System of Equations

**dbakeg00** · November 9th 2010, 07:15 PM

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.

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

$404=432+4c+d$

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

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

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

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

$404=432+4c+d$

$-28=4c+d$

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

Here is where I start getting myself all messed up, any hints on what step to take next?

**Prove It** · November 9th 2010, 07:22 PM

Why don't you just subtract the first equation from the second to eliminate $\displaystyle d$ ?

**pickslides** · November 9th 2010, 07:24 PM

Originally Posted by dbakeg00

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

$404=432+4c+d$ ...(2)

I get,

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

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

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

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

= $-6.5=c$

**dbakeg00** · November 9th 2010, 07:33 PM

I should have seen that!! Thanks

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

Appreciate the help!

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