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Homework Help: Linear equations-help solving for 2 unkown variables

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    It seemed simple at first but, how would you solve x+y=12 also given xy=35. I can only seem to solve it by trial and error. Any solutions, I think I may be missing something that's all.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 24, 2010 #2
    Have you tried solving it by substitution?
  4. Mar 24, 2010 #3


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    Homework Helper

    What Redsummers means by substitution is to solve one of the equations for one of the variables, then substitute the result into the other equation. You then end up with an equation of a single variable that you can solve. Once you know the value of one variable, you can determine the value of the other variable from either equation.
  5. Mar 25, 2010 #4
    Yeah but if i solve for one variable:x+y=12
    xy=35 =y-y,x-x? since im subtracting by the same number, as opposed to a simple problem like 2x+6y=54
    2x+3y=45 I can use simple substitution to figure out 3y=9, and then substitute the answer into one of the equations. I'm sorry I am posting this on a physics forum, it was just the only question I had in mind.
  6. Mar 25, 2010 #5
    You know x+y=12, so you can rearrange that so you define y in terms of x (leaving only y on the left).

    You then substitute your new y into xy=35.

    Hope that helps.
  7. Mar 25, 2010 #6
    hmm, I don't know how to solve this. Anyone who can, with a formula, solve this please try. Though, I don't think its possible, since x and y are interchangeable in this problem, good luck though. using y=35/x and y=12-x only gets you to--35=12y-y^2, which turns out to be: 35=xy
  8. Mar 25, 2010 #7
    You have two equations:

    x+y = 12
    xy = 35

    Now, to do a substitution you will need e.g. to set the x in terms of y, i.e.:

    x = 12 - y

    Now, putting this equation into the other:

    (12 - y)y = -y^2 +12y = 35

    Solving for y:

    [tex]y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    where a=-1, b=12, and c=-35

    And that's definitely not leading you to an imaginary number solution, so it's perfectly doable in [tex]\Re[/tex]. It's not that complex, you could even came up with the two numbers mentally.
  9. Mar 25, 2010 #8
    Lol, I just thought about the quadratic formula soon after this. I can't believe I had forgotten it. I was thinking it was complex, whoops. Thanks Redsummers.
  10. Mar 25, 2010 #9
    Yeah, you gotta use that formula, is like one of the commandments in linear equation religion. You're welcome ^^
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