1)Verificați dacă perechea (3;1)este soluție a sistemului de ecuații

a) x=5-2y
5-2y-y=-2 <=> -3y=-7 |•(-1)
<=> y=7/3
x=5-14/3=(15-14) : 3 = 1/3
=> {y=7/3, x=1/3}.
b) x=3y
2•3y-y=5 <=> 6y-y=5 => 5y=5 => y=1
x=3•1=3
=> { x=3, y=1}.
Sper că te-am ajutat!

[tex]a) \text{ Aplicam metoda reducerii} \\ \\ \left \{ {{x+2y=5 \text{ } \text{ } \text{ } \text{ } \text{ }} \atop {x-y=-2}/*2} \right. \\ \left \{ {{x+\not2y=5 \text{ } \text{ } \text{ }} \atop {2x-\not2y=-4}} \right. \\ \\ \left \{ {{x=5 \text{ } \text{ } \text{ } \text{ } \text{ }} \atop {2x=-4+}} \right. \\ \\ 3x=1 \to x= \frac{1}{3} \\ \\ \left \{ {{x= \frac{1}{3} \text{ } \text{ } \text{ } \text{ } \text{ }} \atop {x+2y=5}} \right. [/tex]
[tex] \left \{ {{x= \frac{1}{3} \text{ } \text{ } \text{ } \text{ } \text{ }} \atop { \frac{1}{3} +2y=5}} \right. \to \left \{ {{x= \frac{1}{3} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \atop {2y= \frac{15-1}{3} }} \right. \to \left \{ {{x= \frac{1}{3} } \atop { y=\frac{ \frac{14}{3} }{2} }} \right. \to \left \{ {{x= \frac{1}{3} } \atop {y= \frac{7}{3} }} \right. \to S= ( \frac{1}{3} , \frac{7}{3} )[/tex]  

[tex]b) \text{Aplicam metoda reducerii} \\ \\ 2x-y=5 \\ x-3y=0 /*(-2) \\ \\ \not2x-y=5 \\ - \not2x+6y=0 + \\ \\ 5y=5 /:5 \to y= 1 \\ \\ y=1 \\ 2x-1=5 \\ \\ y=1 \\ 2x=5+1 \\ \\ y=1 \\ 2x=6 /:2 \\ \\ y=1 \\ x=3 \\ \\ S=(3,1) [/tex]