Solve : ${(6^{- 1} - 8^{- 1})}^{- 1} + {(2^{- 1} - 3^{- 1})}^{- 1}$

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Solution

$\begin{matrix} = {(\frac{1}{6} - \frac{1}{8})}^{- 1} + {(\frac{1}{2} - \frac{1}{3})}^{- 1} = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{(\frac{1}{6} - \frac{1}{8})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{1}{(\frac{1}{2} - \frac{1}{3})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1}{\frac{8 - 6}{48}} ⎞ ⎟ ⎟ ⎟ ⎠ + ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1}{\frac{3 - 2}{6}} ⎞ ⎟ ⎟ ⎟ ⎠ = (\frac{48}{2}) + (\frac{6}{1}) = 24 + 6 = 30 \end{matrix}$
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Similar questions

7 (3 x - 1) - 6 (2 x + 3) = 8 (x - 2) + 1

\frac{1}{2} + \frac{1}{3} + \frac{1}{6}

\frac{x + 1}{2} + \frac{y - 1}{3} = 8

\frac{x - 1}{3} + \frac{y + 1}{2} = 9

\frac{x + 1}{2} + \frac{y - 1}{3} = 8,

\frac{x - 1}{3} + \frac{y + 1}{2} = 9

Question 1(j) :

Solve:

$\frac{1}{2} + \frac{1}{3} + \frac{1}{6}$

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