Let D = x ∈ R 3 : | 2 x 1 − x 2 + 3 x 3 + 1 | + | x 2 + 2 x 3 − 2 | + | 5 x 2 − 3 x 3 | ≤ 10 . Express D as the feasible solution set of a linear system of inequalities (meaning, a system of the form A x ≤ b).How is the feasible solution set represented? Is this problem just a matter of removing the absolute signs and setting up two linear equations as such: 2 x 1 + x 2 − 3 x 3 − 1 ≤ 10 and ( 2 x 1 + x 2 − 3 x 3 − 1 ) ≤ 10?

Question

Let   D  =      x    ∈          R      3        :      |    2    x    1    −    x    2    +    3    x    3    +    1      |    +      |    x    2    +    2    x    3    −    2      |    +      |    5    x    2    −    3    x    3      |    ≤    10  .  Express D as the feasible solution set of a linear system of inequalities (meaning, a system of the form   A  x  ≤  b).How is the feasible solution set represented? Is this problem just a matter of removing the absolute signs and setting up two linear equations as such:   2  x  1  +  x  2  −  3  x  3  −  1  ≤  10 and   (  2  x  1  +  x  2  −  3  x  3  −  1  )  ≤  10?

Sydnee Villegas · Accepted Answer

Your original constraint is  |  2      x    1    −      x    2    +  3      x    3    +  1  |  +  |      x    2    +  2      x    3    −  2  |  +  |  5      x    2    −  3      x    3    |  ≤  10Now consider the case where   5      x    2    −  3      x    3    ≥  0. If we knew that was always true we could write simply  5      x    2    −  3      x    3    ≥  0If we knew   5      x    2    −  3      x    3    ≤  0 then we could write simply  |  2      x    1    −      x    2    +  3      x    3    +  1  |  +  |      x    2    +  2      x    3    −  2  |  −  (  5      x    2    −  3      x    3    )  ≤  10If we have both of these constraints, we capture both possibilities.We can now extend that reasoning for all the       2    3   possible combinations of terms being negative and non-negative, e.g.  +  (  2      x    1    −      x    2    +  3      x    3    +  1  )  +  (      x    2    +  2      x    3    −  2  )  +  (  5      x    2    −  3      x    3    )  ≤  10  −  (  2      x    1    −      x    2    +  3      x    3    +  1  )  +  (      x    2    +  2      x    3    −  2  )  +  (  5      x    2    −  3      x    3    )  ≤  10  +  (  2      x    1    −      x    2    +  3      x    3    +  1  )  −  (      x    2    +  2      x    3    −  2  )  +  (  5      x    2    −  3      x    3    )  ≤  10  −  (  2      x    1    −      x    2    +  3      x    3    +  1  )  −  (      x    2    +  2      x    3    −  2  )  +  (  5      x    2    −  3      x    3    )  ≤  10  +  (  2      x    1    −      x    2    +  3      x    3    +  1  )  +  (      x    2    +  2      x    3    −  2  )  −  (  5      x    2    −  3      x    3    )  ≤  10  +  (  2      x    1    −      x    2    +  3      x    3    +  1  )  −  (      x    2    +  2      x    3    −  2  )  −  (  5      x    2    −  3      x    3    )  ≤  10  −  (  2      x    1    −      x    2    +  3      x    3    +  1  )  −  (      x    2    +  2      x    3    −  2  )  −  (  5      x    2    −  3      x    3    )  ≤  10  −  (  2      x    1    −      x    2    +  3      x    3    +  1  )  +  (      x    2    +  2      x    3    −  2  )  −  (  5      x    2    −  3      x    3    )  ≤  10If thats a bit confusing, consider  |      y    1    |  +  |      y    2    |  ≤  1      y    1    +      y    2    ≤  1  −      y    1    +      y    2    ≤  1      y    1    −      y    2    ≤  1  −      y    1    −      y    2    ≤  1Hopefully you see the connection.

Let D = x ∈ R 3 </msup> : | 2 x 1

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