# Re-arranging and solving for q

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• May 9th 2008, 07:25 AM
jdizzle1985
Re-arranging and solving for q
35q^(-3/5)=-21q^(-8/5)-((-21q^8/5)*35q^(-3/5))

Re-arrange and solve for q. I did it but got some mad answer.
• May 9th 2008, 09:19 AM
masters
How 'bout showing us how far you got!
• May 9th 2008, 10:50 AM
Moo
Hello,

Quote:

Originally Posted by jdizzle1985
35q^(-3/5)=-21q^(-8/5)-((-21q^8/5)*35q^(-3/5))

Re-arrange and solve for q. I did it but got some mad answer.

Let's put it all in one side :

$35q^{-3/5}+21q^{-8/5}+({\color{red}(-21q^{8/5})*35q^{-3/5}})=0$

Look at the red thing

$-21q^{8/5}*35q^{-3/5}=-(21*35)q^{8/5-3/5}=-21*35 q^1=-21 \cdot 35 \cdot q$

Hence the equation is now :

$35q^{-3/5}+21q^{-8/5}-21 \cdot 35 \cdot q=0$

Multiplying everything by $q^{8/5}$, we get :

$35q^1+21q^0-21\cdot 35 \cdot q^{13/5}=0$

$35q+21-21\cdot 35 \cdot q^{13/5}=0$

I got mad answers too (Nerd)
• May 9th 2008, 11:38 AM
blair_alane
Quote:

Originally Posted by Moo
$35q^{-3/5}+21q^{-8/5}+({\color{red}(-21q^{8/5})*35q^{-3/5}})=0$

Look at the red thing

"Look at the red thing?" hahahaha! That was my favorite part! (Clapping)