Question

What the solution to the following system?

Answer 1

The answer to this problem is

A.  x=0, y=1, z=2

hope this helps

Answer 2

Answer:

Choice A is the correct answer.

Step-by-step explanation:

We have given a system of equations.

5x+2y+z  = 4                         eq(1)

x+2z =   4                              eq(2)

2x+y-z  = -1                            eq(3)

We have to find the solution of the system.

From eq(2), we have

x  = 4-2z                                eq(4)

Putting above eq(4) in  eq(3), we have

2(4-2z)+y-z = -1

8-4z+y-z = -1

8-5z+y = -1

y-5z = -1-8

y-5z = -9                 eq(a)

Multiplying above equation by 2 , we have

2y-10z  =  -18                  eq(5)

Putting eq(4) into eq(1),we have

5(4-2z)+2y+z = 4

20-10z+2y+z = 4

2y-9z = -16              eq(6)

Subtracting eq(6) from eq(5), we have

2y-9z-(2y-10z) = -16-(-18)

2y-9z-2y+10z = -16+18

z  = 2

Putting the value of z in eq(4), we have

x  = 4-2(2)  = 4-4

x  = 0

Putting the value of z in eq(a), we have

y-5(2) = -9

y-10 = -9

y = -9+10

y = 1

Hence, the solution is x = 0,y = 1 and z = 2.