Solve the following for x, y and z, 2x + 3y + z = 5, x + 3y + 5z = 8, 4x + y + z = 10

To solve the equations
:

2x+3y+z = 5...(1) x+3y+5z = 8....(2) 4x+y+z =
10.....(3)

(3)-(1):  (4x+y+z) - (2x+3y+z) =  10-5 =
5

2x-2y = 5...(4).

5(3)-(2) :
5(4x+y+z)  -(x+3y+5z)= 5*10 - 8 = 42

19x-2y =
42...(5).

(4)+(5): 21x = 42+5 =
47

x = 47/21. Put this value of x  in
eq(4):

 2(47/21) -2y = 5. So 2y = 94/21 -5 = -11/21. Or y =
-11/42.

Therefore substitute x = 47/21 , y = -11/42 in eq
(1), that is , in  2x+3y+z = 5:

2(47/21)+3(-11/42)+z = 5.
Or z = 5-94/21 +33/42 = (210-188+33)/42.

z =
55/42

Therefore x = 47/22 , y = -11/42 and z =
55/42.