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Thread: [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?

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

    I have to solve this and show validation.

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

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

    $\displaystyle (x+3)(2x+1)=(2x-7)(x+3)$

    $\displaystyle 2x^2+x+6x+3=2x^2+6x-7x-21$

    $\displaystyle 2x^2+x+6x-2x^2-6x+7x=-21-3$

    $\displaystyle 8x=-24$

    $\displaystyle x=-3$

    Validation:

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

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

    $\displaystyle 0=0$
    Last edited by Neenoon; Aug 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?

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

    I have to solve this and show validation.

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

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

    $\displaystyle (x+3)(2x+1)=(2x-7)(x+3)$

    $\displaystyle 2x^2+x+6x+3=2x^2+6x-7x-21$

    $\displaystyle 2x^2+x+6x-2x^2-6x+7x=-21-3$

    $\displaystyle 8x=-24$

    $\displaystyle x=-3$

    Validation:

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

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

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

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

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

    --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 $\displaystyle \frac {x+3}{2x-7}=\frac {2x-1}{x-3}$.

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

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

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

    $\displaystyle (x+3)(x-3)=(2x-1)(2x-7)$

    $\displaystyle x^2-3x+3x-9=4x^2-14x-2x+7$


    $\displaystyle x^2-4x^2+16x=9+7$

    $\displaystyle -3x^2+16x=16$

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

    What am I not getting here?
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  4. #4
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    Factor out 3 and the equation becomes:

    $\displaystyle x^2 - \frac{16x}{3} + \frac{16}{3} = 0$
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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:

    $\displaystyle x^2 - \frac{16x}{3} + \frac{16}{3} = 0$
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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?

    $\displaystyle (x+3)(x-3)=(2x-1)(2x-7)$

    $\displaystyle x^2-3x+3x-9=4x^2-2x-14x+7$

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

    $\displaystyle -3x^2+16x-16=0$

    So, a=-3 b=16 c=-16
    $\displaystyle
    x=-b \pm \frac{\sqrt{b^2-4ac}}{2a}$
    $\displaystyle
    x=-16 \pm \frac{\sqrt{16^2-(4 \bullet -3 \bullet 16)}}{2(-3)}$

    $\displaystyle
    x=-16 \pm \frac{\sqrt{64}}{-6)}$

    $\displaystyle
    x=\frac{-16+8}{-6}$

    and
    $\displaystyle
    x=\frac{-16-8}{-6}$

    $\displaystyle x=\frac{4}{3}\ and\ x=4$
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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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