x+2y -z=3......(1)

2x-y+2z=8......(2)

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

First we will add(3) and (2):

==> 3x +3z = 11........(4)

Now multiply (2) by 2 and then add to (1):

==> 5x +3z = 19.....(5)

Now subtract (4) from (5):

==> 2x = 8

==> **x= 4**

Substitute in (5) :

5(4) + 3z = 19

==> 3z= -1

==> **z= -1/3**

Now substitute in (3):

4+ y -1/3 = 3

==> y= -1+1/3

==> **y= -2/3**

Being a system of linear equations, we'll calculate the determinant of the matrix of the system, which is formed from the coefficients of x, y and z.

We'll identify a = 1, b = 2, c = -1, d = 2, e = -1, f = 2, g = 1, h = 1, i = 1

det (A)=2(-1*1 - 1*2)-2(2*1 + 1)+1*(2*2-1)

det (A) = -6-6+6= -6

Since det A has a value different from zero, that means that the system is determined and it has a single solution.

The system will be solved using Cramer formula.

With this formula, we'll calculate each unknown in this way:

x = det x/ det A

y = det y/ detA

z = det z/ det A

To calculate det x, we'll substitute in the determinant A, the column of the coefficient of the unknown x. So, instead of a, d, g, we'll have 3,8,3.

det x = 3(-1*1 - 1*2) - 8(2*1 + 1)+ 3*(2*2-1)

det x = -9-24+9

det x = -24

x = -24/-6

**x = 4**

Now, we can substitute x in the first and the third equation:

2y - z = 3-4

y + z = 3-4

We'll add these equations and we'll get:

3y = -2

y = -2/3

-2/3 + z = -1

z = -1 + 2/3

z = -1/3

**The solution of the system is: (4 , -2/3 , -1/3).**

Given:

x + 2y - z = 3 ... (1)

2x - y + 2z = 8 ... (2)

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

Adding equations (1) and (3):

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

2x + 3y = 6 ... (4)

Multiplying equation (3) by 2:

2x + 2y + 2z = 6 ... (5)

Subtracting equation (2) from (5):

2x - 2x + 2y + y + 2z - 2z = 6 - 8

3y = -2

y = -2/3

Substituting this value of y in equation (4):

2x + 3*(-2/3) = 6

2x - 2 = 6

2x = 6 + 2 = 8

x = 8/2 = 4

Substituting these values of x and y in equation (3):

4 - 2/3 + z = 3

10/3 + z = 3

z = 3 - 10/3 = -1/3

Answer:

x = 4

y = -2/3

z = -1/3

To solve:

x+2y-z =3.........(1)

2x-y+2z=8........(2)

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

Elimination of z:

Eq(2)+Eq(1)*2 gives: 4x+3y = 14.........(4)

Eq(1)+Eq(3): gives: 2x+3y = 6..........(5)

Elimination of y:

Eq(4)-(5) gives: 2x=8, x = 8/2 = 4. From (5) , 3y = 6-2x =6-2(4) = -2, y=-2/3,

From (3), x+y+z =3, 4 -2/3+z = 3 , z = 3-4+2/3 = -1/3

(x,y,z) = (4,-2/3,-1/3)