Answer
$A=\dfrac{LH}{k(T_1-T_2)}$
Work Step by Step
By multiplying both sides by the $LCD=
L
$, the given equation, $
H=\dfrac{kA(T_1-T_2)}{L}
,$ is equivalent to
\begin{array}{l}\require{cancel}
LH=kA(T_1-T_2)
.\end{array}
Using the properties of equality, then, in terms of $
A
,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{LH}{k(T_1-T_2)}=A
\\\\
A=\dfrac{LH}{k(T_1-T_2)}
.\end{array}
