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Solving systems of equations

  1. Oct 2, 2005 #1
    I need to solve this system of equations.

    7/10S + 1D <=630
    1/2S + 5/6D <= 600
    1S + 2/3D <=708
    1/10S + 1/4D <=135


    I attempted to use the Excel solver to figure it out, but I couldn't understand completely how to work it.

    Any ideas on how I can get started?
     
  2. jcsd
  3. Oct 2, 2005 #2

    hotvette

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    If I read this right, there are just 2 unknowns, S and D. You have 4 equations. They are linearly dependent. That's probably why Excel barfed. With just 2 unknowns, just take 2 of them and solve by substitution. Should be straightforward.

    Whoops, inequalities! Hmmm, don't know.
     
  4. Oct 3, 2005 #3
    I need help with this problem.

    Any input is greatly appreciated.
     
  5. Oct 3, 2005 #4
    Graphing always works when in doubt just graph them and see what kind of a soluiton set you get. Remeber greater then shades up and less than shades down.
     
  6. Oct 3, 2005 #5

    hotvette

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    Yep, just figured it out. In addition to graphing to see what you get, you can also find the interesection of each corner of the region analytically. Consider each equation an equality instead of inequality. You have 4 linear equations of D in S (or S in D). Remembering how to find intersections between 2 lines you can get all of the intersection points that define the solution region. Graphing will help you decide which pairs of equations are appropriate.
     
    Last edited: Oct 3, 2005
  7. Oct 3, 2005 #6
    prove that a set with an uncountable subset is itself uncountable
     
  8. Oct 3, 2005 #7
    So I should rewrite the equations.. setting S = Y and D = X? and then graph it on my TI-86?
     
  9. Oct 3, 2005 #8

    hotvette

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    Yep, but I did it the other way around (substitute X for S and Y for D). I don't see why it should matter which way you do it. For graphing, I used Excel, but that shouldn't matter either. You should find that the intersecting lines define a region that constitutes the solution.
     
  10. Oct 3, 2005 #9
    I got
    y<= /710x + 630
    y<= (-.5x + 600)(6/5)
    y<=(-x+708)(3/2)
    y<=(-1/10x+135(4)

    i graphed it on my TI-86 (don't know how in excel).. and the lines don't all intersect at 1 point
     
  11. Oct 3, 2005 #10

    hotvette

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    Re check the first equation. Slope is wrong.
     
  12. Oct 3, 2005 #11
    whoops, I mistyped it.. its 7/10

    I got 540,252 as my max.

    Looks to be right.
     
  13. Oct 3, 2005 #12

    hotvette

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    Check the sign of the slope.
     
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