(x-y)(x

Adding:

(x+y)(x

(x+y)(x

x

I have not gone any further....