Tuesday, June 15, 2010

`x + y + z + w = 6, 2x + 3y - w = 0, -3x + 4y + z + 2w, = 4, x + 2y - z + w = 0` Solve the system of linear equations and check any solutions...

You may use the reduction method to solve the system, hence, you may multiply the first equation by 3, such that:


`3(x + y + z + w) = 3*6`


`3x + 3y + 3z + 3w = 18`


You may now add the equation `3x + 3y + 3z + 3w = 18` to the third equation -`3x + 4y + z + 2w= 4` , such that:


`3x + 3y + 3z + 3w - 3x + 4y + z + 2w= 18 + 4`


`7y + 4z + 5w = 22`


Adding the first equation to the second yields:


`3x + 4y + z = 6`


Adding the second equation to the last yields:


`3x + 5y - z = 0`


Adding the resulted equations yields:


`6x + 9y = 6 => 2x + 3y = 2`


Multiply the second equation by 2 and add it to the third, such that:


`x + 10y + z = 4`


Add this equation to the `3x + 5y - z = 0` , such that:


`3x + 5y - z + x + 10y + z = 0 + 4`


`4x + 15y = 4`


Consider a system formed by equations `4x + 15y = 4` and `2x + 3y = ` 2, such that:


`-2*(2x + 3y) + 4x + 15y  = -4 + 4`


`-4x - 6y + 4x + 15y = 0`


`9y = 0 => y = 0`


You may replace 0 for y in equation `2x + 3y = 2` , such that:


`2x + 0 = 2 => x = 1`


You may also replace 1 for x and 0 for y in equation `2x + 3y - w = 0` , such that:


`2 - w = 0 => -w = -2 => w = 2`


You may also replace 1 for x, 0 for y and 2 for w in equation `x + y + z + w = 6` , such that:


`1 + 0 + z + 2 = 6 => z = 6 - 3 => z = 3`


Hence, evaluating the solution to the given system, yields that `x =1, y = 0, z = 3, w = 2.`

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