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Math Help - Solving an equation with fractions

  1. #1
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    Solving an equation with fractions

    I have the following equation to solve. I have the answer, but don't know how to get to that answer. Please help:

    Equation

    -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

    The answer is given as:

    x/y = 1/2

    Please help, and write out each step to getting the answer if possible. I know it's probably simple, but I'm very bad at maths.
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  2. #2
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    Quote Originally Posted by parkparky View Post
    I have the following equation to solve. I have the answer, but don't know how to get to that answer. Please help:

    Equation

    -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

    The answer is given as:

    x/y = 1/2

    Please help, and write out each step to getting the answer if possible. I know it's probably simple, but I'm very bad at maths.
    -(\frac {3}{(\frac {1}{2} + \frac{x}{y})}) + \frac {9y}{2x} -6 = 0

    -(3 \cdot \frac {1}{\frac{1}{2} + \frac{x}{y}}) + \frac {9y}{2x} -6 = 0

    -3 \cdot \frac {-1}{\frac{1}{2} + \frac{x}{y}} + \frac {9y}{2x} -6 = 0

    -3 \cdot \frac {-1}{2x(\frac{1}{2} + \frac{x}{y})} + \frac {9y}{(\frac{1}{2} + \frac{x}{y})2x} -6 = 0

    -3 \cdot \frac {-1}{x + \frac{2x^2}{y}} + \frac {9y}{x + \frac{2x^2}{y}} -6 = 0

    -3 \cdot \frac {-1 + 9y}{x + \frac{2x^2}{y}} -6 = 0

    \frac {3 - 27y}{x + \frac{2x^2}{y}} -6 = 0

    3 - 27y = 6(x+ \frac{2x^2}{y})

    3 - 27y = 6x+ \frac{12x^2}{y}

    3(1 - 9y) = 6x(1 + \frac{2x}{y})

    I simplified it the most I could, but I can't seem to find any possible way to finish this. Let's see what the other people can find...
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  3. #3
    Senior Member I-Think's Avatar
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    Substitute

    Greetings.

    We have the equation: -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

    Rewritten as: 9y/2x – [3/(1/2 + x/y)] – 6 = 0

    Let x/y = K, hence y/x = 1/K

    So the equation becomes: 9/2K – [3/(1/2 + K)] – 6 = 0

    Multiply the equation by 2K: 9 – [6K/(1/2 + K)] – 12K = 0

    Multiply the equation by (1/2 + K): 4.5 +9K – 6K – 6K – 12K^2 = 0

    Multiply the equation by 2: 9 + 18K – 12K – 12K – 24K^2 = 0

    Rearrange and simplify: 8K^2 + 2K –3 = 0

    Solve the quadratic equation and you’ll obtain your solution.
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  4. #4
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    Quote Originally Posted by I-Think View Post
    Greetings.

    We have the equation: -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

    Rewritten as: 9y/2x – [3/(1/2 + x/y)] – 6 = 0

    Let x/y = K, hence y/x = 1/K

    So the equation becomes: 9/2K – [3/(1/2 + K)] – 6 = 0

    Multiply the equation by 2K: 9 – [6K/(1/2 + K)] – 12K = 0

    Multiply the equation by (1/2 + K): 4.5 +9K – 6K – 6K – 12K^2 = 0

    Multiply the equation by 2: 9 + 18K – 12K – 12K – 24K^2 = 0

    Rearrange and simplify: 8K^2 + 2K –3 = 0

    Solve the quadratic equation and you’ll obtain your solution.

    Thank you so much. Really helpful. Do you know what topic this kind of question would go under so I can try more exercises like it?

    Thanks xxx
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  5. #5
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    Since the problem involves two variables in one equation, you can't solve for either x or y separately. You can solve for y/x or x/y since x and y always occur together. And I would recommend replacing "y/x" by a single variable: say a= y/x.
    -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

    Then -\frac{3}{1/2+ 1/a}+ (9/2)a- 6= 0.
    Multiply numerator and denominator of that first fraction by 2a to simplify it: -\frac{6a}{a+ 2}- (9/2)a- 6= 0

    Multiply the entire equation by 2(a+2) to get rid of all the denominators: 12a- 9a(a+2)- 12(a+2)= 0
    12a- 9a^2- 18a- 12a- 24= -9a^2- 18a- 24= 0
    That's a quadratic equation which could be solved by the quadratic formula or completing the square.
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  6. #6
    Senior Member I-Think's Avatar
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    Technique: Substitution

    Allo. This topic falls under (Pre-Algebra + Algebra)?...

    Okay, I'm not totally sure what topic this falls under, but to solve questions such as these, we usually substitute another variable to make the question easier to solve.

    In your question, we let x/y=K

    Another example, :x^4 – 3x^2 + 2 = 0
    We let x^2=Z
    So the equation becomes :Z^2 - 3Z + 2 = 0
    Solve for Z and we then determine x.

    Final point, when you see a variable as part of the denominator of one of the terms in the question, multiply both sides of the equation to get rid of it.

    Eg. 6f + 3/f - 2 = 0

    Multiply by f: 6f^2 + 3 - 2f = 0, Solve as normal

    Eg 2. 5/(1+z) -3 + 11z = 0

    Multiply by (1+z): 5- 3-3z +11z+11z^2 = 0
    11z^2 +8z +2 = 0, Solve as normal

    That's techniques I believe are necessary to solve questions of this *mysterious* topic.

    Farewell and adieu. Church beckons.
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