Question

8. Rezolvați prin metoda substituţiei următoarele sisteme:
2(x + y) – 3y =-11
(3(x + 2) - 2y = 2
b)
1 - 4x +3(x + y) =18
|3x + 4(7 – y) = 26
(2x +3(y + x) = -1
[2(x + y) + 4x = 12
c)
d)
3(x + y) - 2y =1
|3x – 2(x + 2y)=-11
(2(7x + y) – 3(2x - y)=-1
e)
(2(3x + 2y)+2(-x - 4y) =
3(2x - 3y) + 5y = -24
Repede imi trebuie pana la 23:40!

Answer 1

$\it a)\ \ 2(x+y) -3y =-11 \Rightarrow 2x+2y-3y=-11 \Rightarrow 2x-y=-11 \Rightarrow \\ \\ \Rightarrow 2x+11=y \Rightarrow y=2x+11\ \ \ \ (*),\ rela\c{\it t}ia\ de\ substitui\c{\it t}ie\\ \\ 3(x+2)-2y=2 \Rightarrow 3x+6-2y=2|_{-6} \Rightarrow 3x-2y=-4\stackrel{(*)}{\Longrightarrow}\\ \\ \Rightarrow 3x-2(2x+11)=-4 \Rightarrow 3x-4x-22=-4 \Rightarrow -x-22=-4|_{+22} \Rightarrow \\ \\ \Rightarrow -x=18|_{\cdot(-1)} \Rightarrow x=-18$

$\it x=-18\ \stackrel{(*)}{\Longrightarrow}\ y=2\cdot(-18)+11=-36+11=-25\\ \\ S=\{(-18,\ -25)\}$