What is the answer to solvefor y,(x+1)(y-1)-2e^{-2x}+1=0 ?

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

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solvefor y,(x+1)(y-1)-2e^{-2x}+1=0

solve for

y, (x + 1) (y - 1) - 2 e^{- 2 x} + 1 = 0

y = \frac{2 e ^{- 2 x} - 1}{x + 1} + 1; x \neq = - 1

Solution steps

(x + 1) (y - 1) - 2 e^{- 2 x} + 1 = 0

Move

2 e^{- 2 x}

to the right side

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

Move

1

to the right side

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

Divide both sides by

x + 1; x \neq = - 1

y - 1 = \frac{2 e ^{- 2 x} - 1}{x + 1}; x \neq = - 1

Move

1

to the right side

y = \frac{2 e ^{- 2 x} - 1}{x + 1} + 1; x \neq = - 1

\frac{13}{40} = 1 - (\frac{x}{7127})

x, - x + 2 y = - 5

x, 3 x - y = 90

x, x (s) = \frac{5}{( s ( s + 2 ) )}

\frac{33}{100} = 1 - (\frac{x}{7466})

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