Question

Daca x+2y=17 si y-3z=4 atunci calculati:
x+3y-3z=?
2x+5y-3z=?
(3x+6y)+(2y-6z)=?
(2x+4y)×(x+4y-6z)=?

Answer 1

x+3y-3z se descompune in x+2y+y-3z => x+3y-3z=17+4=21
2x+5y-3z inmultim ecuatia x+2y=17 cu 2 => 2x+4y=34. pe acelasi principiu al descompunerii de mai sus => 2x+4y+y-3z=34+4=38
(3x+6y)+(2y-6z) inmultim ecuatia x+2y=17 cu 3 si ecuatia y-3z=4 cu 2 si atunci 3x+6y=51 si 2y-6z=8 => (3x+6y)+(2y-6z)=51+8=59