Question

Functia de gradul 1
1. Rezolvati sistemele:

a) $\left \{ {{ \frac{2}{x-y} + \frac{3}{x+y} =-1} \atop { \frac{3}{x-y} - \frac{2}{x+y} = 5}} \right.$

b) $\left \{ {{ 3[x] + 2[y] = -1} \atop { [x] - 3[y] = -4}} \right.$

Answer 1

[tex]a) \left \{ {{\frac{2}{x-y}+\frac{3}{x+y}=-1} \atop {\frac{3}{x-y}-\frac{2}{x+y}=5}} \right. \\ Notam:\frac{1}{x-y}=a, \frac{1}{x+y}=b\\ \left \{ {{2a+3b=-1}|\cdot 3 \atop {3a-2b=5}|\cdot 2} \right. \Leftrightarrow \left \{ {{6a+9b=-3} \atop {6a-4b=10}} \right. \\ Le\ scadem: \ \ \ \ \ 13b=-13\Rightarrow b=-1\\ 2a-3=-1\Rightarrow a=1\\ Asadar:\\ \left \{ {{\frac{1}{x-y}=1} \atop {\frac{1}{x+y}=-1}} \right. \Leftrightarrow \left \{ {{x-y=1} \atop {-x-y=1}} \right.\\ [/tex]
[tex]Le\ adunam: -2y=2\Rightarrow \boxed{y=-1}\\ x-y=1\\ x+1=1\\ \boxed{x=0}\\ \\ b) \left \{ {{3[x]+2[y]=-1} \atop {[x]-3[y]=-4}|\cdot3} \right.\Leftrightarrow \left \{ {{3[x]+2[y]=-1} \atop {3[x]-9[y]=-12}} \right.\\ Le\ scadem:\ \ \ \ \ \ \ \ \ \ 11[y]=11\\ .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [y]=1\Rightarrow y\in[1,2)\\ 3[x]+2=-1\\ 3[x]=-3\\ {[x]=-1\Rightarrow x\in [-1, -2)} [/tex]