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anonymous

  • 5 years ago

Graph the feasible region to the following system of inequalities y-2x≥1 2x+3y≥6

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  1. amistre64
    • 5 years ago
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    with what?

  2. anonymous
    • 5 years ago
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    or how would you go about solving it?

  3. anonymous
    • 5 years ago
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    draw the two lines, the feasible region lies above both of the lines

  4. amistre64
    • 5 years ago
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    graph the lines: y-2x = 1 and 2x +3y = 6

  5. amistre64
    • 5 years ago
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    then determine which point, not on the line; satisfies the equatons

  6. anonymous
    • 5 years ago
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    if the inequalities were less than or equal to would i shade below the lines?

  7. amistre64
    • 5 years ago
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    pick (x=0,y=0) since its not on the line.... if it makes a true statement, sahde that side; if false? shaded the other side

  8. anonymous
    • 5 years ago
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    feasible region can be found by putting (0,0) in the line,

  9. anonymous
    • 5 years ago
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    yes amistre is right

  10. anonymous
    • 5 years ago
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    origin is easiest point to check for the required region

  11. amistre64
    • 5 years ago
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    by true statement i mean, use it in the original inequality :)

  12. anonymous
    • 5 years ago
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    \[y-2x \ge 1\]

  13. anonymous
    • 5 years ago
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    substituting x=0 and y=0 o greater than /equal 1,false

  14. anonymous
    • 5 years ago
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    so the area opposite to origin is shaded

  15. anonymous
    • 5 years ago
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    similar for the second line if the two region coincide, it is the required region, otherwise solution does not exist

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