Question

Using appropriate properties find: $-\dfrac{2}{3}\times \dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times \dfrac{1}{6}$ \[\]

Answer 1

Hint: We use the commutative property of addition that is $a+b=b+a$ for any two number $a,b$ to bring $-\dfrac{2}{3}\times \dfrac{3}{5}-\dfrac{3}{5}\times \dfrac{1}{6}+\dfrac{5}{2}$. We then take $\dfrac{3}{5}$ common using the distributive property of addition and multiplication that is $a\times b+a\times c=a\left( b+c \right)$. We simplify using these properties. \[\]

Complete step by step answer:
We know from the BODMAS rule that when we are given an expression with multiple arithmetic operations and then we have first to simplify the terms with brackets and then order( or power or exponent), division, multiplication, addition, subtraction in sequence. We are asked the question to find the value of
\[-\dfrac{2}{3}\times \dfrac{3}{5}+\dfrac{5}{2}-\dfrac{3}{5}\times \dfrac{1}{6}\]
 We see that we are given 5 rational numbers and 3 operations: multiplication, division and addition. We use the property of commutative of addition to have
\[\Rightarrow -\dfrac{2}{3}\times \dfrac{3}{5}-\dfrac{3}{5}\times \dfrac{1}{6}+\dfrac{5}{2}\]
We use the distributive property of multiplication and addition to take $\dfrac{3}{5}$ common in the above step to have;
\[\Rightarrow \dfrac{3}{5}\left( -\dfrac{2}{3}-\dfrac{1}{6} \right)+\dfrac{5}{2}\]
We use the distributive property of multiplication and addition to take $-1$ common in the above step to have;
 \[\begin{align}
  & \Rightarrow \dfrac{3}{5}\left( -1 \right)\left( \dfrac{2}{3}+\dfrac{1}{6} \right)+\dfrac{5}{2} \\
 & \Rightarrow \dfrac{-3}{5}\left( \dfrac{2}{3}+\dfrac{1}{6} \right)+\dfrac{5}{2} \\
\end{align}\]
We follow the BODMAS rule and solve the bracket first by adding the fractions inside. We have;
\[\begin{align}
  & \Rightarrow \dfrac{-3}{5}\left( \dfrac{2\times 2+1}{6} \right)+\dfrac{5}{2} \\
 & \Rightarrow \dfrac{-3}{5}\left( \dfrac{4+1}{6} \right)+\dfrac{5}{2} \\
 & \Rightarrow \dfrac{-3}{5}\times \dfrac{5}{6}+\dfrac{5}{2} \\
\end{align}\]
We follow the BODAMS rule and solve the multiplication first by cancelling out 5 in the numerator and denominator. We have;
\[\begin{align}
  & \Rightarrow \dfrac{-3}{6}+\dfrac{5}{2} \\
 & \Rightarrow \dfrac{-1}{2}+\dfrac{5}{2} \\
 & \Rightarrow \dfrac{-1+5}{2} \\
 & \Rightarrow \dfrac{4}{2}=2 \\
\end{align}\]

Note:
We know that when we add two rational numbers $\dfrac{a}{b},\dfrac{c}{d}\left( b\ne 0,d\ne 0 \right)$ we fist convert them into like rotational numbers which means rational number with same denominators. When we are cancelling out two numbers during multiplication we should remember that one must be in numerators and one must be in the denominator. The commutative property of multiplication is given by $a\times b=b\times a$.The associative property of addition is given by $\left( a+b \right)+c=a+\left( b+c \right)$ and associative property of multiplication is given by $a\times \left( b\times c \right)=\left( a\times b \right)\times c$.