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Math Help - Solving equation

  1. #1
    Senior Member Mukilab's Avatar
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    Solving equation

    Step by step please

    Solve
    \frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}}=2

    I got to

    4=(y-1)^{2} or 4=y^{2}-2y+1
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    Step by step please

    Solve
    \frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}}=2

    I got to

    4=(y-1)^{2} or 4=y^{2}-2y+1
    Multiply throughout by 2(y-1)^2 to clear the denominator

    2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2

    Simplify

    6 + 3 = 4(y-1)^2

    You could expand and use the quadratic formula but completing the square is more appropriate here IMO

    Divide by 4

    (y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2

    Take the square root of both sides and don't forget the \pm sign
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  3. #3
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Multiply throughout by 2(y-1)^2 to clear the denominator

    2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2

    Simplify

    6 + 3 = 4(y-1)^2

    You could expand and use the quadratic formula but completing the square is more appropriate here IMO

    Divide by 4

    (y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2

    Take the square root of both sides and don't forget the \pm sign

    I simplified this
    2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2

    to

    2(y-1)^2\cdot \frac{3}{(y-1)^2}+3=2


    o__O
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Mukilab View Post
    I simplified this
    2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2

    to

    2(y-1)^2\cdot \frac{3}{(y-1)^2}+3=2


    o__O
    I simplified that bit for you, the answer is 6+3 = 4(y-1)^2

    Multiplying by 2(y-1)^2 is to clear the denominators
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  5. #5
    Super Member bigwave's Avatar
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    another way

    x every term by 2 you have

     <br />
\frac{9}{(y-1)^2} = 4<br />
\Rightarrow<br />
4(y-1)^2 = 9 <br />
\Rightarrow<br />
(y-1)^2 = \frac{9}{4}<br />
    take the square of both sides
     <br />
y-1 = \frac{3}{2}\Rightarrow<br />
y = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}<br />
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  6. #6
    MHF Contributor
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    Quote Originally Posted by Mukilab View Post
    I simplified this
    2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2

    to

    2(y-1)^2\cdot \frac{3}{(y-1)^2}+3=2


    o__O
    Do you see the error you made here, Mukilab?

    You could have cancelled terms on the left, you did cancel terms
    in the middle but you dropped a term on the right.

    Did you understand what e^{i*pi} meant when he mentioned + and - in relation to the square roots?
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  7. #7
    Senior Member Mukilab's Avatar
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    /facepalm
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  8. #8
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    There's no problem,
    no-one is judging,
    but you learn from seeing your errors and correcting them.

    Now, there are 2 answers,

    since (3)(3)=9 and (-3)(-3)=9 for example.

    Of course, they are the same thing really.

    3(-3) means subtract 3 three times,
    while -3(-3) means the opposite of that which is add 3 three times.

    So, square roots have 2 answers.
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  9. #9
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    e^(i*pi)'s Avatar
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    In this case

    <br />
(y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2<br />

    (y-1) = \pm \frac{3}{2}
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