How would one proceed with proving that there exists none ( x , y , z ) such that all the following hold: x + 2 y + 7 z = 0 P ( x + 3 y + 9 z ≥ 0 ) = 1 P ( x + y + 5 z ≥ 0 ) = 1 P ( x + 13 y + 10 z ≥ 0 ) = 1 P ( x + 3 y + 9 z &gt; 0 ) &gt; 0 P ( x + y + 5 z &gt; 0 ) &gt; 0 P ( x + 13 y + 10 z &gt; 0 ) &gt; 0

Question

How would one proceed with proving that there exists none   (  x  ,  y  ,  z  ) such that all the following hold:  x  +  2  y  +  7  z  =  0  P  (  x  +  3  y  +  9  z  ≥  0  )  =  1  P  (  x  +  y  +  5  z  ≥  0  )  =  1  P  (  x  +  13  y  +  10  z  ≥  0  )  =  1  P  (  x  +  3  y  +  9  z  &amp;gt;  0  )  &amp;gt;  0  P  (  x  +  y  +  5  z  &amp;gt;  0  )  &amp;gt;  0  P  (  x  +  13  y  +  10  z  &amp;gt;  0  )  &amp;gt;  0

Lucille Lucas · Accepted Answer

x  =  y  =  z  =  0 is a solution to the first 4 constraints That means that there exists   a  ,  b  ,  c  ≥  0 such that                    x        +        2        y        +        7        z                            =        0                            x        +        3        y        +        9        z                            =        a                            x        +        y        +        5        z                            =        b                            x        +        13        y        +        10        z                            =        c            You must prove that the equations are not linearly dependent, so the only solution is   x  =  y  =  z  =  0 for   a  =  b  =  c  =  0. But then the last three inequalities are not obeyed (strict inequality)

Govindennz34j · Accepted Answer

It looks like   (  x  ,  y  ,  z  )  =  (  1  ,  3  ,  −  1  ) satisfies all the conditions

How would one proceed with proving that there exists none ( x , y , z ) such that al

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