Question

subtract the sum of 2 whole 3/7 and -5 whole 4/7 from the difference of 4/15 and -2/5​

Answer 1

$\huge \bf{ Answer}$

$\tt \red{Sum \: of \: 2\dfrac{3}{7} \: and \: - 5 \dfrac{4}{7} }$

$\implies \bf \dfrac{17}{7} + ( \dfrac{ - 39}{7} )$

$\implies \bf \dfrac{17}{7} - \dfrac{39}{7}$

$\implies \bf \dfrac{17 - 39}{7}$

$\implies \bf \dfrac{ - 22}{7}$

$\tt \red{Difference \: of \: \dfrac{4}{15} \: and \: \dfrac{ - 2}{5} }$

$\implies \bf \dfrac{4}{15} - ( \dfrac{ - 2}{5} )$

$\implies \bf \dfrac{4}{15} + \dfrac{2}{5}$

$\implies \bf \dfrac{4 +6 }{15}$

$\implies \bf \dfrac{10}{15}$

Now, According to Question

$\implies \bf \blue{Difference - Sum = ?}$

Substituting the values

$\implies \bf {\dfrac{10}{15} - (\dfrac{ - 22}{7} })$

$\implies \bf \dfrac{10}{15} + \dfrac{22}{7}$

$\implies \bf \dfrac{70 + 330}{105}$

$\implies \bf \cancel \dfrac{400}{105}$

$\implies \bf \dfrac{80}{21}$

$\huge \bf {Some \ Important \ Identities}$

$\bf \pink {(a+b)^{0} = 1}$

$\bf \pink {(a+b)^{1} = a + b}$

$\bf \pink {(a+b)^{2} = a^{2} + 2ab + b^{2}}$

$\bf \pink {(a-b)^{2} = a^{2} - 2ab + b^{2}}$

$\bf \pink {a^{2} - b^{2} = (a+b)(a-b)}$

$\bf \pink { {a}^{2} + {b}^{2} = (a + b {)}^{2} - 2ab }$

$\bf \pink { {a}^{2} + {b}^{2} = (a - b {)}^{2} + 2ab }$

$\bf \pink { {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2}) }$

$\bf \pink { {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2}) }$

$\bf \pink{(a + b + c {)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}$

$\bf \pink{ {a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }$

Answer 2

Sum of $\longrightarrow{\tt{ 2\dfrac{3}{7} and\: -5 \dfrac{4}{7}}}$ is :

$\longrightarrow{\tt{ \dfrac{17}{7} and \: \dfrac{-39}{7}}}$

$\longrightarrow{\tt{ \dfrac{17 + (-39)}{7}}}$

$\boxed{\tt{\dfrac{-22}{7}}}$

Difference of $\longrightarrow{\tt{ \dfrac{4}{15} and \: \dfrac{-2}{5}}}$ is :

$\longrightarrow{\tt{ \dfrac{4}{15} - \dfrac{-2}{5}}}$

$\longrightarrow{\tt{ LCM \implies 15}}$

$\longrightarrow{\tt{ \dfrac{4 - (-6)}{15}}}$

$\boxed{\tt{\dfrac{10}{15}}}$

So , the difference of sum and difference of given problem :

$\longrightarrow{\tt{ \dfrac{10}{15} - \dfrac{-22}{7}}}$

$\longrightarrow{\tt{ LCM \implies 105}}$

$\longrightarrow{\tt{ \dfrac{70 - (-330)}{105}}}$

$\boxed{\tt{\dfrac{400}{105}}}$

This can be simplified into :

$\large{\boxed{\tt{\dfrac{80}{21}}}}$