We are given a system of three linear equations with three variables, $x$, $y$, and $z$. The system is: $x + 2y + z = 4$ (1) $-x - y + 2z = 2$ (2) $2x + y + 3z = 6$ (3) We need to find the values of $x$, $y$, and $z$ that satisfy all three equations.

1. Problem Description

We are given a system of three linear equations with three variables, $x$, $y$, and $z$. The system is:

$x + 2y + z = 4$ (1)

$-x - y + 2z = 2$ (2)

$2x + y + 3z = 6$ (3)

We need to find the values of $x$, $y$, and $z$ that satisfy all three equations.

2. Solution Steps

First, add equations (1) and (2) to eliminate $x$:

$(x + 2y + z) + (-x - y + 2z) = 4 + 2$

$y + 3z = 6$ (4)

Next, multiply equation (1) by -2 and add it to equation (3) to eliminate $x$:

$-2(x + 2y + z) + (2x + y + 3z) = -2(4) + 6$

$-2x - 4y - 2z + 2x + y + 3z = -8 + 6$

$-3y + z = -2$ (5)

Now, we have a system of two equations with two variables, $y$ and $z$:

$y + 3z = 6$ (4)

$-3y + z = -2$ (5)

Multiply equation (4) by 3:

$3(y + 3z) = 3(6)$

$3y + 9z = 18$ (6)

Add equations (5) and (6) to eliminate $y$:

$(-3y + z) + (3y + 9z) = -2 + 18$

$10z = 16$

$z = \frac{16}{10} = \frac{8}{5}$

Substitute the value of $z$ into equation (4) to find $y$:

$y + 3(\frac{8}{5}) = 6$

$y + \frac{24}{5} = 6$

$y = 6 - \frac{24}{5} = \frac{30}{5} - \frac{24}{5} = \frac{6}{5}$

Substitute the values of $y$ and $z$ into equation (1) to find $x$:

$x + 2(\frac{6}{5}) + \frac{8}{5} = 4$

$x + \frac{12}{5} + \frac{8}{5} = 4$

$x + \frac{20}{5} = 4$

$x + 4 = 4$

$x = 0$

3. Final Answer

$x = 0$

$y = \frac{6}{5}$

$z = \frac{8}{5}$