Question

Fie x,y,z numere reale astfel incat x+3y=11 si 2y+z=5 determinati:
a. x+y-z
b. x+5y+z
c. x-y-2z

Answer 1

 

$\displaystyle\\a)\\\begin{cases}x+3y=11\\2y+z=5~~~\Big|~\times(-1)\end{cases}\\\begin{cases}x+3y=11\\-2y-z=-5\end{cases}\\............................. \text{Adunam ecuatiile.}\\x+3y-2y-z=11-5\\\boxed{x+y-z=6}\\\\b)\\\begin{cases}x+3y=11\\2y+z=5\end{cases}\\............................. \text{Adunam ecuatiile.}\\x+3y+2y+z=11+5\\\boxed{x+5y+z=16}$

$\displaystyle\\c)\\\begin{cases}x+3y=11\\2y+z=5~~~\Big|~\times(-2)\end{cases}\\\begin{cases}x+3y=11\\-4y-2z=-10\end{cases}\\............................. \text{Adunam ecuatiile.}\\x+3y-4y-2z=11-10\\\boxed{x-y-2y=1}$