Question

Quero o resultado da equação{5x+y=39. x-y=3​

Answer 1

Oie, Td Bom?!

{5x + y = 39

{x - y = 3

{y = 39 - 5x

{x - y = 3

• x - y = 3

x - (39 - 5x) = 3

x - 39 + 5x = 3

6x - 39 = 3

6x = 3 + 39

6x = 42

x = 42/6

x = 7

• y = 39 - 5x

y = 39 - 5 . 7

y = 39 - 35

y = 4

S = {(x , y) = (7 , 4)}

Att. Makaveli1996

Answer 2

Resposta:

OLÁ VAMOS A SUA PERGUNTA:⇒⇒

$\left. \begin{cases} { 5x+y=39 } \\ { x-y=3 } \end{cases} \right.$

$5x+y=39,x-y=3$

$\left(\begin{matrix}5&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}39\\3\end{matrix}\right)$

$inverse(\left(\begin{matrix}5&1\\1&-1\end{matrix}\right))\left(\begin{matrix}5&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&1\\1&-1\end{matrix}\right))\left(\begin{matrix}39\\3\end{matrix}\right)$

$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&1\\1&-1\end{matrix}\right))\left(\begin{matrix}39\\3\end{matrix}\right)$

$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&1\\1&-1\end{matrix}\right))\left(\begin{matrix}39\\3\end{matrix}\right)$

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{-1}{5\left(-1\right)-1}&-\frac{1}{5\left(-1\right)-1}\\-\frac{1}{5\left(-1\right)-1}&\frac{5}{5\left(-1\right)-1}\end{matrix}\right)\left(\begin{matrix}39\\3\end{matrix}\right)$

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}&\frac{1}{6}\\\frac{1}{6}&-\frac{5}{6}\end{matrix}\right)\left(\begin{matrix}39\\3\end{matrix}\right)$

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}\times 39+\frac{1}{6}\times 3\\\frac{1}{6}\times 39-\frac{5}{6}\times 3\end{matrix}\right)$

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\\4\end{matrix}\right)$

$\star \: \text{Extraia os elementos x e y da matriz.}$

$\color{blue}{}\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese \:  x=7,y=4 }}}\ \checkmark$

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