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Math Help - Use the method of substitution to solve the system

  1. #1
    Amy
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    Exclamation Use the method of substitution to solve the system

    Use the method of substitution to solve the system
    1. 25Y^2-16X^2=400
    9Y^2-4X^2=36



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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Amy View Post
    Use the method of substitution to solve the system
    1. 25Y^2-16X^2=400
    9Y^2-4X^2=36



    Thanks
    check your question. this system has no real solutions
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  3. #3
    Amy
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    Quote Originally Posted by Jhevon View Post
    check your question. this system has no real solutions
    the question is correct. and you are right that the system has no real roots.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Amy View Post
    Use the method of substitution to solve the system
    1. 25Y^2-16X^2=400
    9Y^2-4X^2=36
    Quote Originally Posted by Amy View Post
    the question is correct. and you are right that the system has no real roots.
    okie dokie. so we use the same method as if it had real roots, no surprises here.

    25y^2 - 16x^2 = 400 ..................(1)

    9y^2 - 4x^2 = 36 .......................(2)

    Solving for y^2 from (2) we see that y^2 = 4 + \frac {4}{9} x^2
    Substitute 4 + \frac {4}{9} x^2 for y^2 in equation (1), we get:

    25 \left( 4 + \frac {4}{9} x^2 \right) - 16x^2 = 400

    \Rightarrow 100 + \frac {100}{9} x^2 - 16x^2 = 400

    \Rightarrow - \frac {44}{9} x^2 = 300

    \Rightarrow x^2 = - \frac {675}{11}

    \Rightarrow x = \pm \sqrt { \frac {675}{11}} ~i \approx 7.8335 ~i

    Now we simply substitue \pm \sqrt { \frac {675}{11}} ~i for x in either of the original equations to find y

    You should get y = \pm \sqrt { \frac {256}{11}} ~i \approx 4.8242 ~i
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