What is the answer to solvefor x,(1/x)-(1/y)-2=0 ?

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

Popular >

solvefor x,(1/x)-(1/y)-2=0

solve for

x, (\frac{1}{x}) - (\frac{1}{y}) - 2 = 0

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

Solution steps

(\frac{1}{x}) - (\frac{1}{y}) - 2 = 0

Simplify

(\frac{1}{x}) : \frac{1}{x}

\frac{1}{x} - \frac{1}{y} - 2 = 0

Multiply both sides by

x

1 - \frac{x}{y} - 2 x = 0

Move

1

to the right side

- \frac{x}{y} - 2 x = - 1

Multiply both sides by

y

- x - 2 x y = - y; y \neq = 0

Factor

- x - 2 x y : - x (1 + 2 y)

- x (1 + 2 y) = - y; y \neq = 0

Divide both sides by

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

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

y, 3 x^{2} y^{4} = 5 x^{3} y^{- 3} z^{2}

\frac{( 1 + x ^{4} + x ^{8} )}{x ^{4}} = 4

\frac{2 ( x + 1 )}{3 + 4 ( x - 2 )} = \frac{x}{2 + 5}

\frac{3 x - 2}{x} = \frac{59 x ^{2} - 4}{x ^{2}}

(\frac{x + 1}{x})^{2} - 2 (\frac{x - 1}{x + 4}) - 11 = 0

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