[tex]\begin{gathered} \text{Given} \\ T=C(8+AB) \end{gathered}[/tex][tex]\begin{gathered} \text{Divide both sides by }C \\ T=C(8+AB) \\ \frac{T}{C}=\frac{C(8+AB)}{C} \\ \frac{T}{C}=\frac{\cancel{C}(8+AB)}{\cancel{C}} \\ \frac{T}{C}=8+AB \\ \\ \text{Subtract both sides by }8 \\ \frac{T}{C}-8=8-8+AB \\ \frac{T}{C}-8=\cancel{8-8}+AB \\ \frac{T}{C}-8=AB \\ \\ \text{Divide both sides by }B \\ \frac{\frac{T}{C}-8}{B}=\frac{AB}{B} \\ \frac{\frac{T}{C}}{B}-\frac{8}{B}=\frac{A\cancel{B}}{\cancel{B}} \\ \frac{T}{CB}-\frac{8}{B}=A \\ \\ \text{By symmetric property of equality, we can swap left and right side of equations} \\ \frac{T}{CB}-\frac{8}{B}=A \\ A=\frac{T}{CB}-\frac{8}{B} \end{gathered}[/tex]
