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Math Help - solve for y help

  1. #1
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    solve for y help

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  2. #2
    Super Member dhiab's Avatar
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    Quote Originally Posted by realistic View Post
    Hello: \begin{array}{l}<br />
\frac{{ - y}}{{ - a}} = \frac{a}{x} \\ <br />
- y = \frac{{ - a^2 }}{x} \\ <br />
\end{array}<br /> <br />
    Solve for y : y = \frac{{a^2 }}{x}

    Solve for y :  x = \frac{{a^2 }}{y}<br />
    THANK YOU
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  3. #3
    Senior Member pacman's Avatar
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    -a/-y = x/a; solve for y?

    cross-multiply,

    -(a)^2 = -xy, cancel the negative sign

    y = a^2/x

    or

    x = a^2/y

    is this what you want?
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  4. #4
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    Quote Originally Posted by pacman View Post
    -a/-y = x/a; solve for y?

    cross-multiply,

    -(a)^2 = -xy, cancel the negative sign

    y = a^2/x

    or

    x = a^2/y

    is this what you want?

    can you explain how you cancelled the negative after cross multi? not sure how this happens
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  5. #5
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    Quote Originally Posted by dhiab View Post
    Hello: \begin{array}{l}<br />
\frac{{ - y}}{{ - a}} = \frac{a}{x} \\ <br />
- y = \frac{{ - a^2 }}{x} \\ <br />
\end{array}<br /> <br />
    Solve for y : y = \frac{{a^2 }}{x}

    Solve for y :  x = \frac{{a^2 }}{y}<br />
    THANK YOU

    ya thanks... can you help me with the second one plz
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  6. #6
    Super Member dhiab's Avatar
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    Quote Originally Posted by realistic View Post
    ya thanks... can you help me with the second one plz
    HELLO :
    \begin{array}{l}<br />
\frac{{ - y}}{{ - a}} = \frac{a}{x} \\ <br />
\left( { - y} \right) \times \left( x \right) = \left( { - a} \right) \times \left( a \right) \\ <br />
\left( { - y} \right) \times \left( x \right) = - a^2 \\ <br />
y \times x = a^2 \\ <br />
\end{array}
    divide by x :


    \frac{{y \times x}}{x} = \frac{{a^2 }}{x}<br />
    conclusion y = \frac{{a^2 }}{x}<br />
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  7. #7
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    Quote Originally Posted by dhiab View Post
    HELLO :
    \begin{array}{l}<br />
\frac{{ - y}}{{ - a}} = \frac{a}{x} \\ <br />
\left( { - y} \right) \times \left( x \right) = \left( { - a} \right) \times \left( a \right) \\ <br />
\left( { - y} \right) \times \left( x \right) = - a^2 \\ <br />
y \times x = a^2 \\ <br />
\end{array}
    divide by x :


    \frac{{y \times x}}{x} = \frac{{a^2 }}{x}<br />
    conclusion y = \frac{{a^2 }}{x}<br />
    could you pleae help me with the second question plz
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  8. #8
    Senior Member furor celtica's Avatar
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    a negative times a negative equals a positive, man did you ever see stand and deliver or wot?
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  9. #9
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    Quote Originally Posted by furor celtica View Post
    a negative times a negative equals a positive, man did you ever see stand and deliver or wot?
    i know that i just cant see this step

    [tex]

    <br />
\begin{array}{l}<br />
\left( { - y} \right) \times \left( x \right) = - a^2 \\<br />
y \times x = a^2 \\<br />
\end{array}<br />

    are you just x it self by a negative?
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  10. #10
    Senior Member furor celtica's Avatar
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    think of it this way
    (-y) x (X) = -a^2
    could be rewritten
    -1 x y x X = -1 x a x a
    -1 is a factor of both sides
    so cut it out
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  11. #11
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    Quote Originally Posted by furor celtica View Post
    think of it this way
    (-y) x (X) = -a^2
    could be rewritten
    -1 x y x X = -1 x a x a
    -1 is a factor of both sides
    so cut it out

    cool i see now clearly

    also could you help out on the second question in the title plz
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  12. #12
    Senior Member furor celtica's Avatar
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    sorry kid i cant understand your writing (is that paint or what?)
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  13. #13
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    Quote Originally Posted by furor celtica View Post
    sorry kid i cant understand your writing (is that paint or what?)

    yah paint

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  14. #14
    Super Member dhiab's Avatar
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    Hello : question 2
     <br /> <br />
\begin{array}{l}<br />
x = k - \frac{{\sqrt {pq} }}{y} \\ <br />
\frac{{\sqrt {pq} }}{y} = k - x \\ <br />
y = \frac{{\sqrt {pq} }}{{k - x}} \\ <br />
\end{array}<br /> <br />
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  15. #15
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    Quote Originally Posted by dhiab View Post
    Hello : question 2
     <br /> <br />
\begin{array}{l}<br /> <br />
\frac{{\sqrt {pq} }}{y} = k - x \\ <br />
y = \frac{{\sqrt {pq} }}{{k - x}} \\ <br />
\end{array}<br /> <br />
    can you show the step you did in the middle of that? plz
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