x+2y+z= 4.....(1)

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

3x+y + 3z=17....(3)

We are going to use the elimination method t solve the system:

Let us first add (2) and (3)

==> 8x + 2z = 18 .....(4)

Now let us multiply (2) by 2 and add to (1)

==> 10x-2y-2z=2

x+2y+z=4

==> 11x -z = 6 . ==> z= 11x-6

Now substitute z in (4)

8x+2z =18

8x+ 2(11x-6) = 18

8x +22x -12= 18

30 x = 30

=> x= 1

==> z= 11x-6 = 11(1)-6= 5

==> y = 5x-z-1= 5(1) - 5 -1 = -1

x + 2y + z = 4 ... (1)

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

3x + y + 3x = 17 ... (3)

Adding equation (1) and (2) we get:

x + 5x + 2y - y + z - z = 4 + 1

6x + y = 5

Multiplying the above equation by 3 we get:

18x + 3y = 15 ... (4)

Multiplying equation (2) by 3 we get

15x - 3y - 3z = 3 ... (5)

Adding equations (3) and (5) we get:

3x + 15x + y - 3y + 3z - 3z = 17 + 3

18x - 2y = 20 ... (6)

Subtracting equations (6) from (4) we get:

18x - 18x + 2y + 3y = 15 - 20

5y = - 5

Therefore:

y = -5/5 = -1

Substituting this value of y in equation 6 we get:

18x - 2(-1) = 20

18x + 2 = 18

18x = 18

Therefore:

x = 18/18 = 1

Substituting these values of x and y in equation (1) we get:

1 + 2(-1) + z = 4

1 - 2 + z = 4

-1 + z = 4

z = 4 + 1 = 5

Answer:

x = 1, y = -1, z = 5