Question

utilizando o método de adiçao resolva: 
1)2x+3y=14   -x+5y=-7

2)x+7y=-2
-x-4y=-1

3)3x-5y=-14
-2x-8y=-2

Answer 1

E aí!

no primeiro, multiplique a equação II, por 2:

$+\begin{cases}2x+3y=14~~(I)\\ -2x+10y=-14~~(II)\end{cases}\\ ~~~~~-------\\ ~~~~~~~~~~~~~~~13y=0\\ ~~~~~~~~~~~~~~~~y=0/13\\ ~~~~~~~~~~~~~~~~~y=0\\\\\\ 2x+3y=14\\ 2x+3\cdot0=14\\ 2x=14\\ x=14/2\\ x=7\\\\\\ \large\boxed{\boxed{S_{x,y}=\{(7,0)\}}}.\\.$

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No segundo sistema basta somar as equações:

$+\begin{cases}~~x+7y=-2\\ -x-4y=-1\end{cases}\\ ~~~~~~-------\\ ~~~~~~~~~~~~~~3y=-3\\ ~~~~~~~~~~~~~~~y=-3/3\\ ~~~~~~~~~~~~~~~y=-1\\\\\\ x+7y=-2\\ x+7\cdot(-1)=-2\\ x-7=-2\\ x=-2+7\\ x=5\\\\\\ \large\boxed{\boxed{S_{x,y}=\{(5,-1)\}}}.\\.$

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No terceiro, multiplique a equação II por 1,5 e depois some ambas as equações:

$+\begin{cases}~~3x-5y=-14~~(I)~~(lembre-se~a~eq~II~esta~mult.~por~1,5)\\ -3x-12y=-3~~(II)\end{cases}\\ ~~~~~~---------\\ ~~~~~~~~~~~-17y~=~-17$
$y= \dfrac{-17}{-17}\\\\ y=1\\\\\\ 3x-5y=-14\\ 3x-5\cdot1=-14\\ 3x-5=-14\\ 3x=-14+5\\ 3x=-9\\\\ x= \dfrac{-9}{~~3}\\\\ x=-3\\\\\\ \large\boxed{\boxed{S_{x,y}=\{(-3,~1)\}}}.\\.$