Question

♥ Hello Friends !!! ♥


Answer this question ⤵⤵


If the ordered pair ( p , q ) satisfies the simultaneous equation (a + b)x + (b + c)y + (c + a) = 0 and
(b + c)x + (c + a)y + (a + b) = 0 such that p and q are in the ratio 1 : 2, then which of the following is correct ?


A) a^2 + 2ac + 3c^2 = 2b^2 + 3ab + bc

B) a^2 + b^2 + c^2 = ab + bc + ca

C) a^2 + 3ac + 3c^2 = 3b^2 + 3ab + bc

D) a^3 + b^3 + c^3 = 3abc


❎ NO SPAM ❎


Best answer will be marked as BRAINLIEST!

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$




$\textsf{It is given that p and q satisfies the equation , } \\ \textsf{it means p = x and q = y .}$




$\textsf{According to question, } \\ \\ \sf \implies p \: : \: q = \: 1 \: : \: 2 \: = \: x \: : \: y \\ \\ \sf \implies \dfrac{p}{q} \: = \: \dfrac{x}{y} \: = \: \dfrac{1}{2}$




$\underline{\textsf{Now,}} \\ \\ \sf \implies (a \: + \: b)x \: + \: (b \: + \: c)y \: + \: ( c \: + \: a ) \: = \: 0 \quad...(1) \\ \\ \textsf{And,} \\ \\ \sf \implies (b \: + \: c)x \: + \: (a \: + \: c)y \: + \: (a \: + \: b) \: = \: \quad...(2)$




$\underline{\textsf{By Cross Multiplication Method : }}$




$\sf Coefficient \: of \: x \quad Coefficient \: of \: y \quad Constant \: term \\ \\ \sf \: \: \: \: \: (a \: + \: b) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (b \: + \: c) \qquad \quad \: \: (a \: + \: c) \\ \\ \sf \: \: \: \: \: (b\: + \: c) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (a \: + \: c) \qquad \quad \: \: (a \: + \: b)$




$\sf \implies \dfrac{x}{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} \: = \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \dfrac{y}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b) } \: = \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf\dfrac{1}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }$




$\underline{\textsf{Now,}} \\ \\ \sf \implies \dfrac{x}{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} \: = \: \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{1}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) } \\ \\ \sf \: \: \therefore \: \: x \: = \: \dfrac{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} {(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }$




$\underline{\textsf{Again,}} \\ \\ \sf \implies\dfrac{y}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b) } \: = \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \dfrac{1}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) } \\ \\ \: \: \sf\therefore \: \: y \: = \: \dfrac{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b)}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }$




$\underline{\textsf{According to question , }} \\ \\ \sf \implies \dfrac{x}{y} \: = \: \dfrac{1}{2} \\ \\ \textsf{Plug the value of x and y , }$




$\sf \implies \dfrac{ \qquad\dfrac{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} { \cancel{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }} \qquad}{\dfrac{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b)}{ \cancel{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c)} } } \: = \: \dfrac{1}{2}$





$\sf \implies \dfrac{ \qquad (b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c) \qquad}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b) } \: = \: \dfrac{1}{2} \\ \\ \\ \sf \implies \dfrac{ \qquad (b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )^{2} \qquad}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)^{2} } \: = \: \dfrac{1}{2}$





$\sf \implies 2\{ (b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )^{2} \} \: = \\ \sf \: \: \: \: \: \: \: \: \: \: \:(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)^{2} \\ \\ \sf \implies2 \{ab \: + \: {b}^{2} \: + \: ac \: + \: bc \: - \: {a}^{2} \: - \: {c}^{2} \: - \: 2ac \} \: = \\ \: \: \: \: \: \: \: \: \: \: \: \sf ab \: + \: ac \: + \: bc \: + \: {c}^{2} \: - \: {a}^{2} \: - \: {b}^{2} \: - \: 2ab$





$\sf \implies2ab \: + \: 2{b}^{2} \: + \:2 ac \: + \: 2bc \: - \:2 {a}^{2} \: - \: 2 {c}^{2} \: - \: 4ac \: = \\ \: \: \: \: \: \: \: \: \: \: \: \sf ab \: + \: ac \: + \: bc \: + \: {c}^{2} \: - \: {a}^{2} \: - \: {b}^{2} \: - \: 2ab$





$\sf \implies2ab \: - \: ab \: + \: 2ab \: + \: 2 {b}^{2} \: + \: {b}^{2} \: + \: 2ac \: - \: 4ac \: - \: ac \\ \sf + \: 2bc \: - \: bc \: - \: 2 {a}^{2} \: + \: {a}^{2} \: - \: 2 {c}^{2} \: - \: {c}^{2} \: = \: 0$




$\sf \implies3ab \: + \: 3 {b}^{2} \: - \: 3ac \: + \: bc \: - \: {a}^{2} \: - \: 3 {c}^{2} \: = \: 0 \\ \\ \sf \implies 3 {b}^{2} \: + \: 3ab \: + \: bc \: = \: 3 {c}^{2} \: + \: {a}^{2} \: + \: 3ac \\ \\ \sf \: \: \therefore \: \: {a}^{2} \: + \: 3ac \: + \: 3 {c}^{2} \: = \: 3 {b}^{2} \: + \: 3ab \: + bc$




