Math question! 10 POINTS best answer!!!

Answer you are looking for is D.

You can solve it through trial and error by plugging the values in for x y and z and see if it works out. Or you can place the leading coefficiants into a matrix and put it in reduced row echelon form.

x+4y-5z=-8 ....given (1)
=> x = -8 - 4y + 5z ...........(1a)
x+4y-4z=-2 ....given (2)
=> x = -2 - 4y + 4z ...........(2a)
x+5y-4z=3 ......given (3)
=> x = 3 - 5y + 4z ........(3a)

From (1a) and (2a), we have
-8 - 4y + 5z = -2 - 4y + 4z
=> 5z - 4z - 4y + 4y = -2 + 8
=> z = 6 .............(4)

From (2a) and (3a), we have
-2 - 4y + 4z = 3 - 5y + 4z
=> - 4y + 5y + 4z - 4z = 3 + 2
=> y = 5 ...........(5)

Substituting the values of y and z from (4) and (5) into equation (1a), we have

x = -8 - 4y + 5z
=> x = - 8 - 4(5) + 5(6)
=> x = -8 - 20 + 30
=> x = 2

Thus x = 2, y = 5, z = 6

Answer (D)

yup they are ALL right now give me da points man