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Thread: System of Equations

  1. #1
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    System of Equations

    I should really know how to solve this seeing as I will be taking Cal II next semester...so this is kind of embarrassing, but I just keep getting stuck for some reason.

    $\displaystyle \frac{-5}{4}=\frac{29}{4}+c+d$

    $\displaystyle 404=432+4c+d$

    Here is what I have until the point where I get stuck...

    $\displaystyle \frac{-5}{4}*\frac{4}{1}=\frac{4}{1}*(\frac{29}{4}+c+d)$

    $\displaystyle -5=29+4c+4d$

    $\displaystyle \frac{-34}{4d}=4c$

    $\displaystyle 404=432+4c+d$

    $\displaystyle -28=4c+d$

    $\displaystyle -28=\frac{-34}{4d}+d$

    Here is where I start getting myself all messed up, any hints on what step to take next?
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  2. #2
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    Why don't you just subtract the first equation from the second to eliminate $\displaystyle \displaystyle d$?
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  3. #3
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    Quote Originally Posted by dbakeg00 View Post
    $\displaystyle \frac{-5}{4}=\frac{29}{4}+c+d$ ...(1)

    $\displaystyle 404=432+4c+d$ ...(2)
    I get,

    (2)-(1) = $\displaystyle 404-\frac{-5}{4}=432-\frac{29}{4}+3c$

    = $\displaystyle \frac{1616}{4}-\frac{-5}{4}=\frac{1728}{4}-\frac{29}{4}+3c$

    = $\displaystyle \frac{1621}{4}=\frac{1699}{4}+3c$

    = $\displaystyle \frac{-78}{4}=3c$

    = $\displaystyle -6.5=c$
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  4. #4
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    I should have seen that!! Thanks

    I ended up getting $\displaystyle c=\frac{-13}{2}$ and $\displaystyle d=-2$

    Appreciate the help!
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