\left\{ \begin{array} { c } { x + 2 y + 3 z = 1 } \\ { 2 x + 3 y + z = 2 } \\ { 3 x + y + 2 z = 3 } \end{array} \right.
Solve for x, y, z
x=1
y=0
z=0
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x=-2y-3z+1
Solve x+2y+3z=1 for x.
2\left(-2y-3z+1\right)+3y+z=2 3\left(-2y-3z+1\right)+y+2z=3
Substitute -2y-3z+1 for x in the second and third equation.
y=-5z z=-\frac{5}{7}y
Solve these equations for y and z respectively.
z=-\frac{5}{7}\left(-5\right)z
Substitute -5z for y in the equation z=-\frac{5}{7}y.
z=0
Solve z=-\frac{5}{7}\left(-5\right)z for z.
y=-5\times 0
Substitute 0 for z in the equation y=-5z.
y=0
Calculate y from y=-5\times 0.
x=-2\times 0-3\times 0+1
Substitute 0 for y and 0 for z in the equation x=-2y-3z+1.
x=1
Calculate x from x=-2\times 0-3\times 0+1.
x=1 y=0 z=0
The system is now solved.
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