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Math Help - solving a system of nonlinear equations

  1. #1
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    solving a system of nonlinear equations

    14x^2-3xy+9y^2=71
    2x+3y=9

    I solve for x in the second equation and get x=-3y+9 /2 but when i plug it back in i am not sure how to square (-3y+9)/2 in the first equation?
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  2. #2
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    Quote Originally Posted by vbarnett326 View Post
    14x^2-3xy+9y^2=71
    2x+3y=9

    I solve for x in the second equation and get x=-3y+9 /2 but when i plug it back in i am not sure how to square (-3y+9)/2 in the first equation?
    14 \cdot \left( \frac{9 - 3y}{2}\right)^2 = 14 \cdot \left(\frac{3(3 - y)}{2}\right)^2 = 14 \cdot \frac{9 (3 - y)^2}{4} = \frac{63}{2} \cdot (9 - 6y + y^2)
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  3. #3
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    i don't understand how you got 63/2.

    and also -3((-3y+9)/2))y equals (9y^2 -27y)/-6y?????? is that correct
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  4. #4
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    Quote Originally Posted by vbarnett326 View Post
    i don't understand how you got 63/2. Mr F says: Did you multiply the numbers? {\color{red}\frac{14 \cdot 9}{4} = \frac{63}{2}}. You're meant to do more than just look at the solution - you're meant to mess in and do it yourself, using what I posted as a guide.

    and also -3((-3y+9)/2))y equals (9y^2 -27y)/-6y?????? is that correct

    Mr F says: Certainly not! Why are you multiplying the bottom by -3 and y as well?? Do you honestly think that something like {\color{red}2 \cdot \frac{4}{3} = \frac{2 \cdot 4}{2 \cdot 3}} ? The answer is (9y^2 -27y)/2
    ..
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