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Sequence question

  1. Aug 6, 2014 #1

    adjacent

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    1. The problem statement, all variables and given/known data
    The last two terms, are added together to produce the next term

    ##-33,x,y,z,88##
    ##\text{Find }x,y\text{ and }z##
    2. Relevant equations
    ##y=x-33##
    ##x+y=z##
    ##y+z=88##
    3. The attempt at a solution

    By substituting the expression for y in the third equation for the first equation, we get:
    ##88-z=x-33 \rightarrow x=121-z##
    Again, by substituting the expression for y in the first equation for the second equation we get:
    ##x+(x-33)=z \rightarrow 2x-33=z##

    So we get two equations,
    ##x=121-z##
    ##2x-33=z##

    Solving both gives the right answers but I remember getting the wrong answer by not including one of the equations there.
    Is there any easy way to solve all three equations at once?
     
    Last edited: Aug 6, 2014
  2. jcsd
  3. Aug 6, 2014 #2

    HallsofIvy

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    I'm not sure what you mean by "solve all three at once". I get this image of just immediately writing down the values for x, y, and z! I could not do that!:tongue:

    But you can rewrite the first equation as y- x= -33. Adding that to x+ y= z gives 2y= z- 33. The last equation, y+ z= 88, is equivalent to y= -z+ 88. Adding those equations, 3y= 55.

    That's about as simple as I can make it.
     
  4. Aug 6, 2014 #3

    adjacent

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    Oh, I must have made some silly mistake in the previous calculations. Thanks :)

    Haha, I mean like how you solve two simultaneous equation, is there a method for solving three simultaneous equations like this? Other than substituting the values?
     
  5. Aug 6, 2014 #4
    This system of linear equations, like any other, can be solved by first writing it in upper-triangular form then using the method of back-substitution. Let's first define the terms we used: A system in upper-triangular form is one like: $$\left\{\begin{array}{rl}4x-3y+2z&=-5\\14y+2z&=18\\-4z&=3\end{array}\right.,\qquad\left\{\begin{array}{rl}
    15x-2y+z&=1\\
    3z&=-8\end{array}\right..$$ As you can see, what's common about them is that if ##x## appears in no equation other than the first one and ##y## appears in no equation after the second... (It is possible that ##y## may not even appear in the second equation as in e.g.2)
    You can easily solve systems in this type using the method of back-substitution. Take the first example, using the third equation you solve for ##z## to get ##z=-\tfrac34##, you substitute that value in the second equation to get ##14y-\tfrac32=18## which can be used to solve for ##y##, and when you get your value for ##y##, you substitute it in the first equation along with your value for ##z## to solve for ##x##.
    What's interesting now is that you can try to transform any system you have in upper-triangular form, then you use back-substitution and you're done. The basic way we proceed with this task consists of applying the elementary operations which are: 1. multiplying both sides of an equation by a nonzero constant, 2. interchanging the order of two equations, 3. adding to one equation a multiple of another.
    I think this will give you enough information to solve any system in 3 variables or more. If you need to know more then just ask away.
     
  6. Aug 7, 2014 #5

    adjacent

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    Thank you so much HakimPhilo!
    I think I can do better with systems of 3 equations now.
     
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