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Math Help - [SOLVED] HW3-QST1B- Albegra equation & Validation -> Please check if right?

  1. #1
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    Question [SOLVED] HW3-QST1B- Albegra equation & Validation -> Please check if right?

    Here is one more that is not right I think but I can't figure anything else?

    \frac {x+3}{2x-7}-\frac {2x-1}{x-3}=0

    I have to solve this and show validation.

    Here is what I did (or tried to do):

    \frac {(x+3)}{(2x-7)}=\frac {(x+3)}{(2x+1)}

    (x+3)(2x+1)=(2x-7)(x+3)

    2x^2+x+6x+3=2x^2+6x-7x-21

    2x^2+x+6x-2x^2-6x+7x=-21-3

    8x=-24

    x=-3

    Validation:

    \frac {-3+3}{2(-3)-7} = \frac {-3+3}{2(-3)+1}

    \frac {0}{-13}=\frac {0}{-5}

    0=0
    Last edited by Neenoon; August 24th 2008 at 07:39 AM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Neenoon View Post
    Here is one more that is not right I think but I can't figure anything else?

    \frac {x+3}{2x-7}-\frac {2x-1}{x-3}=0

    I have to solve this and show validation.

    Here is what I did (or tried to do):

    \color{red}\frac {(x+3)}{(2x-7)}=\frac {(x+3)}{(2x+1)} This is where you went wrong.

    (x+3)(2x+1)=(2x-7)(x+3)

    2x^2+x+6x+3=2x^2+6x-7x-21

    2x^2+x+6x-2x^2-6x+7x=-21-3

    8x=-24

    x=-3

    Validation:

    \frac {-3+3}{2(-3)-7} = \frac {-3+3}{2(-3)+1}

    \frac {0}{-13}=\frac {0}{-5}

    0=0
    Your equation should have been \frac {x+3}{2x-7}=\frac {2x-1}{x-3}.

    We then see that (x+3)(x-3)=(2x-1)(2x-7)

    Try to solve the equation from here. Let us know what you get for x.

    --Chris
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Your equation should have been \frac {x+3}{2x-7}=\frac {2x-1}{x-3}.

    We then see that (x+3)(x-3)=(2x-1)(2x-7)

    Try to solve the equation from here. Let us know what you get for x.

    --Chris
    That's what I tried to do first, but I can't solve it like that... let's see...

    (x+3)(x-3)=(2x-1)(2x-7)

    x^2-3x+3x-9=4x^2-14x-2x+7


    x^2-4x^2+16x=9+7

    -3x^2+16x=16

    And then I'm stuck with that -3x^2 ???

    What am I not getting here?
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  4. #4
    Senior Member Spec's Avatar
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    Factor out 3 and the equation becomes:

    x^2 - \frac{16x}{3} + \frac{16}{3} = 0
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  5. #5
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    I'm sorry but I'm still not getting how am I suppose to get the value of x from that?

    Quote Originally Posted by Spec View Post
    Factor out 3 and the equation becomes:

    x^2 - \frac{16x}{3} + \frac{16}{3} = 0
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  6. #6
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    Haven't you learned how to solve quadratic equations yet? It's not that hard really.

    Quadratic equation - Wikipedia, the free encyclopedia
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  7. #7
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    Here is what I did, tell me if this is right?

    (x+3)(x-3)=(2x-1)(2x-7)

    x^2-3x+3x-9=4x^2-2x-14x+7

    x^2-4x^2-3x+3x+2x+14x-9-7=0

    -3x^2+16x-16=0

    So, a=-3 b=16 c=-16
    <br />
x=-b \pm \frac{\sqrt{b^2-4ac}}{2a}
    <br />
x=-16 \pm \frac{\sqrt{16^2-(4 \bullet -3 \bullet 16)}}{2(-3)}

    <br />
x=-16 \pm \frac{\sqrt{64}}{-6)}

    <br />
 x=\frac{-16+8}{-6}

    and
    <br />
 x=\frac{-16-8}{-6}

    x=\frac{4}{3}\ and\ x=4
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  8. #8
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    Brave! You're correct.

    But you should write the -b in the numerator not outside.
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  9. #9
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    Thank you, will do!
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