Chapter 6 - Section 6.6 - Rational Equations and Problem Solving - Exercise Set - Page 387: 12

$T_2=\dfrac{kAT_1-LH}{kA}$

Using the properties of equality, in terms of $ T_2 ,$ the given equation, $ H=\dfrac{kA(T_1-T_2)}{L} ,$ is equivalent to \begin{array}{l} LH=kA(T_1-T_2) \\\\ LH=kAT_1-kAT_2 \\\\ kAT_2=kAT_1-LH \\\\ T_2=\dfrac{kAT_1-LH}{kA} .\end{array}