What is 10x+y+z=1,2x+10y+z=1,2x+y+10z=12 ?

The solution to 10x+y+z=1,2x+10y+z=1,2x+y+10z=12 is x=-1/53 ,y=-8/477 ,z= 575/477

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10x+y+z=1,2x+10y+z=1,2x+y+10z=12

10 x + y + z = 1, 2 x + 10 y + z = 1, 2 x + y + 10 z = 12

x = - \frac{1}{53}, y = - \frac{8}{477}, z = \frac{575}{477}

Solution steps

10 x + y + z = 1 2 x + 10 y + z = 1 2 x + y + 10 z = 12

Isolate

x

for

10 x + y + z = 1 : x = \frac{1 - y - z}{10}

Substitute

x = \frac{1 - y - z}{10}

[2 \cdot \frac{1 - y - z}{10} + 10 y + z = 1 2 \cdot \frac{1 - y - z}{10} + y + 10 z = 12]

Simplify

[\frac{49 y + 4 z + 1}{5} = 1 \frac{4 y + 49 z + 1}{5} = 12]

Isolate

y

for

\frac{49 y + 4 z + 1}{5} = 1 : y = \frac{- 4 z + 4}{49}

Substitute

y = \frac{- 4 z + 4}{49}

[\frac{4 \cdot \frac{- 4 z + 4}{49} + 49 z + 1}{5} = 12]

Simplify

[\frac{477 z + 13}{49} = 12]

Isolate

z

for

\frac{477 z + 13}{49} = 12 : z = \frac{575}{477}

For

y = \frac{- 4 z + 4}{49}

Substitute

z = \frac{575}{477}

y = \frac{- 4 \cdot \frac{575}{477} + 4}{49}

\frac{- 4 \cdot \frac{575}{477} + 4}{49} = - \frac{8}{477}

y = - \frac{8}{477}

For

x = \frac{1 - y - z}{10}

Substitute

y = - \frac{8}{477}, z = \frac{575}{477}

x = \frac{1 - ( - \frac{8}{477} ) - \frac{575}{477}}{10}

\frac{1 - ( - \frac{8}{477} ) - \frac{575}{477}}{10} = - \frac{1}{53}

x = - \frac{1}{53}

The solutions to the system of equations are:

x = - \frac{1}{53}, y = - \frac{8}{477}, z = \frac{575}{477}

3 x + y = 1, 5 x - y = - 9

5 x + 2 y = - 34, 4 x - 3 y = - 41

x - 7 - 3 x - 2 y = - 12, y = 5

5 (y + 2) x = 125 y, x = 7

\frac{1}{x} + \frac{1}{y} = 1, \frac{2}{x} - \frac{3}{y} = - 13

What is 10x+y+z=1,2x+10y+z=1,2x+y+10z=12 ?
The solution to 10x+y+z=1,2x+10y+z=1,2x+y+10z=12 is x=-1/53 ,y=-8/477 ,z= 575/477