What is 2x+y+4z=12,8x-3y+2z=23,4x+11y-z=33 ?

The solution to 2x+y+4z=12,8x-3y+2z=23,4x+11y-z=33 is x= 47/14 ,y= 13/7 ,z= 6/7

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2x+y+4z=12,8x-3y+2z=23,4x+11y-z=33

2 x + y + 4 z = 12, 8 x - 3 y + 2 z = 23, 4 x + 11 y - z = 33

x = \frac{47}{14}, y = \frac{13}{7}, z = \frac{6}{7}

Solution steps

2 x + y + 4 z = 12 8 x - 3 y + 2 z = 23 4 x + 11 y - z = 33

Isolate

x

for

2 x + y + 4 z = 12 : x = \frac{12 - y - 4 z}{2}

Substitute

x = \frac{12 - y - 4 z}{2}

[8 \cdot \frac{12 - y - 4 z}{2} - 3 y + 2 z = 23 4 \cdot \frac{12 - y - 4 z}{2} + 11 y - z = 33]

Simplify

[- 7 y - 14 z + 48 = 23 9 y - 9 z + 24 = 33]

Isolate

y

for

- 7 y - 14 z + 48 = 23 : y = - \frac{14 z - 25}{7}

Substitute

y = - \frac{14 z - 25}{7}

[9 (- \frac{14 z - 25}{7}) - 9 z + 24 = 33]

Simplify

[\frac{- 189 z + 225}{7} + 24 = 33]

Isolate

z

for

\frac{- 189 z + 225}{7} + 24 = 33 : z = \frac{6}{7}

For

y = - \frac{14 z - 25}{7}

Substitute

z = \frac{6}{7}

y = - \frac{14 \cdot \frac{6}{7} - 25}{7}

- \frac{14 \cdot \frac{6}{7} - 25}{7} = \frac{13}{7}

y = \frac{13}{7}

For

x = \frac{12 - y - 4 z}{2}

Substitute

y = \frac{13}{7}, z = \frac{6}{7}

x = \frac{12 - \frac{13}{7} - 4 \cdot \frac{6}{7}}{2}

\frac{12 - \frac{13}{7} - 4 \cdot \frac{6}{7}}{2} = \frac{47}{14}

x = \frac{47}{14}

The solutions to the system of equations are:

x = \frac{47}{14}, y = \frac{13}{7}, z = \frac{6}{7}

\frac{x}{2} + \frac{y}{3} = 2, \cdot \frac{x}{4} + \frac{y}{2} = 2

3 x + y = - 17, 5 x + 2 y = - 29

3 x - 2 y = 5, \frac{( 2 x )}{3} + \frac{y}{2} = - \frac{7}{9}

y = \frac{5}{7}, x = 1

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

What is 2x+y+4z=12,8x-3y+2z=23,4x+11y-z=33 ?
The solution to 2x+y+4z=12,8x-3y+2z=23,4x+11y-z=33 is x= 47/14 ,y= 13/7 ,z= 6/7