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Homework Help: Solve system of three equations using elimination

  1. Nov 27, 2005 #1
    How can I use elimination to solve a system of three equations that are 3-dimensional using elimination? When I do it the point I find doesn't work for all three equations.
    Last edited: Nov 27, 2005
  2. jcsd
  3. Nov 27, 2005 #2


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    You did exactly the right thing when you checked your solution with the orginal equations. As to why you get the wrong answer, it would be impossible to say unless you showed us what you did. My guess is that you made an algebraic error somewhere in your work.
  4. Nov 27, 2005 #3


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    You use elimination for three equations exactly the way you would for two. Use two of the equations together to eliminate one unknown. Then use one of those equations together with the third equation to eliminate that same unknown. Now you have two equations in two unknowns. Use those to eliminate a second unknown leaving you with one equation in one unknown. Solve that, then substitute back into the equations to find the other two. If that solution doesn't work for all three equations, then you are doing something wrong!

    Example: x+ y+ z= 0, x- 2y- z= 0, x+ y- z= 6.

    If we add both sides of x+ y+ z= 0 and x- 2y-z= 0, we eliminate z and get 2x- y= 0. If we add both sides of x+ y+ z= 0 to x+ y- z= 6, we again eliminate z and get 2x+ 2y= 6 which is the same as x+ y= 3. We now have the two equations 2x-y= 0 and x+ y= 3. If we add both sides of 2x- y= 0 and x+ y= 3 we eliminate y and have 3x= 3. Divide both sides of that last equation and we get x= 1. Put that into either 2x- y= 0, so that 2- y= 0, or x+ y= 3, so that 1+ y= 3, and we get y= 2. Put x= 1, y= 2 into any of the original equations and we get 1+ 2+ z= 0 or 1- 4- z= 0 or 1+ 2- z= 6, all of which lead to z= -3. The point (1, 2, -3) satisfies all three equations.
  5. Nov 27, 2005 #4
    Thanks for the example. I can't seem to get the correct y and x coordinates with my equations:

    (1) 3x + 6y + 9z + 18 = 0, (2) x + 4y + 8z + 2 = 0, (3) x + 3y + 4z + 6 = 0

    Subtract (2) and (3) to get y + 4z - 4 = 0, and subtract (1) and (3)*3 to get -3y - 3z = 0 or y = -z

    (-z) + 4z -4 = 0
    z = 3/4

    But y and x depend on which equations a plug my known values into...
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