FlamingoOhOh FlamingoOhOh
  • 25-08-2021
  • Mathematics
contestada

Solve the system using the elimination method.

2x + 2y + 5z =−1

2x − y + z =2

2x + 4y − 3z =14

What x,y, and z?​

Respuesta :

sadovnichenkoant sadovnichenkoant
  • 25-08-2021

Answer:

x=151/64   y=9/8 z=-51/32

Step-by-step explanation:

2x+2y+5z- (2x-y+z)=-1-2

3y+4z=-3

2x+2y+5z- (2x+4y-3z)=-1-14

-2y+8z=-15

We have two equations 3y+4z=-3 and -2y+8z=-15 (We get it due to elimination method)

3y+4z=-3 (2)

-2y+8z=-15 (-3)

6y+8z=-6

6y-24z=45

(6y+8z)- (6y-24z)= -6-45

32z=-51

z=-51/32

3y-204/32=-3

3y= -96/32+204/32

3y=108/32

y=36/32=9/8

2x+2*9/8+5*(-51/32)=-1

2x+9/4-255/32=-1

2x+72/32-255/32=-1

2x-183/32=-32/32

2x=151/32

x=151/64

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