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Math Help - Linear and non-linear equations

  1. #1
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    Linear and non-linear equations

    I can't find the right answer to this question... Is it just me, or are the solutions given wrong?

    x^2 + 3x + 5y = 20
    x + 3y = 1

    Given solutions are x = 3 or x = -9/2

    Where did I go wrong?

    X = 1 3y ---- (1)
    X^2 + 3x + 5y = 20 --- (2)
    Sub (1) into (2), (1 3y)^2 + 3(1 3y) + 5y = 20
    1 6y + 9y^2 + 3 9y + 5y = 20
    9y^2 10y 16 = 0
    (9y + 8 ) (y 2) = 0
    y = -8/9 or y = 2

    Sub y = -8/9 in (1),
    x = 1 - 3(-8/9)
    = 11/3

    Sub y = 2 in (1),
    x = 1 - 3(2)
    = -5

    Are the given solutions correct? And are what I worked the right answers? If not, please tell me where I made a mistake. I'm not too sure if I substituted the y values correctly...
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Linear and non-linear equations

    Quote Originally Posted by spirit79 View Post
    I can't find the right answer to this question... Is it just me, or are the solutions given wrong?

    x^2 + 3x + 5y = 20
    x + 3y = 1

    Given solutions are x = 3 or x = -9/2

    Where did I go wrong?

    X = 1 3y ---- (1)
    X^2 + 3x + 5y = 20 --- (2)
    Sub (1) into (2), (1 3y)^2 + 3(1 3y) + 5y = 20
    1 6y + 9y^2 + 3 9y + 5y = 20
    9y^2 10y 16 = 0
    (9y + 8 ) (y 2) = 0
    y = -8/9 or y = 2

    Sub y = -8/9 in (1),
    x = 1 - 3(-8/9)
    = 11/3

    Sub y = 2 in (1),
    x = 1 - 3(2)
    = -5

    Are the given solutions correct? And are what I worked the right answers? If not, please tell me where I made a mistake. I'm not too sure if I substituted the y values correctly...
    Your answer is correct, you can also test your results by subbing them into the original equation.

    Spoiler:
    For (-5,2) : (-5)^2 + 3(-5) + 5(2) = 20 \Leftrightarrow 25 - 15+10 = 20 which is correct as is (-5) + 3(2) = 1 so (-5,2) is a valid solution

    For (11/3 , -8/9):  \left(\dfrac{11}{3}\right)^2 + 3\left(\dfrac{11}{3}\right) + 5\left(\dfrac{-8}{9}\right) = 20 so (11/3 , 8/9) is a valid solution.
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  3. #3
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    Re: Linear and non-linear equations

    Quote Originally Posted by e^(i*pi) View Post
    Your answer is correct, you can also test your results by subbing them into the original equation.

    Spoiler:
    For (-5,2) : (-5)^2 + 3(-5) + 5(2) = 20 \Leftrightarrow 25 - 15+10 = 20 which is correct as is (-5) + 3(2) = 1 so (-5,2) is a valid solution

    For (11/3 , -8/9):  \left(\dfrac{11}{3}\right)^2 + 3\left(\dfrac{11}{3}\right) + 5\left(\dfrac{-8}{9}\right) = 20 so (11/3 , 8/9) is a valid solution.
    Thanks for your clarification...!

    I have another question, but I'm not too keen to start another thread topic on it seeing that the questions are related...

    Solve:

    5x + 3y + 7 = 0
    x^2 - 4y + 3 = 3y^2

    I tried, but got the wrong answers...?

    5x = -3y 7
    25x^2 = 9y^2 42y + 49 --- (1)
    X^2 4y + 3 = 3y^2
    25x^2 100y + 75 = 75y^2 --- (2)
    Sub (1) into (2),
    9y^2 42y + 49 100y + 75 = 75y^2
    9y^2 42y 100y 75y^2 + 49 + 75 = 0
    -142y 75y^2 + 124 = 0
    66y^2 + 142y 124 = 0
    2 (33y^2 + 71y 62) = 0
    (11y + 31) (3y 2) = 0
    Y = -31/11 or y = 2/3
    When y = -31/11,
    5x = -3 (-31/11) 7
    X = 16/55
    When y = 2/3,
    5x = -3 (2/3) 7
    X = -9/5

    The answer key states y = -1 29/33 or y = 1, x = 3/11 or x = -2.
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