6. Distributivity of multiplication over addition and subtraction

 

$a ×$ ($b + c$) $=$ ($a × b$) $+$ ($a × c$)

 

Let $a = \frac{-9}{2}$, $b = \frac{-5}{4}$ and $c = \frac{1}{3}$.

 

Then, $\frac{-9}{2} \times \left(\frac{-5}{4} + \frac{1}{3} \right)$ $=$ $\left(\frac{-9}{2} \times \frac{-5}{4} \right) +$ $\left(\frac{-9}{2} \times \frac{1}{3} \right)$

 

Let us verify the property.

 

Consider the LHS.

 

LHS $= \frac{-9}{2} \times \left(\frac{-5}{4} + \frac{1}{3} \right)$

 

$= \frac{-9}{2} \times \left(\frac{-15 + 4}{12} \right)$ [LCM of $3$ and $4$ is $12$]

 

$= \frac{-9}{2} \times \left(\frac{-11}{12} \right)$

 

$= \frac{99}{24}$

 

$= \frac{33}{8}$ ---- ($1$)

 

Now, consider the RHS.

 

RHS $=$ $\left(\frac{-9}{2} \times \frac{-5}{4} \right) +$ $\left(\frac{-9}{2} \times \frac{1}{3} \right)$

 

$= \frac{45}{8} + \left(\frac{-9}{6} \right)$

 

$= \frac{45}{8} - \frac{3}{2}$

 

$= \frac{45 - 12}{8}$ [LCM of $2$ and $8$ is $8$]

 

$= \frac{33}{8}$ ---- ($2$)

 

From equations ($1$) and ($2$), we have:

 

$\frac{-9}{2} \times \left(\frac{-5}{4} + \frac{1}{3} \right)$ $=$ $\left(\frac{-9}{2} \times \frac{-5}{4} \right) +$ $\left(\frac{-9}{2} \times \frac{1}{3} \right)$

 

This implies that $a ×$ ($b + c$) $=$ ($a × b$) $+$ ($a × c$).