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

$x = \dfrac{3}{4}, y = \dfrac{8}{9}$ and $z = -5$

To prove:

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

Taking LHS:

$x \times (y - z)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8}{9} - (-5)\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8}{9} - \dfrac{-5}{1}\Big)$

LCM of 9 and 1 is 3 x 3 = 9.

$=\dfrac{3}{4} \times \Big(\dfrac{8 \times 1}{9 \times 1} - \dfrac{-5 \times 9}{1 \times 9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8}{9} - \dfrac{-45}{9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8 - (-45)}{9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{8 + 45}{9}\Big)\\[1em] =\dfrac{3}{4} \times \Big(\dfrac{53}{9}\Big)\\[1em] =\Big(\dfrac{3 \times 53}{4 \times 9}\Big)\\[1em] =\Big(\dfrac{159}{36}\Big)\\[1em] =\Big(\dfrac{53}{12}\Big)$

Taking RHS:

$x \times y - x \times z\\[1em] =\dfrac{3}{4} \times \dfrac{8}{9} - \dfrac{3}{4} \times -5\\[1em] = \dfrac{3 \times 8}{4 \times 9} - \dfrac{3 \times -5}{4 \times 1}\\[1em] =\dfrac{24}{36} - \dfrac{-15}{4}\\[1em] =\dfrac{2}{3} - \dfrac{-15}{4}$

LCM of 3 and 4 is 2 x 2 x 3 = 12

$=\dfrac{2 \times 4}{3 \times 4} - \dfrac{-15 \times 3}{4 \times 3}\\[1em] =\dfrac{8}{12} - \dfrac{-45}{12}\\[1em] =\dfrac{8 - (-45)}{12}\\[1em] =\dfrac{8 + 45}{12}\\[1em] =\dfrac{53}{12}$

∴ LHS = RHS

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