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`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 algebraically.

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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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