Question

como resolver este sistemas de ecuaciones

3x+2z+5=0

2y-3z-12=0

x+y-1=0

Answer 1

Respuesta:

3x+2z+5=0          

2y-3z-12=0

x+y-1=0

y = 1 - x

2(1 - x) -3z -12 = 0

2 - 2x - 3z - 12 = 0

-2x -3z = 12 - 2

-2x - 3z = 10

-3z = 10 + 2x

z = (10 + 2x)/ -3

3x + 2 ((10 + 2x) / -3 ) + 5 = 0

3x - 20/3 - 4x/3 = -5

5x/3 = -5 + 20/3

5x/3 = 5/3

5x = 3 (5/3)

5x = 15/3

5x = 5

x = 5/5

x= 1

z = (10 + 2(1))/-3

z= 10 + 2 / -3

z= 12/-3

z= -4

y = 1 - 1

y= 0