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

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

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

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

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

Solution steps

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

Isolate

x

for

x + y + z = 1 : x = 1 - y - z

Substitute

x = 1 - y - z

[2 (1 - y - z) - 2 y + z = 3 - 3 (1 - y - z) + y + 2 z = 0]

Simplify

[- 4 y - z + 2 = 3 4 y + 5 z - 3 = 0]

Isolate

y

for

- 4 y - z + 2 = 3 : y = - \frac{z + 1}{4}

Substitute

y = - \frac{z + 1}{4}

[4 (- \frac{z + 1}{4}) + 5 z - 3 = 0]

Simplify

[4 z - 4 = 0]

Isolate

z

for

4 z - 4 = 0 : z = 1

For

y = - \frac{z + 1}{4}

Substitute

z = 1

y = - \frac{1 + 1}{4}

- \frac{1 + 1}{4} = - \frac{1}{2}

y = - \frac{1}{2}

For

x = 1 - y - z

Substitute

y = - \frac{1}{2}, z = 1

x = 1 - (- \frac{1}{2}) - 1

Simplify

x = \frac{1}{2}

The solutions to the system of equations are:

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

\frac{1}{2} \cdot x + \frac{1}{3} \cdot y = 3, x + \frac{1}{2} \cdot y = 5

- \frac{1}{8} \cdot x + y = \frac{33}{4}, - \frac{1}{9} \cdot x - \frac{1}{10} \cdot y = - \frac{47}{30}

9 x + 3 y = 33, 3 x + 9 y = 51

16 x + 11 y = 15, 411 x + 16 y = 89

3 x + y = 9, x - 2 y = - 4

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