Resolva as equações em R.1) 2 x + 6 = x + 18 2) 5 x – 3 = 2 x + 9 3) 3 (2x-3) + 2 (x + 1) = 3x + 184) 2x + 3 (x – 5) = 4x + 95) 2 (x + 1) – 3 (2x –5)

Resolva as equações em R.1) 2 x + 6 = x + 18 2) 5 x – 3 = 2 x + 9 3) 3 (2x-3) + 2 (x + 1) = 3x + 184) 2x + 3 (x – 5) = 4x + 95) 2 (x + 1) – 3 (2x –5) = 6x – 3 6) 3x – 5 = x – 27) 3x – 5 = 138) 3x + 5 = 2 9) x – (2x – 1) = 23 10) 2x – (x – 1) = 5 – (x – 3)

Matemática

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Colegio Fazer Crescer

Francisco Coelho

16.04.2019

💡 5 Respostas

Andre Smaira

30.09.2019

Para resolver as equações segue-se os seguintes passos:

\(1)\)

: desenvolva os membros entre parênteses, caso haja necessidade;

\(2)\)

: isole as partes literais de um lado; e os números do outro, efetue os cálculos. Por fim

\(3)\)

: Passe o coeficiente numérico da parte literal dividindo os membros de tal forma que se encontre, enfim, o valor do conjunto verdade. Da teoria á prática:

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$1){\text{ }}2{\text{ }}x{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}x{\text{ }} + {\text{ }}18 \to x = 12\therefore \boxed{V = \{ 12\} }$

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$2){\text{ }}5{\text{ }}x{\text{ }}-{\text{ }}3{\text{ }} = {\text{ }}2{\text{ }}x{\text{ }} + {\text{ }}9 \to 3x = 12 \to x = 4\therefore \boxed{V = \{ 4\} }$

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$3){\text{ }}3{\text{ }}\left( {2x - 3} \right){\text{ }} + {\text{ }}2{\text{ }}\left( {x{\text{ }} + {\text{ }}1} \right){\text{ }} = {\text{ }}3x{\text{ }} + {\text{ }}18 \to 6x - 9 + 2x + 2 = 3x + 18 \to 5x = 25 \to x = 5\therefore \boxed{V = \{ 5\} }$

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$4){\text{ }}2x{\text{ }} + {\text{ }}3{\text{ }}\left( {x{\text{ }}-{\text{ }}5} \right){\text{ }} = {\text{ }}4x{\text{ }} + {\text{ }}9 \to 2x + 3x - 15 = 4x + 9 \to x = 24\therefore \boxed{V = \{ 24\} }$

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$5){\text{ }}2{\text{ }}\left( {x{\text{ }} + {\text{ }}1} \right){\text{ }}-{\text{ }}3{\text{ }}\left( {2x{\text{ }}-5} \right){\text{ }} = {\text{ }}6x{\text{ }}-{\text{ }}3 \to 2x + 2 - 6x + 15 = 6x - 3 \to - 10x = - 20 \to x = 2\therefore \boxed{V = \{ 2\} }$

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$6){\text{ }}3x{\text{ }}-{\text{ }}5{\text{ }} = {\text{ }}x{\text{ }}-{\text{ }}2 \to 2x = 3 \to x = \dfrac{3}{2}\therefore \boxed{V = \{ \dfrac{3}{2}\} }$

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$7){\text{ }}3x{\text{ }}-{\text{ }}5{\text{ }} = {\text{ }}13 \to 3x = 18 \to x = 6\therefore \boxed{V = \{ 6\} }$

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$8){\text{ }}3x{\text{ }} + {\text{ }}5{\text{ }} = {\text{ }}2{\text{ }} \to 3x = 3 \to x = 1\therefore \boxed{V = \{ 1\} }$

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$9){\text{ }}x{\text{ }}-{\text{ }}\left( {2x{\text{ }}-{\text{ }}1} \right){\text{ }} = {\text{ }}23{\text{ }} \to x - 2x + 1 = 23 \to - x = 22 \to x = - 22\therefore \boxed{V = \{ - 22\} }$

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$10){\text{ }}2x{\text{ }}-{\text{ }}\left( {x{\text{ }}-{\text{ }}1} \right){\text{ }} = {\text{ }}5{\text{ }}-{\text{ }}\left( {x{\text{ }}-{\text{ }}3} \right) \to 2x - x + 1 = 5 - x + 3 \to 2x = 7 \to x = \dfrac{7}{2}\therefore \boxed{V = \{ \dfrac{7}{2}\} }$