What is 2x+4y+3z=31,0x-8y-9z=4,x+4y-3z=2 ?

The solution to 2x+4y+3z=31,0x-8y-9z=4,x+4y-3z=2 is x= 161/9 ,y=-31/12 ,z= 50/27

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2x+4y+3z=31,0x-8y-9z=4,x+4y-3z=2

2 x + 4 y + 3 z = 31, 0 x - 8 y - 9 z = 4, x + 4 y - 3 z = 2

x = \frac{161}{9}, y = - \frac{31}{12}, z = \frac{50}{27}

Solution steps

2 x + 4 y + 3 z = 31 0 \cdot x - 8 y - 9 z = 4 x + 4 y - 3 z = 2

Isolate

x

for

2 x + 4 y + 3 z = 31 : x = \frac{31 - 4 y - 3 z}{2}

Substitute

x = \frac{31 - 4 y - 3 z}{2}

[0 \cdot \frac{31 - 4 y - 3 z}{2} - 8 y - 9 z = 4 \frac{31 - 4 y - 3 z}{2} + 4 y - 3 z = 2]

Simplify

[- 8 y - 9 z = 4 \frac{4 y - 9 z + 31}{2} = 2]

Isolate

y

for

- 8 y - 9 z = 4 : y = - \frac{4 + 9 z}{8}

Substitute

y = - \frac{4 + 9 z}{8}

[\frac{4 ( - \frac{4 + 9 z}{8} ) - 9 z + 31}{2} = 2]

Simplify

[\frac{- 27 z + 58}{4} = 2]

Isolate

z

for

\frac{- 27 z + 58}{4} = 2 : z = \frac{50}{27}

For

y = - \frac{4 + 9 z}{8}

Substitute

z = \frac{50}{27}

y = - \frac{4 + 9 \cdot \frac{50}{27}}{8}

- \frac{4 + 9 \cdot \frac{50}{27}}{8} = - \frac{31}{12}

y = - \frac{31}{12}

For

x = \frac{31 - 4 y - 3 z}{2}

Substitute

y = - \frac{31}{12}, z = \frac{50}{27}

x = \frac{31 - 4 ( - \frac{31}{12} ) - 3 \cdot \frac{50}{27}}{2}

\frac{31 - 4 ( - \frac{31}{12} ) - 3 \cdot \frac{50}{27}}{2} = \frac{161}{9}

x = \frac{161}{9}

The solutions to the system of equations are:

x = \frac{161}{9}, y = - \frac{31}{12}, z = \frac{50}{27}

23 p + 27 q = 10, 27 p + 23 q = 100

2 x + y = - 1, 0 x - 3 y = 2

3 x - 4 y = 20, x - 12 y = - 20

2 x - y = 90, x - 3 y = - 40

2 x - 3 y = 30, 3 x - 4 y - 5 = 0

What is 2x+4y+3z=31,0x-8y-9z=4,x+4y-3z=2 ?
The solution to 2x+4y+3z=31,0x-8y-9z=4,x+4y-3z=2 is x= 161/9 ,y=-31/12 ,z= 50/27