A 3x3 system by elimination

Statement

Solve the linear system in three unknowns: $\begin{cases} x + 2y + z = 9\\ 2x - y + 3z = 4\\ 3x + y - 2z = 7 \end{cases}$

Solution

Isolate $x$ from the first: $x = 9 - 2y - z$.

Substitute into the second: 2(9 - 2y - z) - y + 3z = 4 \implies 18 - 4y - 2z - y + 3z = 4 \implies -5y + z = -14. \tag{A} Substitute into the third: $3(9 - 2y - z) + y - 2z = 7 \implies 27 - 6y - 3z + y - 2z = 7 \implies -5y - 5z = -20,$ i.e. $y + z = 4$. \tag{B}

From (A), $z = 5y - 14$; inserting it into (B): $y + 5y - 14 = 4 \implies 6y = 18 \implies y = 3$. Then $z = 5\cdot 3 - 14 = 1$ and $x = 9 - 6 - 1 = 2$. Check: $2+6+1=9$, $4-3+3=4$, $6+3-2=7$. Correct. $\boxed{x = 2, \quad y = 3, \quad z = 1}$