Algebraic Rearrangement

• Aug 9th 2011, 08:15 AM
maloneb
Algebraic Rearrangement
Hi, I'm new to this site and I hope you guys can help. I'm not actually pre university but an engineer, yet I'm having some problems rearranging an equation and was hoping someone could help. The equation is as follow:

((A/B)-1)*(((A/C)^2)-1)=2*(A/C)*((D*E)/F)

and the equation needs rearranging to find A. Can anyone help?

Thanks
• Aug 9th 2011, 08:21 AM
Siron
Re: Algebraic Rearrangement
• Aug 9th 2011, 10:21 PM
rdtedm
Re: Algebraic Rearrangement
Quote:

Originally Posted by maloneb
Hi, I'm new to this site and I hope you guys can help. I'm not actually pre university but an engineer, yet I'm having some problems rearranging an equation and was hoping someone could help. The equation is as follow:

((A/B)-1)*(((A/C)^2)-1)=2*(A/C)*((D*E)/F)

and the equation needs rearranging to find A. Can anyone help?

Thanks

I start first by distrubuting the exponents to get:

Next, combine like bases and simplify:

Multiply both sides by A^3 to get:

BC^2 = (2A^4DE)/(CF)

Multiply both sides by (CF) to get:

BC^3F = 2A^4DE

Divide both sides by 2DE to get:

(BC^3F)/(2DE) = A^4

Take the fourth root of each side to get:

A = fourth_root[(BC^3F)/(2DE)]
• Aug 9th 2011, 10:22 PM
rdtedm
Re: Algebraic Rearrangement
Quote:

Originally Posted by rdtedm
I start first by distrubuting the exponents to get:

Next, combine like bases and simplify:

Multiply both sides by A^3 to get:

BC^2 = (2A^4DE)/(CF)

Multiply both sides by (CF) to get:

BC^3F = 2A^4DE

Divide both sides by 2DE to get:

(BC^3F)/(2DE) = A^4

Take the fourth root of each side to get:

A = fourth_root[(BC^3F)/(2DE)]

I assumed here, that "-1" meant the inverse.. if its actually minus one, then obviously this is incorrect.
• Aug 9th 2011, 10:49 PM
maloneb
Re: Algebraic Rearrangement
Quote:

Originally Posted by rdtedm
I assumed here, that "-1" meant the inverse.. if its actually minus one, then obviously this is incorrect.

No, -1 is not inverse, it is actually -1
• Aug 10th 2011, 08:12 AM
SammyS
Re: Algebraic Rearrangement
It's a cubic equation in A. Have fun !!!