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Math Help - quadratical equation

  1. #1
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    quadratical equation

    hello this is my first post on this forum and i hope i am in the right category,
    i have a two equations i just cant solve, so please,please,please help me.

    (2x+3)/(x-1)=2+(5/(x-1))


    x-3x+1-(9/4)-3*(3/2)+1=(x-(3/2))
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  2. #2
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    Quote Originally Posted by beethoven View Post
    hello this is my first post on this forum and i hope i am in the right category,
    i have a two equations i just cant solve, so please,please,please help me.

    (2x+3)/(x-1)=2+(5/(x-1))


    x-3x+1-(9/4)-3*(3/2)+1=(x-(3/2))

    HI

    \frac{2x+3}{x-1}=\frac{2(x-1)+5}{x-1}

    Multiply both sides by x-1

    2x+3=2x-2+5

    Continue here .

    (2) x^2-3x+1-\frac{9}{4}-3(\frac{3}{2})+1=(x-\frac{3}{2})^2

    x^2-3x+1-\frac{9}{4}-\frac{9}{2}+1=x^2-3x+\frac{9}{4}

    Multiply both sides by 4

    4x^2-12x+4-9-18+4=4x^2-12x+9

    -12x-23=-12x

    Are you sure this is the correct equation , or else any value of real x would satisfy .
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  3. #3
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    yes the equation is false,herse the rigth one:

    x^2-3x+1-(-3/2)=(x-(3/2))^2
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  4. #4
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    Quote Originally Posted by beethoven View Post
    hello this is my first post on this forum and i hope i am in the right category,
    i have a two equations i just cant solve, so please,please,please help me.

    (2x+3)/(x-1)=2+(5/(x-1))


    ...
    \dfrac{2x+3}{x-1}=2+\dfrac5{x-1}
    Multiply both sides by (x - 1):

    2x+3 = 2(x-1) + 5~\implies~ 0=0

    That means: This equation is true for all x\in \mathbb{R} \setminus \{1\}
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  5. #5
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    Quote Originally Posted by beethoven View Post
    yes the equation is false,herse the rigth one:

    x^2-3x+1-(-3/2)=(x-(3/2))^2
    x^2-3x+\frac{5}{2} = x^2-3x+\frac{9}{4}

    Multiply by 4

    (4x^2-12x+10)-(4x^2-12x+9) = 0

    10 - 9 \neq 0 which means there are no solutions

    You sure that's correct?
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