1. **Stating the problem:** Solve the system of linear equations: $$\begin{cases} x + 2y + z = 4 \\ 2x - y + 3z = 1 \\ -x + 3y + 2z = 7 \end{cases}$$ 2. **Formula and method:** We will use the method of substitution or elimination to solve for $x$, $y$, and $z$. The goal is to reduce the system step-by-step to find each variable. 3. **Step 1: Add the first and third equations to eliminate $x$:** $$ (x + 2y + z) + (-x + 3y + 2z) = 4 + 7 $$ $$ \cancel{x} + 2y + z - \cancel{x} + 3y + 2z = 11 $$ $$ 5y + 3z = 11 $$ 4. **Step 2: Express $x$ from the first equation:** $$ x = 4 - 2y - z $$ 5. **Step 3: Substitute $x$ into the second equation:** $$ 2(4 - 2y - z) - y + 3z = 1 $$ $$ 8 - 4y - 2z - y + 3z = 1 $$ $$ 8 - 5y + z = 1 $$ $$ -5y + z = 1 - 8 $$ $$ -5y + z = -7 $$ 6. **Step 4: Now we have two equations with $y$ and $z$:** $$ \begin{cases} 5y + 3z = 11 \\ -5y + z = -7 \end{cases} $$ 7. **Step 5: Add these two equations to eliminate $y$:** $$ (5y + 3z) + (-5y + z) = 11 + (-7) $$ $$ \cancel{5y} + 3z - \cancel{5y} + z = 4 $$ $$ 4z = 4 $$ $$ z = \frac{4}{4} = 1 $$ 8. **Step 6: Substitute $z=1$ into $-5y + z = -7$:** $$ -5y + 1 = -7 $$ $$ -5y = -8 $$ $$ y = \frac{-8}{-5} = \frac{8}{5} $$ 9. **Step 7: Substitute $y=\frac{8}{5}$ and $z=1$ into $x = 4 - 2y - z$:** $$ x = 4 - 2 \times \frac{8}{5} - 1 $$ $$ x = 4 - \frac{16}{5} - 1 $$ $$ x = \frac{20}{5} - \frac{16}{5} - \frac{5}{5} $$ $$ x = \frac{20 - 16 - 5}{5} = \frac{-1}{5} $$ 10. **Final solution:** $$ x = -\frac{1}{5}, \quad y = \frac{8}{5}, \quad z = 1 $$ This completes the solution of the first system of equations.