\left\{ \begin{array} { l } { x + 2 y + 3 z = 0 } \\ { 2 x + 3 y + 5 z = 1 } \\ { 2 y + z = 2 } \end{array} \right.
Solve for x, y, z
x=6
y=3
z=-4
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x=-2y-3z
Solve x+2y+3z=0 for x.
2\left(-2y-3z\right)+3y+5z=1
Substitute -2y-3z for x in the equation 2x+3y+5z=1.
y=-z-1 z=2-2y
Solve the second equation for y and the third equation for z.
z=2-2\left(-z-1\right)
Substitute -z-1 for y in the equation z=2-2y.
z=-4
Solve z=2-2\left(-z-1\right) for z.
y=-\left(-4\right)-1
Substitute -4 for z in the equation y=-z-1.
y=3
Calculate y from y=-\left(-4\right)-1.
x=-2\times 3-3\left(-4\right)
Substitute 3 for y and -4 for z in the equation x=-2y-3z.
x=6
Calculate x from x=-2\times 3-3\left(-4\right).
x=6 y=3 z=-4
The system is now solved.
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