Question

resolva a equação pelo método da adição {2x + 3y = -4 { x - 3y = 11

Answer 1

2x+3y= -4  (I)
x-3y= 11  (II)

(I)  +  (II)
2x+3y=-4
+
 x-3y=11
________
3x=7
x=7/3

Substituindo em (II)
x-3y=11
(7/3)-3y=11  m.m.c.(1 , 3)=3
7-9y=33
-9y=33-7
-9y=26  multiplicando por -1
9y=-26
y=-26/9

Answer 2

$\begin{array}{l}\mathsf{Resolvendo~por~matrizes~o~seguinte~sistema:}\\\\\\\mathsf{S=}\left\{\begin{matrix}\mathsf{2x+3y=-4}\\\\\\\mathsf{x-3y=11} \end{matrix}\right.\\\\\\\textsf{Reescrevendo na forma de matriz incompleta, isto \'e, sem os termos}\\\textsf{independentes, teremos o seguinte:}\end{array}$


$\begin{array}{l}\mathsf{M=\left[\begin{array}{ccc}2&~~3\\1&-3\end{array}\right] }~~~~\textsf{Agora vamos calcular o determinante dessa matriz.}\\\\\textsf{O determinante de uma matriz de ordem n = 2 pode ser obtido multipl\underline{i}}\\\textsf{cando os elementos da diagonal principal e depois subtraindo desse valor}\\\textsf{o produto dos elementos da diagonal secund\'aria. Portanto:}\end{array}$

$\begin{array}{l}\mathsf{det~M=\begin{vmatrix}2~~~~~~~3\\1~~~-3\end{vmatrix}}=\mathsf{[2\cdot(-3)]-3\cdot 1=[-6]-3=-6-3=-9}\\\\\\\mathsf{Como~~det~M \neq 0~o~sistema~\'e~poss\'ivel~e~possui~\'unica~solu\c{c}\~ao.}\\\\\\\textsf{Pelo teorema de Crammer a solu\c{c}\~ao do sistema pode ser obtida da}\\\textsf{seguinte maneira:}\end{array}$


$\begin{array}{l}\mathsf{x=\dfrac{det~M_1}{det~M}~~~~onde~det~M_1~\'e~o~determinante~da~matriz~M~substituindo}\\\\\textsf{os elementos da primeira coluna pelos termos independentes (-4, 11).}\\\\\\\mathsf{y=\dfrac{det~M_2}{det~M}~~onde~det~M_2~\'e~o~determinante~da~matriz~M~substituindo}\\\\\textsf{os elementos da segunda coluna pelos termos independentes (-4, 11)}\\\\\\\mathsf{Portanto~vamos~calcular~det~M_1~e~det~M_2~(e~tamb\'em~x~e~y).}\end{array}$


$\begin{array}{l}\mathsf{det~M_1=\begin{vmatrix}\mathsf{-4~~~~~~~3}\\\mathsf{11~~~~-3}\end{vmatrix}=\mathsf{[-4\cdot(-3)]-33=[12]-33=12-33=-21}}\\\\\\\mathsf{x= \dfrac{det~M_1}{det~M}~\Longrightarrow~\dfrac{-21}{-9}=\dfrac{7}{3} }\end{array}$


$\begin{array}{l}\mathsf{det~M_2=\begin{vmatrix}\mathsf{2~~-4}\\\mathsf{1~~~~11}\end{vmatrix}}=\mathsf{(2\cdot 11)-(-4\cdot 1)=22-(-4)=22+4=26}\\\\\\\mathsf{y=\dfrac{det~M_2}{det~M}~\Longrightarrow~\dfrac{26}{-9}=-\dfrac{26}{9} }\end{array}$


$\textsf{Ent\~ao:}$


$\mathsf{Solu\c{c}\~ao~para~o~sistema:}\\\\\\\mathsf{S=\left\{\begin{matrix}\mathsf{x=\dfrac{7}{3}}\\\\\\\mathsf{y=-\dfrac{26}{9} } \end{matrix}\right.}$