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Math Help - simultaneous equations

  1. #1
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    simultaneous equations

    Not sure if I am quite getting this correct?

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

    I am trying to solve the above and compare the results to a graph I plotted with a straight line and a parabola. I am looking at where the graph intersects and at the moment there seems to be slight differneces between my values from the graph and the above.

    2x - 3
    -x^2 + 4x +5

    2x - 2 = x - 1
    2 2

    2(-1) - 3 = -2 -3 = -5

    4(-1) + 5 = -4 + 5 = 1

    (x + 1) (x - 5)

    x = - 1 or x = + 5

    These x values conside with previous work I have done and don't seem wrong, please advise.

    By comparison to the graph(s), the straight line graph has points (0, 1.5) and the parabola has points (0, 5).

    so somewhere I sem to be not quite right?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: simultaneous equations

    If the equations are simultaneous and if y=2x-3 and y=-x^2+4x+5 then:
    2x-3=-x^2+4x+5

    Solve this equation.
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  3. #3
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    Re: simultaneous equations

    Quote Originally Posted by Siron View Post
    If the equations are simultaneous and if y=2x-3 and y=-x^2+4x+5 then:
    2x-3=-x^2+4x+5

    Solve this equation.
    Is this a follow the BIDMAS rule to solve it?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: simultaneous equations

    If the two equations are equal to y then they're equal to each other. Do you understand that?
    Can you solve it? It's just a quadratic equation you've to solve. Put all the terms to one side and use the quadratic formula or factore.
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  5. #5
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    Re: simultaneous equations

    Quote Originally Posted by Siron View Post
    If the two equations are equal to y then they're equal to each other. Do you understand that?
    Can you solve it? It's just a quadratic equation you've to solve. Put all the terms to one side and use the quadratic formula or factore.
    I don't seem to get it, this is what I have tried?

    2x - 3 = -x^2 + 4x + 5
    2x = -x^2 + 4x + 5 + 3
    2x = -x^2 + 4x + 8

    then

    2x + x^2 + 4x + 8
    6x + x^2 + 8 = y

    Using Ax^2 + BX + c = 0

    a = 6, b = 1, c = 8

    using the quadratic formula

    - b + or - square root b^2 - 4ac / 2a

    - 1 + or - square root 1^2 - 4 x 6 x 8 / 2 x 6

    = square root 1 - 192

    square root of 191 = 13.82 which is divided by 12

    = 1.15 + 1 = 2.15 or

    1.15 - 1 = 0.15

    I believe neither of the above roots are correct, the x- values I think should be -1 and 5 as previously?
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  6. #6
    Member anonimnystefy's Avatar
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    Re: simultaneous equations

    hi David Green

    you got the coefficients a,b and c wrong:

    6x + x^2 + 8 = 0

    we can write this as:

    x^2 + 6x + 8 = 0

    the coefficient with x^2 is 1,so a=1

    the coefficient with x is 6,so b=6

    the last coefficient is 8,so c=8.

    can you solve now?
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  7. #7
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    Re: simultaneous equations

    Quote Originally Posted by anonimnystefy View Post
    hi David Green

    you got the coefficients a,b and c wrong:

    6x + x^2 + 8 = 0

    we can write this as:

    x^2 + 6x + 8 = 0

    the coefficient with x^2 is 1,so a=1

    the coefficient with x is 6,so b=6

    the last coefficient is 8,so c=8.

    can you solve now?
    Thanks I will have another go?
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  8. #8
    Member anonimnystefy's Avatar
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    Re: simultaneous equations

    hi

    your welcome.let me know what you got.
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  9. #9
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    Re: simultaneous equations

    No the results are still incorrect?

    Using the formula

    -6 + or - square root 6^2 - 4 x 1 x 8
    2 x 1

    = square root of 4 = 2

    2 / 2 = 1

    1 + 6 = 7

    1 - 6 = - 5

    7 is not correct, somewhere I still am not getting the picture?
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: simultaneous equations

    If 2x-3=-x^2+4x+5
    \Leftrightarrow 2x-3+x^2-4x-5=0
    \Leftrightarrow x^2-2x-8=0

    D=(-2)^2-4(1)(-8)=4+32=36=6^2 so we have two different solutions:
    x_1=\frac{2-6}{2}=-2
    x_2=\frac{2+6}{2}=4

    Check:
    y=2(-2)-3 \Leftrightarrow y=-7 and y=2(4)-3=5
    So two solutions (-2,-7) and (4,5)
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  11. #11
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    Re: simultaneous equations

    Thanks for all your helps much appreciated.

    I am just going to study what you have done and will let you know when I understand it now. My Graph coordinates now look wrong?
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  12. #12
    MHF Contributor Siron's Avatar
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    Re: simultaneous equations

    You're welcome!
    Take a look here for the graph:
    2x-3=-x^2+4x+5 - Wolfram|Alpha
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