Solving for $T$ in this Algebra problem I'm taking Physics E&M and I'm trying to solve the following equation for $T$. I feel like the algebra I'm doing is correct but I cannot get the solution in the book which is $53.1 $C > $K_{br}$ = brass = $109.0 \frac{W}{mK}$ > > $K_{cu}$ = copper = $385.0 \frac{W}{mK}$ > > $A=.000500m^2$ > > $L_{br}=.200m$ > > $L_{cu}=.800m$ > > $T_H=100$C > > $T_C=0$C $K_{br}A\frac{T_H-T}{L_{br}} = K_{cu}A\frac{T-T_c}{L_{cu}}$ $L_{cu}K_{br}A\frac{T_H-T}{L_{br}}=K_{cu}AT-T_c$ $L_{cu}K_{br}AT_{H}-T=L_{br}K_{cu}AT-T_c$ $L_{cu}K_{br}AT_H=L_{br}K_{cu}AT-T_c+T$ $L_{cu}K_{br}AT_H+T_c=L_{br}K_{cu}AT+T$ $L_{cu}K_{br}AT_H+T_c=T(L_{br}K{cu}A+1)$ $\frac{L_{cu}K_{br}AT_H+T_c}{L_{br}K_{cu}A+1} = T$ The correct answer is **53.1** C but I get 31.480 Clearly there is a problem with my Algebra somewhere. Can anyone see it?

Assuming your first equation in the white part (after the blue one) is correct:

$$\begin{align}&K_{br}A\frac{T_H-T}{L_{br}}=K_{cu}A\frac{T-T_c}{L_{cu}}\\\\{}\\\ &K_{br}L_{cu}(T_H-T)=K_{cu}L_{br}(T-\overbrace{T_c}^{=0})\\\\{}\\\ &(K_{cu}L_{br}+K_{br}L_{cu})T=K_{br}L_{cu}T_H\\\\{}\\\ &T=\frac{K_{br}L_{cu}T_H}{K_{cu}L_{br}+K_{br}L_{cu}}\end{align}$$

Now input the corresponding values and check we get what you say we should.