# System of Equations

• Nov 9th 2010, 07:15 PM
dbakeg00
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?
• Nov 9th 2010, 07:22 PM
Prove It
Why don't you just subtract the first equation from the second to eliminate $\displaystyle \displaystyle d$?
• Nov 9th 2010, 07:24 PM
pickslides
Quote:

Originally Posted by dbakeg00
$\displaystyle \frac{-5}{4}=\frac{29}{4}+c+d$ ...(1)

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

I get,

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

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

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

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

= $\displaystyle -6.5=c$
• Nov 9th 2010, 07:33 PM
dbakeg00
I should have seen that!! Thanks

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

Appreciate the help!