What is the answer to solvefor y,y^{-1}=x(3e^x+c) ?

The answer to solvefor y,y^{-1}=x(3e^x+c) is y= 1/(x(3e^x+c));\quad x\ne 0,x\ne ln(-c/3)

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solvefor y,y^{-1}=x(3e^x+c)

solve for

y, y^{- 1} = x (3 e^{x} + c)

y = \frac{1}{x ( 3 e ^{x} + c )}; x \neq = 0, x \neq = ln (- \frac{c}{3})

Solution steps

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

Refine

\frac{1}{y} = x (3 e^{x} + c)

Multiply both sides by

y

1 = x (3 e^{x} + c) y

Switch sides

x (3 e^{x} + c) y = 1

Divide both sides by

x (3 e^{x} + c); x \neq = 0, x \neq = ln (- \frac{c}{3})

y = \frac{1}{x ( 3 e ^{x} + c )}; x \neq = 0, x \neq = ln (- \frac{c}{3})

\frac{x + 1}{x} = \frac{x}{x + 1}

\frac{x ^{2} - 4 x}{x - 4} = 0

\frac{5}{x + 3} = 0

- \frac{3}{4 x} - \frac{3}{4 x} = 4

\frac{1}{x - 2} - 5 = \frac{16}{x - 2}

What is the answer to solvefor y,y^{-1}=x(3e^x+c) ?
The answer to solvefor y,y^{-1}=x(3e^x+c) is y= 1/(x(3e^x+c));\quad x\ne 0,x\ne ln(-c/3)