Question

Determine o conjunto-solução dos sistemas:

$a) \: x + y = \frac{5}{6}$
$2x - y = \frac{2}{3}$

$b) \: \frac{x}{4} + \frac{y}{2} = 9$
$\frac{x}{2} - \frac{3y}{5} = - 4$
​

Answer 1

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$\tt a)~\begin{cases}\sf x+y=\dfrac{5}{6}\\\sf 2x-y=\dfrac{2}{3}\end{cases}\\\underline{\sf preparando~o~sistema~temos:}\\\sf x+y=\dfrac{5}{6}\cdot 6\\\sf 6x+6y=5\\\sf 2x-y=\dfrac{2}{3}\cdot3\\\sf 6x-3y=2\\\underline{\sf o~sistema~equivalente~\acute e:}\\\begin{cases}\sf 6x+6y=5\\\sf 6x-3y=2\end{cases}\\\underline{\sf multiplicando~a~2^{\underline a}~equac_{\!\!,}\tilde ao~por~2~temos:}\\\begin{cases}\sf6x+6y=5\\\sf 6x-3y=2\cdot(2)\end{cases}$

$+\underline{\begin{cases}\sf6x+\diagdown\!\!\!\!\!\!6y=5\\\sf12x-\diagdown\!\!\!\!\!\!6y=4\end{cases}}\\\sf 18x=9\\\sf x=\dfrac{9\div9}{18\div9}\\\sf x=\dfrac{1}{2}\\\sf 6x+6y=5\\\sf 6\cdot\left(\dfrac{1}{2}\right)+6y=5\cdot 2\\\sf 6+12y=10\\\sf 12y=10-6\\\sf 12y=4\\\sf y=\dfrac{4\div4}{12\div4}\\\sf y=\dfrac{1}{3}\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\left\{\dfrac{1}{2},\dfrac{1}{3}\right\}}}}}$

$\tt b)~\begin{cases}\sf\dfrac{x}{4}+\dfrac{y}{2}=9\\\sf\dfrac{x}{2}-\dfrac{3y}{5}=-4\end{cases}\\\underline{\sf preparando~o~sistema~temos:}\\\sf \dfrac{x}{4}+\dfrac{y}{2}=9\cdot4\\\sf x+2y=36\\\sf \dfrac{x}{2}-\dfrac{3y}{5}=-4\cdot10\\\sf 5x-6y=-40\\\underline{\sf o~sistema~equivalente~\acute e:}\\\begin{cases}\sf x+2y=36\\\sf 5x-6y=-40\end{cases}\\\underline{\sf multiplicando~a~1^{\underline a}~equac_{\!\!,}\tilde ao~por~3~temos:}}\\\begin{cases}\sf x+2y=36\cdot(3)\\\sf5x-6y=-40\end{cases}$

$+\underline{\begin{cases}\sf 3x+\diagdown\!\!\!\!\!\!6y=108\\\sf5x-\diagdown\!\!\!\!\!\!6y=-40\end{cases}}\\\sf 8x=68\\\sf x=\dfrac{68\div4}{8\div4}\\\sf x=\dfrac{17}{2}\\\sf x+2y=36\\\sf\dfrac{17}{2}+2y=36\cdot(2)\\\sf 17+4y=72\\\sf 4y=72-17\\\sf 4y=55\\\sf y=\dfrac{55}{4}\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\left\{\dfrac{17}{2},\dfrac{55}{4}\right\}}}}}$