Question

Verify the following property
$\begin{align}
  & a)-\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=\left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23} \\
 & b)-\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right) \\
\end{align}$

Answer 1

Hint: Now to verify the properties we will first calculate the LHS of the equation and then calculate RHS of the equation. Hence we can compare and say that the two are equal. To solve expressions on LHS or RHS we will use the BODMAS rule.

Complete step-by-step solution:
Now consider the first given equation $-\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=\left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}$
This property is associative property which means $a\times \left( b\times c \right)=\left( a\times b \right)\times c$ .
Now first consider LHS of equation which is $-\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)$
To find the value of this expression we will use BODMAS.
Hence we will first solve brackets
Now we know that $\left( \dfrac{p}{q}\times \dfrac{r}{s} \right)=\dfrac{p\times r}{q\times s}$ .
hence we get $\dfrac{5}{13}\times \dfrac{6}{23}=\dfrac{5\times 6}{13\times 23}$
Now $5 \times 6 = 30 $ and $13 \times 23 = 299$
Hence we have
$\begin{align}
  & -\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=-\dfrac{3}{20}\times \left( \dfrac{30}{299} \right) \\
 & \Rightarrow -\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=-\dfrac{3}{2}\times \left( \dfrac{3}{299} \right) \\
\end{align}$
Now again using the property $\left( \dfrac{p}{q}\times \dfrac{r}{s} \right)=\dfrac{p\times r}{q\times s}$
$-\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=-\dfrac{3\times 3}{299\times 2}$
Now $3 \times 3 = 9$ and $299 \times 2 = 598$
Hence we get $-\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=-\dfrac{9}{598}.....................\left( 1 \right)$
Now consider RHS of the equation.
$\left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}$
Now we will again solve this by BODMAS
Now as we know $\left( \dfrac{p}{q}\times \dfrac{r}{s} \right)=\dfrac{p\times r}{q\times s}$ we get
$-\dfrac{3}{20}\times \dfrac{5}{13}=-\dfrac{3\times 5}{20\times 13}$
Now we know $3 \times 5 = 15$ and $20 \times 13 = 260.$
Hence we get.
$\left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}=\left( -\dfrac{15}{260} \right)\times \dfrac{6}{23}$
Now again using $\left( \dfrac{p}{q}\times \dfrac{r}{s} \right)=\dfrac{p\times r}{q\times s}$ we get
$\left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}=-\dfrac{15\times 6}{260\times 23}$
Now we know $15 \times 6 = 90$ and $260 \times 23 = 5980.$
Hence we get
$\begin{align}
  & \left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}=-\dfrac{90}{5980} \\
 & \Rightarrow \left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}=-\dfrac{9}{598}.........................\left( 2 \right) \\
\end{align}$
Hence from equation (1) and equation (2) we get
$-\dfrac{3}{20}\times \left( \dfrac{5}{13}\times \dfrac{6}{23} \right)=\left( -\dfrac{3}{20}\times \dfrac{5}{13} \right)\times \dfrac{6}{23}$

(b) Now consider the equation $-\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)$
This property is nothing but $a\times \left( b-c \right)=\left( a\times b \right)-\left( a\times c \right)$ and called distributive property.
First let us consider LHS of the equation
$-\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)$
Now we will use the BODMAS rule, hence first solve the brackets.
Now in the bracket, we have subtraction of two fractions. Now we cannot directly subtract fractions. We will first have to equate their denominator. Hence we will first equate it to LCM of 12 and 21 which is 252.
Now consider
$\left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \dfrac{5\times 21}{12\times 21}-\dfrac{\left( -20 \right)\times 12}{21\times 12} \right)$
Now $21\times 12 = 252$, $5 \times 21 = 105,$ and $(-20) \times 12 = -240.$
Hence we get
$\left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \dfrac{105}{252}-\dfrac{-240}{252} \right)$
Now we know that $\left( \dfrac{p}{q}-\dfrac{r}{q} \right)=\dfrac{p-r}{q}$
Hence we get
$\begin{align}
  & \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \dfrac{105-\left( -240 \right)}{252} \right) \\
 & \Rightarrow \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \dfrac{105+240}{252} \right) \\
 & \Rightarrow \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \dfrac{345}{252} \right) \\
 & \Rightarrow -\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=-\dfrac{3}{10}\times \left( \dfrac{345}{252} \right) \\
 & \Rightarrow -\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=-\left( \dfrac{3\times 345}{10\times 252} \right) \\
\end{align}$
Now $3 \times 345 = 1035$ and $252 \times 10 = 2520.$
Hence we get
$-\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=-\dfrac{1035}{2520}$
On simplifying the fraction by dividing numerator and denominator by 45 we get
$-\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=-\dfrac{23}{56}..............\left( 3 \right)$
Now consider RHS of the equation. $\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)$
Now we know that $\left( \dfrac{p}{q}\times \dfrac{r}{s} \right)=\dfrac{p\times r}{q\times s}$
Hence we get
$\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( -\dfrac{3\times 5}{10\times 12} \right)-\left( -\left( \dfrac{3\times \left( -20 \right)}{10\times 21} \right) \right)$
Now $3 \times 5 = 15, 10 \times 12 = 120, 3 \times (-20) = -60, 10 \times 21 = 210.$
Hence we get
$\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( -\dfrac{15}{120} \right)+\left( \dfrac{-60}{210} \right)$
Now taking LCM of 120 and 210 we get.
$\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( -\dfrac{15\times 7}{120\times 7} \right)+\left( \dfrac{-60\times 4}{210\times 4} \right)$
Now $15 \times 7 = 105, -60 \times 4 = -240$ and $120 \times 7 = 210 \times 4 = 840.$
$\begin{align}
  & \left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( -\dfrac{105}{840} \right)+\left( \dfrac{-240}{840} \right) \\
 & \Rightarrow \left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( \dfrac{-105}{840} \right)+\left( \dfrac{-240}{840} \right) \\
\end{align}$
Now we have $\left( \dfrac{p}{q}+\dfrac{r}{q} \right)=\dfrac{p+r}{q}$
Hence we get
$\begin{align}
  & \left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( \dfrac{-105-240}{840} \right) \\
 & \Rightarrow \left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( \dfrac{-345}{840} \right) \\
\end{align}$
Now simplifying the fraction by dividing numerator and denominator by 15 we get.
$\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)=\left( -\dfrac{23}{56} \right)......................\left( 4 \right)$
Hence from equation (3) and equation (4) we get.
$-\dfrac{3}{10}\times \left( \dfrac{5}{12}-\dfrac{\left( -20 \right)}{21} \right)=\left( \left( -\dfrac{3}{10} \right)\times \dfrac{5}{12} \right)-\left( -\left( \dfrac{3}{10} \right)\times \left( \dfrac{-20}{21} \right) \right)$

Note: Note that when we multiply fractions we can multiply directly and $\left( \dfrac{p}{q}\times \dfrac{r}{s} \right)=\dfrac{p\times r}{q\times s}$ but when we add fractions we must check if their denominators are same, if not then take LCM of both denominators and make them same by multiplying appropriate number. Then for addition of fraction with same denominator we have $\left( \dfrac{p}{q}+\dfrac{r}{q} \right)=\dfrac{p+r}{q}$