Question

determine o valor de x+y​

Answer 1

$\left(\begin{array}{ccc}3&-1\\1&3\end{array}\right)~.~\left(\begin{array}{ccc}x\\y\end{array}\right)~=~4\,.\,\left(\begin{array}{ccc}1\\2\end{array}\right)\\\\\\\left(\begin{array}{ccc}3~.~x\,-\,1~.~y\\1~.~x\,-\,3~.~y\end{array}\right)~=~\left(\begin{array}{ccc}4~.~1\\4~.~2\end{array}\right)\\\\\\\left(\begin{array}{ccc}3x\,-\,y\\x\,-\,3y\end{array}\right)~=~\left(\begin{array}{ccc}4\\8\end{array}\right)$

$\left\{\begin{array}{ccc}3x\,-\,y&=&4\\x\,-\,3y&=&8\end{array}\right\\\\\\Isolando~y~na~primeira~equacao~e~substituindo\!-\!o~na~segunda\\\\\\3x-y~=~4~~~\rightarrow~~~\boxed{y~=~3x-4}\\\\\\x-3y~=~8\\\\\\x-3\,.\,(3x-4)~=~8\\\\\\x-9x+12~=~8\\\\\\-8x~=\,-4\\\\\\\boxed{x~=~\frac{1}{2}}$

$Substituindo~o~valor~de~"x"\\\\\\y~=~3x-4\\\\\\y~=~3\,.\,\frac{1}{2}-4\\\\\\y~=~\frac{3}{2}-4\\\\\\\boxed{y~=\,-\frac{5}{2}}$

$Sendo~assim,~x+y~\acute{e}~igual~a:\\\\\\x+y~=~\frac{1}{2}~+\,\left(-\frac{5}{2}\right)\\\\\\x+y~=\,-\frac{4}{2}\\\\\\\boxed{x+y~=\,-2}$

Answer 2

Resposta:

3x-y=4  ==>y=3x-4   (i)

x+3y=8    (ii)

(i)   em (ii)

x+3*(3x-4)=8

x+9x-12=8

10x=20  ==>x=2

Usando (i)   ==> y=3*2-4 = 2

x+y   =  2+2 = 4