# mathSolve for x, y and z, given the following equations: x + y + z = 6 2x - y + 3z = 9 -x + 2y + 2z = 9

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We have to solve

x + y + z = 6 ...(1)

2x - y + 3z = 9 ...(2)

-x + 2y + 2z = 9 ...(3)

(2) + (3)

=> x + y + 5z = 18

but from (1)

x + y = 6 - z

=> 6 - z + 5z = 18

=> 4z = 12

=> z = 3

substitute in (1)

=> x + y = 3

=> x = 3 - y

substitute z and x in (2)

=> 2(3 - y) - y + 9 = 9

=> 6 - 2y - y = 0

=> -3y = -6

=> y = 2

x = 3 - 2 = 1

**The values of x, y and z are (1 , 2, 3)**

x+y+z=6 (1)

2x-y+3z=9 (2)

-x+2y+2z=9 (3)

We'll start by adding 1 and 3 together to eliminate x.

x + y + z - x + 2y + 2z = 6 + 9

We'll combine like terms:

3y + 3z = 15 (4)

We'll multiply the equation (1) by –2 and we'll add (1) and (2) together to eliminate the x.

-2x - 2y - 2z + 2x - y + 3z = -12 + 9

We'll combine like terms:

-3y + z = -3 (5)

We'll add (4) and (5):

3y + 3z -3y + z = 15 - 3

4z = 12

z = 3

y + z = 5

y = 5 - z

y = 5 - 3

y = 2

x + 5 = 6

x = 6 - 5

x = 1

**The solution of the system is {1 ; 2 ; 3}.**