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

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

To prove:

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

Taking LHS:

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

LCM of 5 and 3 is 3 x 5 = 15

$=\Big(\dfrac{4 \times 3}{5 \times 3} + \dfrac{-2 \times 5}{3 \times 5}\Big) \times -4\\[1em] =\Big(\dfrac{12}{15} + \dfrac{-10}{15}\Big) \times -4\\[1em] =\Big(\dfrac{12 + (-10)}{15}\Big) \times -4\\[1em] =\Big(\dfrac{2}{15}\Big) \times -4\\[1em] =\Big(\dfrac{2 \times -4}{15 \times 1}\Big)\\[1em] =\Big(\dfrac{-8}{15}\Big)$

Taking RHS:

$x \times z + y \times z\\[1em] =\dfrac{4}{5} \times -4 + \dfrac{-2}{3} \times -4\\[1em] = \dfrac{4 \times -4}{5 \times 1} + \dfrac{-2 \times -4}{3 \times 1}\\[1em] =\dfrac{-16}{5} + \dfrac{8}{3}\\[1em]$

LCM of 5 and 3 is 3 x 5 = 15

$=\dfrac{-16 \times 3}{5 \times 3} + \dfrac{8 \times 5}{3 \times 5}\\[1em] =\dfrac{-48}{15} + \dfrac{40}{15}\\[1em] =\dfrac{-48 + 40}{15}\\[1em] =\dfrac{-8}{15}$

∴ LHS = RHS

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