[itex]\displaystyle \frac{T_3-298}{298T_3}=\frac{T_3}{298T_3}-\frac{298}{298T_3}[/itex]smaan said:Homework Statement
ln(0.0693/0.0475) = (70435/8.314)((T3-298)/(T3*298))
Solve for T3
Homework Equations
The Attempt at a Solution
The fact that there are 2 T3's is really throwing me off, I feel that I am doing something algebraically wrong.
0.377 = 8471.9((T3-298)/(T3*298))
How would I solve for T3?
In order to solve for two unknown variables in a simple algebra problem, you need to have two equations with two variables. You can then use substitution or elimination to solve for one variable, and then plug that value into the other equation to solve for the second variable.
No, in order to solve for two of the same unknown variable, you need to have two separate equations with two different variables. This allows you to have enough information to solve for both variables.
Typically, it is easier to solve for the variable that has a coefficient of 1 or -1. If both variables have coefficients other than 1 or -1, you can choose to solve for the variable that has the smallest coefficient, or the variable that is easier to isolate.
Yes, the same method of using two equations with two variables can be applied to solve for two unknown variables in any algebra problem, as long as the equations are independent and not redundant.
There are no specific shortcuts or tricks to solving for two unknown variables in a simple algebra problem. However, it is important to familiarize yourself with different algebraic techniques, such as substitution and elimination, to make the process more efficient.