Question

Prin substitutie si reducere la fiecare!!
Cine face,îi ofer 5 stele,coroana,inima si ma si abonez daca vreti!!!
Va rog mult!!!​

Answer 1

Răspuns:

1)

substitutie:

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

3x + 2y = - 10

3(-3y/2) + 2y = -10

-9y/2 + 2y = -10

-9y/2 + 4y/2 = -10

-5y/2 = -10/1

-5y = (2)(-10)

-5y = -20

y = 4

x = -3(4)/2

x = -12/2

x = -6

reducere:

2x + 3y = 0   (×3)

3x + 2y = - 10 (×2)

6x + 9y = 0

6x + 4y = -20

5y = 20

y = 4

2x + 3(4) = 0

2x + 12 = 0

2x = -12

x = -6

-----------------------------------------------------------------

2)

substitutie:

-2x - y = 2    => -2x - 2 = y

3x + 4y = 11

3x + 4(-2x - 2) = 11

3x - 8x - 8 = 11

- 5x - 8 = 11

-5x = 11 + 8

-5x = 19

x = -3,8

y = -2(-3,8) - 2

y = 7,6 - 2

y = 5,6

reducere:

-2x - y = 2  (×3)

3x + 4y = 11   (×2)

-6x - 3y = 6

6x + 8y = 22

5y = 28

y = 5,6

-2x - 5,6 = 2

-2x = 2 + 5,6

-2x = 7,6

x = -3,8

------------------------------------------------------------------

3)

substitutie:

2x - 3y = 1

x - 2y = -2      => x = -2 + 2y

2(-2 + 2y) - 3y = 1

-4 + 4y - 3y = 1

y = 1 + 4

y = 5

x = -2 + 2(5)

x = -2 + 10

x = 8

reducere:

2x - 3y = 1

x - 2y = -2     (×2)

2x - 3y = 1

2x - 4y = -4

y = 5

x - 2(5) = -2

x - 10 = -2

x = -2 + 10

x = 8