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Math Help - Non-Linear Simultaneous Equation

  1. #1
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    Question Non-Linear Simultaneous Equation

    Hello,

    I've just started on Non-linears and am Stuck on my first problem.

    Heres the equations:

    x^2 + y^2 =13
    y = x+1

    This is how far I've got, I would appreciate it if someone could show me where I'm going wrong.

    • x^2 + (x+1)(x+1) = 13
    • x^2 + (x^2+x+x+1) =13
    • 2x^2 + 2x + 1 = 13
    • 2x^2 + 2x -12 = 0
    • x^2 + x -6 = 0


    Now I'm stuck.
    I've done this over and over and keep getting x^2=3 yet I know it should be x=2
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  2. #2
    Member jacs's Avatar
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    That looks good so far, now you need to factorise the quadratic to solve for x, you should get two solutions:
    x^2+x-6=0
    (x - 2)(x + 3) = 0
    x = 2, x = -3
    then from there substitute both in to find your respective y values
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  3. #3
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    Thumbs up

    Nice, of course it is! Thanks.

    The answers should come out as x = 2 and y = -3

    However if I put this in to the second equation y=x+1

    Then y = 2+1 which is 3, not -3.
    What's the deal with that?
    I have a serious issue with +s and -s.

    Are the answers I was given wrong?

    If I put it in the first:
    x^2 + y^2 = 13 so;
    • 2^2 + -3^2 = 13
    • 4 + -9 = -3 ...which isn't 13!


    I know +9 makes 13, but WHY does -3^2 = 9?! as -3^3 = -27.
    I think I'm missing some serious basic maths.
    If I am, any suggestions on things I should practice or learn?

    I was stuck straight in the top set at school, and was rarely taught the basics, just throw straight in to the difficult things.

    EDIT:
    OK now I understand a - * - = +

    That still doesn't answer why "y = 2+1 is 3, not -3.
    Last edited by sxsniper; January 19th 2011 at 03:58 AM.
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  4. #4
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    Quote Originally Posted by sxsniper View Post
    Nice, of course it is! Thanks.

    The answers should come out as x = 2 and y = -3
    NO, x= 2 and y= -3 can't be the correct answer: it does not satisfy the second equation y= x+ 1.

    I suspect you have misunderstood what is given as the two correct answers. From the quadratic equation you got, x= 2 or x= -3. Then from y= x+ 1, y= 2+1= 3 or y= -3+ 1= -2. The two answers to the problem are "x= 2 and y= 3" and "x= -3 and y= -2".

    However if I put this in to the second equation y=x+1

    Then y = 2+1 which is 3, not -3.
    What's the deal with that?
    x= 2, y= -3 is NOT a correct answer!

    [quote]I have a serious issue with +s and -s.

    Are the answers I was given wrong?

    If I put it in the first:
    x^2 + y^2 = 13 so;
    • 2^2 + -3^2 = 13
    • 4 + -9 = -3 ...which isn't 13!

    You have written the second equation incorrectly- all of y is squared:
    2^2+ (-3)^2= (2)(2)+ (-3)(-3)= 4+ 9= 13

    Oh, by the say, 4- 9 is not -3, either, it is -5.

    I know +9 makes 13, but WHY does -3^2 = 9?! as -3^3 = -27
    I think I'm missing some serious basic maths.
    Again, it is NOT "-3^2"- the negative is as much part of y as the "3" is: (-3)^2= (-3)(-3)= +9.

    If I am, any suggestions on things I should practice or learn?

    I was stuck straight in the top set at school, and was rarely taught the basics, just throw straight in to the difficult things.

    EDIT:
    OK now I understand a - * - = +

    That still doesn't answer why "y = 2+1 is 3, not -3.
    I hope you don't mean exactly what you are writing here. If you have 2 banana cream pies on the table and put one more banana cream pie on the table, you have three banana cream pies, not negative three!
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  5. #5
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    That's my point, on the answers page, it says -3
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  6. #6
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    Thumbs up

    Ahhh, you know I've been trying to solve this for about 8 hours flat now.

    Bloody answers page must be wrong, I got 3 plenty of times! And was scratching my eye balls out trying to figure out WHY oh WHY am I not getting -3.

    Thank yooou.
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  7. #7
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    Want some practice? Try:

    a^2 + b^2 + c^2 = 142
    b = a + 1
    c = 2a - 1
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