What is x+2y+3z=2,-x+4y+z=1,4x+y+z=5 ?

The solution to x+2y+3z=2,-x+4y+z=1,4x+y+z=5 is x= 43/38 ,y= 21/38 ,z=-3/38

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x+2y+3z=2,-x+4y+z=1,4x+y+z=5

x + 2 y + 3 z = 2, - x + 4 y + z = 1, 4 x + y + z = 5

x = \frac{43}{38}, y = \frac{21}{38}, z = - \frac{3}{38}

Solution steps

x + 2 y + 3 z = 2 - x + 4 y + z = 1 4 x + y + z = 5

Isolate

x

for

x + 2 y + 3 z = 2 : x = 2 - 2 y - 3 z

Substitute

x = 2 - 2 y - 3 z

[- (2 - 2 y - 3 z) + 4 y + z = 1 4 (2 - 2 y - 3 z) + y + z = 5]

Simplify

[6 y + 4 z - 2 = 1 - 7 y - 11 z + 8 = 5]

Isolate

y

for

6 y + 4 z - 2 = 1 : y = \frac{- 4 z + 3}{6}

Substitute

y = \frac{- 4 z + 3}{6}

[- 7 \cdot \frac{- 4 z + 3}{6} - 11 z + 8 = 5]

Simplify

[\frac{- 38 z - 21}{6} + 8 = 5]

Isolate

z

for

\frac{- 38 z - 21}{6} + 8 = 5 : z = - \frac{3}{38}

For

y = \frac{- 4 z + 3}{6}

Substitute

z = - \frac{3}{38}

y = \frac{- 4 ( - \frac{3}{38} ) + 3}{6}

\frac{- 4 ( - \frac{3}{38} ) + 3}{6} = \frac{21}{38}

y = \frac{21}{38}

For

x = 2 - 2 y - 3 z

Substitute

y = \frac{21}{38}, z = - \frac{3}{38}

x = 2 - 2 \cdot \frac{21}{38} - 3 (- \frac{3}{38})

2 - 2 \cdot \frac{21}{38} - 3 (- \frac{3}{38}) = \frac{43}{38}

x = \frac{43}{38}

The solutions to the system of equations are:

x = \frac{43}{38}, y = \frac{21}{38}, z = - \frac{3}{38}

5 x - 2 y = 4, x + 8 y = 26

7 x + 6 y = 38, 3 x + 5 y = 1750

x + y = 4, 8 x - 3 y = 4

2 x + 3 y = 0, 3 x + 5 y = 2

2 x - 5 y + 4 = 2, x + y - 8 = 0

What is x+2y+3z=2,-x+4y+z=1,4x+y+z=5 ?
The solution to x+2y+3z=2,-x+4y+z=1,4x+y+z=5 is x= 43/38 ,y= 21/38 ,z=-3/38