What is the answer to solvefor y,x^2=((x+3y))/((x-3y)) ?

The answer to solvefor y,x^2=((x+3y))/((x-3y)) is y=-(x-x^3)/(3(x^2+1));\quad x\ne i,x\ne-i

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solvefor y,x^2=((x+3y))/((x-3y))

solve for

y, x^{2} = \frac{( x + 3 y )}{( x - 3 y )}

y = - \frac{x - x ^{3}}{3 ( x ^{2} + 1 )}; x \neq = i, x \neq = - i

Solution steps

x^{2} = \frac{( x + 3 y )}{( x - 3 y )}

Simplify

\frac{( x + 3 y )}{( x - 3 y )} : \frac{x + 3 y}{x - 3 y}

x^{2} = \frac{x + 3 y}{x - 3 y}

Multiply both sides by

x - 3 y

x^{2} (x - 3 y) = x + 3 y

Expand

x^{2} (x - 3 y) : x^{3} - 3 y x^{2}

x^{3} - 3 y x^{2} = x + 3 y

Move

x^{3}

to the right side

- 3 y x^{2} = x + 3 y - x^{3}

Move

3 y

to the left side

- 3 y x^{2} - 3 y = x - x^{3}

Factor

- 3 y x^{2} - 3 y : - 3 y (x^{2} + 1)

- 3 y (x^{2} + 1) = x - x^{3}

Divide both sides by

- 3 (x^{2} + 1); x \neq = i, x \neq = - i

y = - \frac{x - x ^{3}}{3 ( x ^{2} + 1 )}; x \neq = i, x \neq = - i

\frac{x ^{2} + 2}{5 x} = \frac{2}{7}

\frac{- 15}{4 - 17 x} = 9 + 10 x

\frac{( x ^{2} + 3 )}{( x - 3 )} - \frac{( x + 6 )}{( 3 - x )} = 1

(x^{4})^{2} = \frac{2 ^{12}}{x ^{5}}

\frac{( x + 1 ) ^{2} + 14}{4 x + 5} = n

What is the answer to solvefor y,x^2=((x+3y))/((x-3y)) ?
The answer to solvefor y,x^2=((x+3y))/((x-3y)) is y=-(x-x^3)/(3(x^2+1));\quad x\ne i,x\ne-i