What is ((x^{-1}+y^{-1}))/(2x^{-1)} ?

The solution to ((x^{-1}+y^{-1}))/(2x^{-1)} is (y+x)/(2y)

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simplify ((x^{-1}+y^{-1}))/(2x^{-1)}

simplify

\frac{( x ^{- 1} + y ^{- 1} )}{2 x ^{- 1}}

\frac{y + x}{2 y}

Solution steps

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

Simplify

x^{- 1} + y^{- 1} : \frac{y + x}{x y}

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

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

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

Simplify

x y \cdot 2 x^{- 1} : 2 y

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

2 x^{2} + 10 x = 28

(26 + 8) (26 - 8)

(2^{8})^{2}

∣ 3 n + 8 ∣ - 4 = - 10

What is ((x^{-1}+y^{-1}))/(2x^{-1)} ?
The solution to ((x^{-1}+y^{-1}))/(2x^{-1)} is (y+x)/(2y)