What is the answer to solvefor y,x= 2/(y^2+1) ?

The answer to solvefor y,x= 2/(y^2+1) is y=sqrt(2/x-1),y=-sqrt(2/x-1);\quad x\ne 0

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solvefor y,x= 2/(y^2+1)

solve for

y, x = \frac{2}{y ^{2} + 1}

y = \frac{2}{x} - 1, y = - \frac{2}{x} - 1; x \neq = 0

Solution steps

x = \frac{2}{y ^{2} + 1}

Multiply both sides by

y^{2} + 1

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

Solve

x (y^{2} + 1) = 2 : y = \frac{2}{x} - 1, y = - \frac{2}{x} - 1; x \neq = 0

y = \frac{2}{x} - 1, y = - \frac{2}{x} - 1; x \neq = 0

\frac{10}{6} = \frac{6}{x - 7}

I, A = \frac{9 R}{I}

0 = \frac{2 t ^{2}}{t + 1} - 15

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

y, - y^{- 1} = - x^{2} + 3

What is the answer to solvefor y,x= 2/(y^2+1) ?
The answer to solvefor y,x= 2/(y^2+1) is y=sqrt(2/x-1),y=-sqrt(2/x-1);\quad x\ne 0