Verify that $(x + y) \times z = x \times z + y \times z$, if

$x = 2, y = \dfrac{4}{5}$ and $z = \dfrac{3}{-10}$

To prove:

$(x + y) \times z = x \times z + y \times z$

Taking LHS:

$(x + y) \times z\\[1em] =\Big(2 + \dfrac{4}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{2}{1} + \dfrac{4}{5}\Big) \times \dfrac{3}{-10}$

LCM of 1 and 5 is 5.

$=\Big(\dfrac{2 \times 5}{1 \times 5} + \dfrac{4 \times 1}{5 \times 1}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{10}{5} + \dfrac{4}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{10 + 4}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{14}{5}\Big) \times \dfrac{3}{-10}\\[1em] =\Big(\dfrac{14 \times 3}{5 \times -10}\Big)\\[1em] =\Big(\dfrac{42}{-50}\Big)\\[1em] =\Big(\dfrac{-21}{25}\Big)$

Taking RHS:

$x \times z + y \times z\\[1em] =2 \times \dfrac{3}{-10} + \dfrac{4}{5} \times \dfrac{3}{-10}\\[1em] = \dfrac{2 \times 3}{1 \times -10} + \dfrac{4 \times 3}{5 \times -10}\\[1em] =\dfrac{6}{-10} + \dfrac{12}{-50}\\[1em] =\dfrac{-3}{5} + \dfrac{-6}{25}$

LCM of 5 and 25 is 5 x 5 = 25

$=\dfrac{-3 \times 5}{5 \times 5} + \dfrac{-6 \times 1}{25 \times 1}\\[1em] =\dfrac{-15}{25} + \dfrac{-6}{25}\\[1em] =\dfrac{-15 + (-6)}{25}\\[1em] =\dfrac{-21}{25}$

∴ LHS = RHS

$(x + y) \times z = x \times z + y \times z$