Question

Rezolvati prin metoda reducerii urmatoarele sisteme:{2x+3(x+1)=8 5(x+y)-y=2 {5x+3(y-1)=7 7(x+1)+2y=10

Answer 1

{2x+3x+3=8    =>5x+3=8 =>5x= 8-3 =>x=5/5 =>x=1
5x+5y-y=2       =>5x+4y=2=> 5*1+4y=2 => 4y=2-5 =>y= -3/4

{5x+3y-3=7 =>5x+3y=7+3 => 5x+3y=10 |(-2)
7x+7+2y=10 =>7x+2y= 10-7 =>7x+2y= 3 | 3
=>
-10x-6y=-20 =>{ 11x=-11=> x= -1
21x+6y=9
5x+3y=10 => 5*(-1) +3y =10 => -5+ 3y= 10
3y= 10+5
y= 15/3
y= 5