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Math Help - Intersection Problem

  1. #1
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    Intersection Problem

    I have two operations too chose from. Each operation is essentially a base added too a coefficient and a variable.

    Since each operation appears to be linear and the two operations exist within the same system I decided to determine were the lines intercept.

    This is not a problem out of a math book.

    (500+.2142y)(2.06x)+(500+.2142y)(1-x)+(500+.6y)(2.06x)+(500+.6y)(1-x)

    And

    (872+.3858y)(1-x)+(872+.3858y)(2.06x)

    Since I want to find the point where x and y are equal to each other I set the two as equal to each other. After Reducing and simplifying I get;

    1.180548yx+924.32x+.3858y+872=.86305yx+1414.04x+.8 142y+1334

    I decided it would be quicker to work the problem out with out all the long coefficients too see if I could get at a solution at all. For simplicities sake I've been writing it out on paper as;

    xy+x+y+10=3xy+2x+3y+7
    There are no fractions so I start by factoring.
    x(y+1)+y+10=x(3y+2)3y+7
    Then move terms over
    x(y+1)+y-3y+10-7=x(3y+2)
    Combine like terms
    x(y+1)-2y+3=x(3y+2)

    This is were i get stuck. If I divide away the 3y+2 on the right hand side and move it too the left, I still have an x sitting over on the left side.
    Or;
    x(y+1)-2y+3/(3y+2)=x

    I can't just divide out the x on the left, since that would leave me with a 1 on the other side, X/X.
    At first I thought it was a no solution type problem, but that doesn't seem to make sense. Have I just set the whole thing up wrong?
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  2. #2
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    Quote Originally Posted by bkbowser View Post
    x(y+1)-2y+3=x(3y+2)
    This is were i get stuck.
    Get the x terms together:
    x(3y + 2) - x(y + 1) = 3 - 2y
    Factor out the x:
    x(3y + 2 - y - 1) = 3 - 2y
    Simplify:
    x(2y + 1) = 3 - 2y
    Isolate the x:
    x = (3 - 2y) / (1 + 2y)
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