Solve 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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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}.**

We have to solve for x, y and z using

x + y + z = 6 … (1)

2x –y + 3z = 9 … (2)

-x + 2y + 2z = 9 … (3)

Add (1) and (3)

=> 3y + 3z = 15

=> y + z = 5

Substitute this in (1)

=> x + 5 = 6

=> x = 1

substitute this in (2)

=> 2 – y + 3z = 9

=> 2 – (5 – z) + 3z = 9

=> 2 – 5 + z + 3z = 9

=> -3 + 4z = 9

=> 4z = 12

=> z = 3

y + z = 5

=> y = 5 – 3

=> y = 2

**Therefore x = 1, y = 2 and z = 3.**

To Solve for x,y,z

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

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

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

(2)+(1): 3x+4z = 15.....(4)

2(2)+(3): 4x-x 6z+2z = 18+9 = 27.

3x+8z = 27.....(5).

(5)-(4): 4z = 27-15 = 12.

z = 12/4 = 3. Put z= 3 in (4): 3x+4*3 = 15.

=> 3x= 15-12 -= 3. So x= 3/3 = 1.

Put x= 1 and z = 3 in (1): 1+y+3 =6. So y = 6-4 = 2.

So x= 1.y=2 and z = 3.

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