Practice Solving Literal Equations

Formulas may contain multiple variables, along with known numbers and letters that stand for known constants like ‍$$\pi$$.

We can highlight a certain variable in the formula by treating the formula as an equation where we want to solve for that variable.

1. In this case, we need to solve the equation $$F=\frac{9}{5}C+32$$ for C.

   $$F=\frac{9}{5}C+32$$

   $$F-32=\frac{9}{5}C$$

   $$\frac{5}{9} \cdot (F-32)$$
   

   This is the result of rearranging the formula to highlight the measure in degrees Celsius:

   $$C=\frac{5}{9} \cdot (F-32)$$

2. In this case, we need to solve the equation $$U=\frac{W}{O}$$ for O.

   $$U=\frac{W}{O}$$

   $$U \cdot =W$$

   $$O=\frac{W}{U}$$

   This is the result of rearranging the formula to highlight the hourly output per worker:
   

   $$O=\frac{W}{U}$$

3. In this case, we need to solve the equation

   $$A=\frac{1}{2}(b+c)h$$ for h.

   $$A=\frac{1}{2}(b+c)h$$

   $$\frac{2A}{b+c}=h$$

   This is the result of rearranging the formula to highlight the height:

   $$h=\frac{2A}{b+c}$$

4. In this case, we need to solve the equation $$S = \frac{D}{T}$$ for T.

   $$S = \frac{D}{T}$$

   $$S \cdot T = D$$

   $$T = \frac{D}{S}$$

   This is the result of rearranging the formula to highlight time:

   $$T = \frac{D}{S}$$