# algebraic equation

• September 3rd 2012, 12:16 PM
willowthewisp
algebraic equation
I have an equation I need to solve. I have an answer but I don't have much confidence in it.

here is the problem:

"20 kg/m^3 = 1027 kg/m^3 + [.15 kg/m^3 (5 degrees C) + .78 kg/m^3(s-35)]. solve for s"

the answer I got is 94.16. please tell me if I am off and also please advise on how you got there. Many thanks.
• September 3rd 2012, 12:37 PM
skeeter
Re: algebraic equation
Quote:

Originally Posted by willowthewisp
I have an equation I need to solve. I have an answer but I don't have much confidence in it.

here is the problem:

"20 kg/m^3 = 1027 kg/m^3 + [.15 kg/m^3 (5 degrees C) + .78 kg/m^3(s-35)]. solve for s"

the answer I got is 94.16. please tell me if I am off and also please advise on how you got there. Many thanks.

I'd say you were way off since I substituted 94.16 for s in the right side of your equation and got nowhere close to the 20 on the left.

I also question your use of units ... what happens to the units of Celsius temp? Maybe you should post the original problem in its entirety.
• September 3rd 2012, 12:49 PM
willowthewisp
Re: algebraic equation
the original equation is - "density = 1027 kg/m^3 + [a (temp - 10 degrees C) + b (S-35)]" solve for S

the constant (a) is .15 kg/m^3 and the constant (b) is .78 kg/m^3. based on a previous problem, I know density is 20 and the temperature is 15 degree C.

I hope that clarifies things a bit.
• September 3rd 2012, 01:00 PM
HallsofIvy
Re: algebraic equation
No, that's not possible. If a has units of kg/m^3, a(temp- 10 degrees C) would have units of "kg(degrees C)/m^3" and cannot be added to "1027 kg/m^3".
• September 3rd 2012, 01:09 PM
skeeter
Re: algebraic equation
Quote:

Originally Posted by willowthewisp
the original equation is - "density = 1027 kg/m^3 + [a (temp - 10 degrees C) + b (S-35)]" solve for S

the constant (a) is .15 kg/m^3 and the constant (b) is .78 kg/m^3. based on a previous problem, I know density is 20 and the temperature is 15 degree C.

I hope that clarifies things a bit.

Then $s \approx -1257$ , for whatever units that variable represents ... a physical quantity/parameter ?

I still take issue with the units ... how do units of density, $\frac{kg}{m^3} \cdot (^\circ C)$ become $\frac{kg}{m^3}$ ?
• September 3rd 2012, 01:38 PM
willowthewisp
Re: algebraic equation
thanks. S is for water salinity. I'm not sure about the units either.