Question

1: resolva os sistemas abaixo usando o método adição: 2x + y = 60 3x + y = 600​

Answer 1

Resposta:

$2x+y=~60~equacao~I\\ 3x+y=600~equacao~II\\ \\ multiplique~uma~das~equacoes~por~(-1)\\ \\ equacao~I\\ \\ 2x+y=60~~.~(-1)=\\ \\ -2x-y=-60\\ \\ ~~~~~~~~~~-2x-y=-60~equacao~I\\ +~~~~~~~~~~3x+y=600~equacao~II\\ ~~~~~~~~~~--------\\ ~~~~~~~~~~~~~~~x+0=540\\ \\ x=540\\ \\ substitua~(x=540)~em~uma~das~equacoes\\ \\ 3x+y=600\\ \\ 3.(540)+y=600\\ \\ 1620+y=600\\ \\ y=600-1620\\ \\ y=1020\\ \\ \\ S~~\left \{ {{y=-1020;~~x=540} {}} \}$

Answer 2

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$\begin{cases}\sf 2x+y=60\\\sf3x+y=600\end{cases}\\\underline{\rm multiplicando~a~1^{\underline a}~equac_{\!\!,}\tilde ao~por~-1~temos:}\\\begin{cases}\sf2x+y=60\cdot(-1)\\\sf3x+y=600\end{cases}\\\begin{cases}\sf-2x-\diagup\!\!\!\!y=-60\\\sf3x+\diagup\!\!\!\!y=600\end{cases}\\\sf x=540\checkmark\\\sf 2x+y=60\\\sf 2\cdot540+y=60\\\sf 1080+y=60\\\sf y=60-1080\\\sf y=-1020\checkmark\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\{540,-1020\}}}}}\checkmark$