How do you solve the system $x + y - z = 2$ $x + y - z = 2$ , $x - y - z = 3$ $x - y -z = 3$ , $x - y - z = 4$ $x - y - z = 4$ ?

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How do you solve the system $x + y - z = 2$ $x + y - z = 2$ , $x - y - z = 3$ $x - y -z = 3$ , $x - y - z = 4$ $x - y - z = 4$ ?

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Answer 1 · 2015-11-21T19:40:54

$The proposed question has a conflict, thus must be wrong$ $color(red)("The proposed question has a conflict, thus must be wrong")$
Left hand side of Equation(2) is the same as the left hand side of Equation (3) but they are stated as being equal to different values!!!

Explanation:

Given
$x+y-z=2........................................(1)$
$color(red)(x-y-z=color(blue)(3)........................................(2))$
$color(red)(x-y-z=color(green)(4)........................................(3))$

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How do you solve the system $x + y - z = 2$ $x + y - z = 2$ , $x - y - z = 3$ $x - y -z = 3$ , $x - y - z = 4$ $x - y - z = 4$ ?

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How do you solve the system x + y - z = 2x+y−z=2x + y - z = 2, x - y -z = 3x−y−z=3x - y -z = 3, x - y - z = 4x−y−z=4x - y - z = 4?

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How do you solve the system $x + y - z = 2$ $x + y - z = 2$ , $x - y - z = 3$ $x - y -z = 3$ , $x - y - z = 4$ $x - y - z = 4$ ?