Question

Rezolvati prin metoda substitutiei următoarele sisteme de ecuații:​

Answer 1

Salutare!

1) 3x-y= -4  => 3x=y-4 => x=(y-4)/3

x-3y= - 4

(y-4)/3-y= -4

y-4 - 9y=-12

-8y=-8

y=1 => x=-1

2)10x + 27y = 10 => 10x =10-27y => x=(10-27y)/10

-25x + 12y = -25

-25*(10-17)/10 +12y = -25

-50+85y +24y = -50 => y=0 => x=1

3)4x+9y=21 =>4x = 21-9y => x=(21-9y)/4

12x+15y=51

12*(21-9y)/4 +15y = 51

63-27y+15y =51

-12y= -12 => y=-1 => x=15/2

4)1-4x=y => -4x=y-1 => x = (-y-1)/4

2-y-x-y = 0

2- y - (-y-1)/4 - y =0

8-4y + y -1 -4y =0

-7y = -7 => y=1 => x=0

5)8x- y = -15 => 8x= -15 + y => x = (-15+y)/8

-x + 8y = -6

(15+y)/8 + 8y = -6

15 -y +64y = -48

63y = -63

y=-1 => x=-2

6)2x-3y=4 => 2x=3y + 4 => x=(3y+4)/2

3y +4 -3y = 4 => y=0 => x=2

Mult succes!