Question

Rezolvati prin metoda substitutiei sistemul :

6x-2y-2=0
-2x+9y-16=0

Answer 1

Metoda I:

$\left \{ {{6x-2y-2=0} \atop {-2x+9y-16=0}} \right.\\\left \{ {{6x-2y=2\\\\} \atop {-2x+9y=16 (*3)}} \right. \\\left \{ {{6x-2y=2} \atop {-6x+27=48}} \right. \\---------\\0+25y=50\\y=2\\6x-2y=2\\6x-2*2=2\\6x-4=2\\6x=6\\x=1$

Metoda II:

$\left \{ {{6x-2y-2=0} \atop {-2x+9y-16=0}} \right. \\\left \{ {{6x-2y=2} ->2(3x-y)=2=>3x-y=1 => 3x=1+y => x=\frac{1+y}{3} \atop{\\-2x+9y=16\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\}} \right. \\\\\\\\-2*(1+y):3 +9y=16\\(-2-2y):3+9y=16\\-2-2y+27y=48\\25y=50\\y=2\\-2x+9*2=16\\-2x+18=16\\-2x=-2\\x=1$