Exercise 953Algebra › Equations and radicalsHigh schoolAdded 3 Nov 2026

A 3×3 linear system

#linear system#3 equations#elimination

Statement

Solve the system
$\begin{cases}x+y+z=6\\ 2x-y+z=3\\ x+2y-z=2\end{cases}$

Hint

Eliminate $z$ using the first equation, then solve the resulting $2\times2$ system.

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Eliminate $z$ . $z=6-x-y$ . Substituting into the other two:
$2x-y+(6-x-y)=3\Rightarrow x-2y=-3,$
$x+2y-(6-x-y)=2\Rightarrow 2x+3y=8.$

Solve the $2\times2$ system.
$x-2y=-3,\quad 2x+3y=8.$
From the first, $x=2y-3$ . Substituting: $4y-6+3y=8$ , so $7y=14$ , $y=2$ , $x=1$ .

$z$ . $z=6-1-2=3$ .

Conclusion. $(x,y,z)=(1,2,3)$ .