Verify associative law
a+ (b+c) =(a+b) +c
Where a= 7/5, b= -2/5, c= -1/3​

1 thought on “Verify associative law<br />a+ (b+c) =(a+b) +c<br />Where a= 7/5, b= -2/5, c= -1/3​”

1. ☯GIVEN :

   • a = [tex] \bf {\dfrac{7}{5}} [/tex]

   • b= [tex]\bf {\dfrac{-2}{5}} [/tex]

   • c = [tex] \bf{\dfrac{-1}{3}} [/tex]

   ☯TO DO :

   • Verify associative law :[tex] \bf {a+(b+c) =(a+b) +c}[/tex] .

   ➲SOLUTION :

   ★[tex] \bf{LHS = a+(b+c)}[/tex]

   • Substituting the given values :

   [tex] : \longrightarrow\sf{LHS = a + (b + c)} \\ \\[/tex]

   [tex] : \longrightarrow\sf{LHS = \dfrac{7}{5} + \left( \dfrac{ – 2}{5} + \dfrac{ -1 }{3} \right)} \\ \\ [/tex]

   [tex]: \longrightarrow\sf{LHS = \dfrac{7}{5} + \left( \dfrac{ – 6 + ( – 5)}{15} \right)} \\ \\ [/tex]

   [tex]: \longrightarrow\sf{LHS = \dfrac{7}{5} + \left( \dfrac{ – 6 – 5}{15} \right)} \\ \\ [/tex]

   [tex] : \longrightarrow\sf{LHS = \dfrac{7}{5} + \dfrac{ – 11}{15} } \\ \\ [/tex]

   [tex] : \longrightarrow\sf{LHS = \dfrac{21 + ( – 11)}{15} } \\ \\[/tex]

   [tex] : \longrightarrow\sf{LHS = \dfrac{21 – 11}{15} } \\ \\[/tex]

   [tex]: \longrightarrow\sf{LHS = \dfrac{\cancel{10}}{ \cancel{15}} } \\ \\ [/tex]

   [tex] : \longrightarrow \underline{\boxed{ \purple {\bf{LHS = \dfrac{2}{3}}} }} [/tex]

   ★[tex] \bf{RHS = (a+b) +c}[/tex]

   • Substituting the given values :

   [tex] : \longrightarrow\sf{RHS =( a + b ) + c} \\ \\[/tex]

   [tex] : \longrightarrow\sf{RHS = \left( \dfrac{ 7}{5} + \dfrac{ -2}{5} \right) + \dfrac{ – 1}{3} } \\ \\ [/tex]

   [tex]: \longrightarrow\sf{RHS = \left( \dfrac{ 7 + ( – 2)}{5} \right) + \dfrac{ – 1}{3} } \\ \\ [/tex]

   [tex]: \longrightarrow\sf{RHS = \left( \dfrac{ 7 – 2}{5} \right) + \dfrac{ – 1}{3}} \\ \\ [/tex]

   [tex] : \longrightarrow\sf{RHS = \left( \cancel{\dfrac{ 5}{5}} \right) + \dfrac{ – 1}{3}} \\ \\ [/tex]

   [tex] : \longrightarrow\sf{RHS = 1+ \dfrac{ – 1}{3} } \\ \\ [/tex]

   [tex] : \longrightarrow\sf{RHS = \dfrac{3 + ( – 1)}{3} } \\ \\[/tex]

   [tex] : \longrightarrow\sf{RHS = \dfrac{3 – 1}{3} } \\ \\[/tex]

   [tex] : \longrightarrow \underline{\boxed{ \purple {\bf{RHS = \dfrac{2}{3}}} }} [/tex]

   Here,

   [tex] : \longrightarrow \underline{\huge{\boxed{ \green {\bf{LHS = RHS }}}}} [/tex]

   [tex]\huge {\pink {\therefore}}[/tex] Verified.

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