What is 2x+3y+z=0,x+y+2z=1,x-y-z=-1 ?

The solution to 2x+3y+z=0,x+y+2z=1,x-y-z=-1 is x=-1/3 ,y=0,z= 2/3

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2x+3y+z=0,x+y+2z=1,x-y-z=-1

2 x + 3 y + z = 0, x + y + 2 z = 1, x - y - z = - 1

x = - \frac{1}{3}, y = 0, z = \frac{2}{3}

Solution steps

2 x + 3 y + z = 0 x + y + 2 z = 1 x - y - z = - 1

Isolate

x

for

2 x + 3 y + z = 0 : x = \frac{- 3 y - z}{2}

Substitute

x = \frac{- 3 y - z}{2}

[\frac{- 3 y - z}{2} + y + 2 z = 1 \frac{- 3 y - z}{2} - y - z = - 1]

Simplify

[\frac{- y + 3 z}{2} = 1 \frac{- 5 y - 3 z}{2} = - 1]

Isolate

y

for

\frac{- y + 3 z}{2} = 1 : y = - 2 + 3 z

Substitute

y = - 2 + 3 z

[\frac{- 5 ( - 2 + 3 z ) - 3 z}{2} = - 1]

Simplify

[5 - 9 z = - 1]

Isolate

z

for

5 - 9 z = - 1 : z = \frac{2}{3}

For

y = - 2 + 3 z

Substitute

z = \frac{2}{3}

y = - 2 + 3 \cdot \frac{2}{3}

- 2 + 3 \cdot \frac{2}{3} = 0

y = 0

For

x = \frac{- 3 y - z}{2}

Substitute

y = 0, z = \frac{2}{3}

x = \frac{- 3 \cdot 0 - \frac{2}{3}}{2}

\frac{- 3 \cdot 0 - \frac{2}{3}}{2} = - \frac{1}{3}

x = - \frac{1}{3}

The solutions to the system of equations are:

x = - \frac{1}{3}, y = 0, z = \frac{2}{3}

x + 3 y = 1, - x + y = 9

1 x - 6 y = 12, y + 3 x = 12, y = 12 - 2 x

5 x + 3 y = 65, 6 x - 10 y = 10

- x + 3 y = - 5, 2 x - 5 y = 9

\frac{0.6}{x} + \frac{2}{y} = 5, \frac{8}{x} - \frac{1}{y} = 3

What is 2x+3y+z=0,x+y+2z=1,x-y-z=-1 ?
The solution to 2x+3y+z=0,x+y+2z=1,x-y-z=-1 is x=-1/3 ,y=0,z= 2/3