Solve T_{1}=9/5T_{c}+32

$$T _ { 1 } = \frac { 9 } { 5 } T _ { c } + 32$$

Solve for T_c

$$T_{c}=\frac{5\left(T_{1}-32\right)}{9}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{9}{5}T_{c}+32=T_{1}$$

Subtract 32 from both sides.

$$\frac{9}{5}T_{c}=T_{1}-32$$

Divide both sides of the equation by \frac{9}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{9}{5}T_{c}}{\frac{9}{5}}=\frac{T_{1}-32}{\frac{9}{5}}$$

Dividing by \frac{9}{5} undoes the multiplication by \frac{9}{5}.

$$T_{c}=\frac{T_{1}-32}{\frac{9}{5}}$$

Divide T_{1}-32 by \frac{9}{5} by multiplying T_{1}-32 by the reciprocal of \frac{9}{5}.

$$T_{c}=\frac{5T_{1}-160}{9}$$

Solve for T_1

$$T_{1}=\frac{9T_{c}}{5}+32$$