What is 2x+y-3z=4,x+2y-6z=0,x-y+z=0 ?

The solution to 2x+y-3z=4,x+2y-6z=0,x-y+z=0 is x= 8/3 ,y= 14/3 ,z=2

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2x+y-3z=4,x+2y-6z=0,x-y+z=0

2 x + y - 3 z = 4, x + 2 y - 6 z = 0, x - y + z = 0

x = \frac{8}{3}, y = \frac{14}{3}, z = 2

Solution steps

2 x + y - 3 z = 4 x + 2 y - 6 z = 0 x - y + z = 0

Isolate

x

for

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

Substitute

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

[\frac{4 - y + 3 z}{2} + 2 y - 6 z = 0 \frac{4 - y + 3 z}{2} - y + z = 0]

Simplify

[\frac{3 y - 9 z + 4}{2} = 0 \frac{- 3 y + 5 z + 4}{2} = 0]

Isolate

y

for

\frac{3 y - 9 z + 4}{2} = 0 : y = \frac{9 z - 4}{3}

Substitute

y = \frac{9 z - 4}{3}

[\frac{- 3 \cdot \frac{9 z - 4}{3} + 5 z + 4}{2} = 0]

Simplify

[2 (- z + 2) = 0]

Isolate

z

for

2 (- z + 2) = 0 : z = 2

For

y = \frac{9 z - 4}{3}

Substitute

z = 2

y = \frac{9 \cdot 2 - 4}{3}

Simplify

y = \frac{14}{3}

For

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

Substitute

y = \frac{14}{3}, z = 2

x = \frac{4 - \frac{14}{3} + 3 \cdot 2}{2}

\frac{4 - \frac{14}{3} + 3 \cdot 2}{2} = \frac{8}{3}

x = \frac{8}{3}

The solutions to the system of equations are:

x = \frac{8}{3}, y = \frac{14}{3}, z = 2

5 x - 3 y = - 5, 2 x + 4 y = 24

7 x + 8 y = \frac{1}{2}, 4 x + 12 y = \frac{29}{56}

2 x - y = 20, 2 x + y = 48

49 x - 57 y = 17, 257 x - 49 y = 252

- 7 x + y = - 19, - 2 x + 3 y = - 19

What is 2x+y-3z=4,x+2y-6z=0,x-y+z=0 ?
The solution to 2x+y-3z=4,x+2y-6z=0,x-y+z=0 is x= 8/3 ,y= 14/3 ,z=2