Question

Ajudeemmm pooor favooor

Answer 1

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$\large\green{\boxed{\rm~~~\red{ 1)}~\blue{ N\tilde{a}o~s\tilde{a}o~equivalentes. }~~~}}$

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$\large\green{\boxed{\rm~~~\red{ 2)~A)}~\blue{ N\tilde{a}o~\acute{e}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$

$\large\green{\boxed{\rm~~~\red{ 2)~B)}~\blue{ \acute{E}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$

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$\large\green{\boxed{\rm~~~\red{ 3)}~\blue{ \acute{E}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$

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$\bf\large\green{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad}}$

$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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☺lá novamente, Chay. Vamos a mais um exercício❗ Acompanhe a resolução abaixo. ✌

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1)$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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☔ Vamos fazer esta verificação comparando o conjunto solução de ambos os sistemas é igual. Vamos encontrar o conjunto solução através da um método simples de isolar uma variável e substituir na outra equação.

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➡ I) $\sf\blue{ x - y = 2 }$

➡ $\sf\blue{ x = 2 + y }$

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➡ II) $\sf\blue{ 2 \cdot (2 + y) + y = 1 }$

➡ $\sf\blue{ 4 + 4y + y = 1 }$

➡ $\sf\blue{ 5y = -3 }$

➡ $\sf\blue{ y = \dfrac{-3}{5} }$

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➡ I) $\sf\blue{ x = 2 + \dfrac{-3}{5} }$

➡ $\sf\blue{ x = \dfrac{10}{5} - \dfrac{3}{5} }$

➡ $\sf\blue{ x = \dfrac{7}{5} }$

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$\large\gray{\boxed{\rm\blue{ S_1 = \{ \dfrac{7}{5}, \dfrac{-3}{5} \} }}}$

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➡ I) $\sf\blue{ x = -y }$

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➡ II) $\sf\blue{ 3 \cdot (-y) + 2y = 5 }$

➡ $\sf\blue{ -3y + 2y = 5 }$

➡ $\sf\blue{ -y = 5 }$

➡ $\sf\blue{ y = -5 }$

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➡ I) $\sf\blue{ x = -(-5) }$

➡ $\sf\blue{ x = 5 }$

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$\large\gray{\boxed{\rm\blue{ S_2 = \{ 5, -5 \} }}}$

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❌$\sf\blue{ S_1 \neq S_2 }$  ❌

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$\large\green{\boxed{\rm~~~\red{ 1)}~\blue{ N\tilde{a}o~s\tilde{a}o~equivalentes. }~~~}}$ ✅

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2)$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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☔ Inicialmente vamos somar as duas equações, onde encontraremos a equação (já na sua forma simplificada)

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$\large\gray{\boxed{\rm\blue{ 2x + y = 6 }}}$

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Ⓐ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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➡ $\sf\blue{ 2 \cdot (3) + (-15) \red{\overbrace{=}^{\large?}} 6}$

❌ $\sf\blue{ -9 \neq 6 }$ ❌

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$\large\green{\boxed{\rm~~~\red{ 2)~A)}~\blue{ N\tilde{a}o~\acute{e}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$ ✅

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Ⓑ$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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➡ $\sf\blue{ 2 \cdot (-2) + (10) \red{\overbrace{=}^{\large?}} 6 }$

✅ $\sf\blue{ 6 = 6 }$ ✅

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$\large\green{\boxed{\rm~~~\red{ 2)~B)}~\blue{ \acute{E}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$ ✅

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3)$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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☔ De forma semelhante podemos somar ambas as equações do sistema para verificar se a terna ordenada (1, 2, 3) é solução para a reta resultante

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$\large\gray{\boxed{\rm\blue{ 4x + y - 3z = -3 }}}$

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➡ $\sf\blue{ 4 \cdot (1) + (2) - 3 \cdot (3) \red{\overbrace{=}^{\large?}} -3 }$

✅ $\sf\blue{ -3 = -3 }$  ✅

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$\large\green{\boxed{\rm~~~\red{ 3)}~\blue{ \acute{E}~soluc_{\!\!\!,}\tilde{a}o. }~~~}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$