Math, asked by bhagyalingala123, 5 months ago

solve the system of homogeneous equations x+y+w=0,y+x=0,x+y+z+w=0,x+y+2z=0​

Answers

Answered by studarsani18018
3

Answer:

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Step-by-step explanation:

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Answered by payalchatterje
1

Answer:

Required value of x,y,z,w are -k,k,0,0 where k is a natural number.

Step-by-step explanation:

Given equations are

x + y + w = 0...(1) \\ y + x = 0....(2) \\ x + y + z + w = 0.....(3) \\ x + y + 2z = 0....(4)

From equation (2) and (1),

x + y + w = 0 \\ 0 + w = 0 \\ w = 0

and from equation (2) and (4),

x + y + 2z = 0 \\ 0 + 2z = 0 \\ 2z = 0 \\ z = 0

and from equation (2),

x + y = 0 \\ x =  - y

Let, y = k

So,

x =  - k

Where k is a natural number.

This is a problem of Algebra.

Some important Algebra formulas:

{(x + y)}^{2}  =  {x}^{2}  + 2xy +  {y}^{2} \\  {(x  -  y)}^{2}  =  {x}^{2}   -  2xy +  {y}^{2} \\  {(x  + y)}^{3}  =  {x}^{3}  + 3 {x}^{2} y + 3x {y}^{2}  +  {y}^{3}  \\   {(x   -  y)}^{3}  =  {x}^{3}   -  3 {x}^{2} y + 3x {y}^{2}   -  {y}^{3} \\  {x}^{3}  +  {y}^{3}  =  {(x  +  y)}^{3}  - 3xy(x + y) \\ {x}^{3}   -  {y}^{3}  =  {(x   -   y)}^{3}   +  3xy(x  -  y) \\  {x}^{2}  -  {y}^{2}  = (x + y)(x - y) \\    {x}^{2}  +  {y}^{2}  =  {(x - y)}^{2}   + 2xy \\ {x}^{2}   -  {y}^{2}  =  {(x   + y)}^{2}  - 2xy \\  {x}^{3}  -  {y}^{3}  = (x - y)( {x}^{2}  + xy +  {y}^{2} ) \\ {x}^{3}   +   {y}^{3}  = (x  + y)( {x}^{2}   -  xy +  {y}^{2} )

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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