What is x+y+z=3,2x+4y+z=4,3x-y-z=1 ?

The solution to x+y+z=3,2x+4y+z=4,3x-y-z=1 is x=1,y=0,z=2

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x+y+z=3,2x+4y+z=4,3x-y-z=1

x + y + z = 3, 2 x + 4 y + z = 4, 3 x - y - z = 1

x = 1, y = 0, z = 2

Solution steps

x + y + z = 3 2 x + 4 y + z = 4 3 x - y - z = 1

Isolate

x

for

x + y + z = 3 : x = 3 - y - z

Substitute

x = 3 - y - z

[2 (3 - y - z) + 4 y + z = 4 3 (3 - y - z) - y - z = 1]

Simplify

[2 y - z + 6 = 4 - 4 y - 4 z + 9 = 1]

Isolate

y

for

2 y - z + 6 = 4 : y = \frac{z - 2}{2}

Substitute

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

[- 4 \cdot \frac{z - 2}{2} - 4 z + 9 = 1]

Simplify

[- 6 z + 13 = 1]

Isolate

z

for

- 6 z + 13 = 1 : z = 2

For

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

Substitute

z = 2

y = \frac{2 - 2}{2}

Simplify

y = 0

For

x = 3 - y - z

Substitute

y = 0, z = 2

x = 3 - 0 - 2

Simplify

x = 1

The solutions to the system of equations are:

x = 1, y = 0, z = 2

2 x + 6 y = 3, 2 - 2 x + 4 y = 18

a + b = 2, a + c = 3, b + c = 4

5 x + 2 y = 10, - 3 x + 3 y = 5

x + 4 y = - 4, 3 y - 5 x = - 1

7 x - 6 y = 12, 5 x + 3 y = 11

What is x+y+z=3,2x+4y+z=4,3x-y-z=1 ?
The solution to x+y+z=3,2x+4y+z=4,3x-y-z=1 is x=1,y=0,z=2