Question

Rezolvati prin metoda substitutiei urmatoarele sisteme:
1.y=2
2x+3y-9=0
2. x=y+2
2x+3y-4=0
3. 2x=3y
x+6y=5
4. 0.5x-5y=-4
2x-y=3
5. 5u+3v-5=0
u+3/5v-1=0
6. 4x-3y=0
x+2y=0

Answer 1

1. 2x+ 3*2-9= 0 => 2x+6-9=0 => 2x=3 => x=3/2. 

2. x=y+2
2x+3y-4=0 => 2(y+2) +3y-4=0 => 2y+4+3y-4=0 => 5y=0 => y=0 . deci x=2

3.2x=3y
x+6y=5 => x+2*3y = 5 => x+2*2x = 5 => x+4x=5 => 5x=5 => x=1. deci 2*1=3y => y=2/3

4.0.5x-5y = -4
2x-y=3 => y=2x-3 

0.5x-5y=-4 => 0.5x - 5(2x-3) = -4 => 0.5x -10x +15 = -4 => 0.5x - 10x = 11 => 9.5x= 11 => x= 1.15

5.5u+3v-5=0
u+3/5v-1=0 => u= 1-3/5v

5(1-3/5v) + 3v-5 = 0 = > 5- 3v+3v-5=0, adevarat pentru orice v

6. 4x-3y=0
x+2y=0 => x=-2y

4(-2y) -3y=0 => -8y-3y=0=> -11y=0 => y=0, deci x =0.