Question

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If a = -25, b = 20 cm and c = -10, verify that:
(i) a + (b + c) = (a + b) + c
(ii) a × (b + c) = a × b + a × c​

Answer 1

Answer~

Given :-

• a= -25
• b= 20
• c= -10

To find them:-

• (i) a + (b + c) = (a + b) + c
• (ii) a × (b + c) = a × b + a × c

Solution:-

(i) a + (b + c) = (a + b) + c

→(-5)+ {20+(-10)} = {(-5) + 20} + (-10)

→(-5) + 10 = 15 + (-10)

→5 = 5

✅Hence, verifed

Solution:-

(ii) a × (b + c) = a × b + a × c

→(-5) × {20 + (-10)} = (-5) × 20 + (-5) × (-10)

→(-5) × 10 = (-100) + 50

→(-50) = (-50)

✅Hence,verified

Answer 2

Answer:

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Question :}}}}}}\end{gathered}$

If a = -25, b = 20 and c = -10, verify that:

• $\dashrightarrow$ (i) a + (b + c) = (a + b) + c
• $\dashrightarrow$ (ii) a × (b + c) = a × b + a × c

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}\end{gathered}$

$\bigstar \: {\underline{\underline{\bf{\purple{Here}}}}}$

• $\dashrightarrow$ a = -25,
• $\dashrightarrow$ b = 20
• $\dashrightarrow$ c = -10,

$\begin{gathered}\end{gathered}$

$\bigstar \: \underline{\underline{\bf{\purple{(i) \: a + (b + c) = (a + b) + c}}}}$

• Substituting the values

${: \longmapsto{ - 25 + \bigg(20 + (- 10) \bigg) = \bigg(( - 25) + 20\bigg) - 10}}$

${: \longmapsto{\bigg( - 25 + 10 \bigg) = \bigg( - 5- 10} \bigg)}$

${: \longmapsto{\bigg(25 - 10 \bigg) = \bigg( 5 + 10} \bigg)}$

${: \longmapsto{\bigg(15 \bigg) = \bigg(15} \bigg)}$

$\dag{\underline{\boxed{\sf{\red{LHS=RHS }}}}}$

• Hence Verified!

$\begin{gathered}\end{gathered}$

$\bigstar \: \underline{\underline{\bf{\purple{(ii) \: a \times (b + c) = a \times b + a \times c}}}}$

• Substituting the values

${: \implies{\sf{ - 25 \times \bigg(20 + ( - 10) \bigg) = \bigg(( - 25) \times 20\bigg) + \bigg(( - 20) \times ( - 10) \bigg)}}}$

${: \implies{\sf{ - 25 \times \bigg(10 \bigg) = \bigg(( - 25) \times 20\bigg) + \bigg(( \cancel{- } 25) \times ( \cancel{-} 10) \bigg)}}}$

${: \implies{\sf{ - 250 = \bigg( - 500 \bigg) + \bigg(25\times10 \bigg)}}}$

${: \implies{\sf{ - 250 = - 500 + (250)}}}$

${: \implies{\sf{ - 250 = -250}}}$

$\dag{\underline{\boxed{\sf{\red{LHS=RHS }}}}}$

• Hence Verified!

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Additional Information :}}}}}}\end{gathered}$

$\begin{gathered}\small\boxed{\begin{array}{c} \bf{\dag}\:\underline{\rm{Some\:important\:algebric\:identities\:::}} \\\\ \small{\bigstar}\:\rm{ (A+B)^{2} = A^{2} + 2AB + B^{2}} \\\\ \small{\bigstar}\rm\: {(A-B)^{2} = A^{2} - 2AB + B^{2}} \\\\ \small{\bigstar}\rm\:{A^{2} - B^{2} = (A+B)(A-B)}\\\\ \small{\bigstar}\rm\:{(A+B)^{2} = (A-B)^{2} + 4AB}\\\\ \small{\bigstar}\rm\: {(A-B)^{2} = (A+B)^{2} - 4AB}\\\\ \small{\bigstar} \rm\:{(A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}}\\\\ \small{\bigstar}\rm\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\ \small\bigstar\rm\: {A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})} \end{array}}\end{gathered}$

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