We are given the system of linear equations:

3x + y + 2z = 3 \tag{1} 2x - 3y - z = -3 \tag{2} x + 2y + z = 4 \tag{3}

Step 1: Solve for one variable using one of the equations

From equation (3), we can solve for $x$:

x = 4 - 2y - z \tag{4}

Step 2: Substitute $x = 4 - 2y - z$ into equations (1) and (2)

Substitute equation (4) into equations (1) and (2).

Substituting into equation (1):

$3(4 - 2y - z) + y + 2z = 3$ $12 - 6y - 3z + y + 2z = 3$ Simplify: $12 - 5y - z = 3$ $-5y - z = 3 - 12$ -5y - z = -9 \tag{5}

Substituting into equation (2):

$2(4 - 2y - z) - 3y - z = -3$ $8 - 4y - 2z - 3y - z = -3$ Simplify: $8 - 7y - 3z = -3$ $-7y - 3z = -3 - 8$ -7y - 3z = -11 \tag{6}

Step 3: Solve the system of two equations (5) and (6)

We now have the system: -5y - z = -9 \tag{5} -7y - 3z = -11 \tag{6}

We can use substitution or elimination to solve this system. Let's use the elimination method.

Eliminate $z$:

Multiply equation (5) by 3 to align the $z$-terms:

-15y - 3z = -27 \tag{7}

Now subtract equation (6) from equation (7):

$(-15y - 3z) - (-7y - 3z) = -27 - (-11)$ $-15y + 7y = -27 + 11$ $-8y = -16$ $y = 2$

Step 4: Substitute $y = 2$ into equation (5)

Substitute $y = 2$ into equation (5):

$-5(2) - z = -9$ $-10 - z = -9$ $-z = -9 + 10$ $-z = 1$ $z = -1$

Step 5: Substitute $y = 2$ and $z = -1$ into equation (4)

Substitute $y = 2$ and $z = -1$ into equation (4):

$x = 4 - 2(2) - (-1)$ $x = 4 - 4 + 1$ $x = 1$

Final Solution:

The solution to the system of equations is:

$x = 1, \quad y = 2, \quad z = -1$

Verification:

Substitute $x = 1$, $y = 2$, and $z = -1$ into the original equations:

1. $3(1) + 2 + 2(-1) = 3 + 2 - 2 = 3$ (True)
2. $2(1) - 3(2) - (-1) = 2 - 6 + 1 = -3$ (True)
3. $1 + 2(2) + (-1) = 1 + 4 - 1 = 4$ (True)

Thus, the solution is correct.

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Tip: When solving systems of equations, always check your solution by substituting it back into the original equations to confirm that all equations hold true.