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Simultaneously Solve a Linear and Cubic equation

  1. Jul 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve the following equations Simultaneously:
    y=x^3 and y=3x-2

    My question is, how do I solve this? I've tried everything


    3. The attempt at a solution
    I've tried to isolate x in the linear equation and substitute it into the cubic. I don't know if that's wrong but it didn't work for me. This is what I did:

    isolate x in linear equation: x= -2/3-y/3
    substitute this into cubic equation: y=(-2/3-y/3)^3
    I then expanded this and it didn't work.

    I also tried: 3x-2=x^3
    therefore: x^3-3x+2=0
    this didn't work either! please help!
     
  2. jcsd
  3. Jul 29, 2011 #2
    This way is a good start; do you know about the rational root theorem?
     
  4. Jul 30, 2011 #3

    SteamKing

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    By inspection, there is at least one integer solution between 0 and 2 to the two equations.
    There is also another integer solution between -3 and -1.
     
  5. Jul 30, 2011 #4
    I would prefer you told me the methods for solving the equations rather than the answer. I already know the answer by using my graphics calculator. However this is of no use to me as I want to solve it the proper way without a calculator.

    I didn't. But I Googled it and I do know now. Is there any way to do these simultaneously like you can with a quadratic and linear equation?
     
  6. Jul 30, 2011 #5

    ehild

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    Try to group the terms so as something can be factor out.

    x3-3x-2=(x3-x)-2(x-1).

    The rational root theorem is also very useful. It states: For a polynomial equation
    anxn+an-1xn-1+....a0=0,

    if p is an integer factor of the constant term a0, and q is an integer factor of the leading coefficient an, the rational solution is of the form ±p/q.

    In this case, a3=1 and a0=-2, so q=1, and the possible p values are 1 and 2. So you have 4 possible values for x to try if they are roots or not. If one of them (x1) is a root, you can divide the equation by the factor of (x-x1) and you get a quadratic equation to solve.


    ehild
     
  7. Jul 30, 2011 #6
    Believe it or not, inspection is actually the first thing you should always try when given a problem like this. Inspection is an entirely proper technique.

    There is some meta-reasoning involved. They can't actually want to make you solve a random cubic, since that would be too difficult. There might be a simple solution. So you mentally plug in the simplest numbers ... 0, 1, -1, 2, etc. And if you do this, you find a solution.

    Once you have one solution, you can do a polynomial division to reduce the cubic to a quadratic, which you know how to solve using the quadratic formula.

    So even though there are a few more technical things you can and should try; the very first thing you should do is plug in the most obvious simple numbers and see if a solution pops out.
     
    Last edited: Jul 30, 2011
  8. Jul 30, 2011 #7

    PeterO

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    That last expression can be easily factorised into 3 linear factors, and I am sure that was supposed to be the next step in the solution.
     
  9. Jul 31, 2011 #8

    NascentOxygen

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    I can see that x=1 will make the LHS = 0, so x=1 is a solution.

    Now, take out the factor (x-1).

    In your working for this, you divide x3-3x+2 by (x-1)
     
    Last edited: Jul 31, 2011
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