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

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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.

We have 3 equations to solve for x, y and z

2x + 3y + z = 5…(1)

x + 3y + 5z = 8…(2)

4x + y + z = 10…(3)

(1) – (2)

=> 2x + 3y + z - x - 3y - 5z = 5 - 8

=> x - 4z = -3

=> x - 4z = - 3…(4)

1- 3*(3)

=> 2x + 3y + z – 3*(4x + y + z) = 5 - 30

=> 2x + 3y + z – 12x -3y - 3z = -25

=> -10x -2z = -25

=> 10x + 2z = 25…(5)

(4) + 2*(5)

=> x - 4z + 20x + 4z = -3 + 50

=> 21x = 47

=> x = 47/21

Use this in (4)

47/21 - 4z = - 3

=> z = [3 + 47/21]/4

=> z = 55/42

Substitute x = 47/21 and z = 55/42 in (2)

=> 47/21 + 3y + 5*(55/42) = 8

=> y = [8 - 47/21 - 5*(55/42)]/3

=> y = -11/42

**Therefore x= 47/21, y = -11/42 and z = 55/42**