$\boxed{\mathsf{The \: required \: answer \: is \: (c).}}$

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$




$\textsf{It is given that p and q satisfies the equation , } \\ \textsf{it means p = x and q = y .}$




$\textsf{According to question, } \\ \\ \sf \implies p \: : \: q = \: 1 \: : \: 2 \: = \: x \: : \: y \\ \\ \sf \implies \dfrac{p}{q} \: = \: \dfrac{x}{y} \: = \: \dfrac{1}{2}$




$\underline{\textsf{Now,}} \\ \\ \sf \implies (a \: + \: b)x \: + \: (b \: + \: c)y \: + \: ( c \: + \: a ) \: = \: 0 \quad...(1) \\ \\ \textsf{And,} \\ \\ \sf \implies (b \: + \: c)x \: + \: (a \: + \: c)y \: + \: (a \: + \: b) \: = \: \quad...(2)$




$\underline{\textsf{By Cross Multiplication Method : }}$




$\sf Coefficient \: of \: x \quad Coefficient \: of \: y \quad Constant \: term \\ \\ \sf \: \: \: \: \: (a \: + \: b) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (b \: + \: c) \qquad \quad \: \: (a \: + \: c) \\ \\ \sf \: \: \: \: \: (b\: + \: c) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (a \: + \: c) \qquad \quad \: \: (a \: + \: b)$




$\sf \implies \dfrac{x}{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} \: = \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \dfrac{y}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b) } \: = \\ \\ \: \: \: \: \: \: \: \: \: \: \: \sf\dfrac{1}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }$




$\underline{\textsf{Now,}} \\ \\ \sf \implies \dfrac{x}{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} \: = \: \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{1}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) } \\ \\ \sf \: \: \therefore \: \: x \: = \: \dfrac{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} {(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }$




$\underline{\textsf{Again,}} \\ \\ \sf \implies\dfrac{y}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b) } \: = \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \dfrac{1}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) } \\ \\ \: \: \sf\therefore \: \: y \: = \: \dfrac{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b)}{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }$




$\underline{\textsf{According to question , }} \\ \\ \sf \implies \dfrac{x}{y} \: = \: \dfrac{1}{2} \\ \\ \textsf{Plug the value of x and y , }$




$\sf \implies \dfrac{ \qquad\dfrac{(b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c )} { \cancel{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c) }} \qquad}{\dfrac{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b)}{ \cancel{(a \: + \: b)(a \: + \: c) \: - \: (b \: + \: c)(b \: + \: c)} } } \: = \: \dfrac{1}{2}$





$\sf \implies \dfrac{ \qquad (b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )(a \: + \: c) \qquad}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)(a \: + \: b) } \: = \: \dfrac{1}{2} \\ \\ \\ \sf \implies \dfrac{ \qquad (b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )^{2} \qquad}{(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)^{2} } \: = \: \dfrac{1}{2}$





$\sf \implies 2\{ (b \: + \: c )(a \: + \: b ) \: - \: (a \: + \: c )^{2} \} \: = \\ \sf \: \: \: \: \: \: \: \: \: \: \:(a \: + \: c )(b \: + \: c) \: - \: (a \: + \: b)^{2} \\ \\ \sf \implies2 \{ab \: + \: {b}^{2} \: + \: ac \: + \: bc \: - \: {a}^{2} \: - \: {c}^{2} \: - \: 2ac \} \: = \\ \: \: \: \: \: \: \: \: \: \: \: \sf ab \: + \: ac \: + \: bc \: + \: {c}^{2} \: - \: {a}^{2} \: - \: {b}^{2} \: - \: 2ab$





$\sf \implies2ab \: + \: 2{b}^{2} \: + \:2 ac \: + \: 2bc \: - \:2 {a}^{2} \: - \: 2 {c}^{2} \: - \: 4ac \: = \\ \: \: \: \: \: \: \: \: \: \: \: \sf ab \: + \: ac \: + \: bc \: + \: {c}^{2} \: - \: {a}^{2} \: - \: {b}^{2} \: - \: 2ab$





$\sf \implies2ab \: - \: ab \: + \: 2ab \: + \: 2 {b}^{2} \: + \: {b}^{2} \: + \: 2ac \: - \: 4ac \: - \: ac \\ \sf + \: 2bc \: - \: bc \: - \: 2 {a}^{2} \: + \: {a}^{2} \: - \: 2 {c}^{2} \: - \: {c}^{2} \: = \: 0$




$\sf \implies3ab \: + \: 3 {b}^{2} \: - \: 3ac \: + \: bc \: - \: {a}^{2} \: - \: 3 {c}^{2} \: = \: 0 \\ \\ \sf \implies 3 {b}^{2} \: + \: 3ab \: + \: bc \: = \: 3 {c}^{2} \: + \: {a}^{2} \: + \: 3ac \\ \\ \sf \: \: \therefore \: \: {a}^{2} \: + \: 3ac \: + \: 3 {c}^{2} \: = \: 3 {b}^{2} \: + \: 3ab \: + bc$




$\boxed{\mathsf{The \: required \: answer \: is \: (c).}}$