Practice Solving Literal Equations

1. The following formula gives the temperature's measure in degrees Fahrenheit F, where ‍C is the measure in degrees Celsius:

   \(F=\frac{9}{5}C+32\)

   Rearrange the formula to highlight the measure in degrees Celsius.

2. The following formula is used in economics to find a factory's unit labor cost ‍U, where ‍O is the hourly output per worker and ‍W is the hourly compensation per worker.

   \(U = \frac{W}{O}\)

   Rearrange the formula to highlight the hourly output per worker.

3. The following formula gives the area ‍A of a trapezoid, where ‍b and ‍c are the bases of the trapezoid and ‍h is the height:

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

   Rearrange the formula to highlight the height.

4. The following formula gives an object's average speed ‍S, where ‍T is the time it takes the object to travel a distance ‍D.

   \(S=\frac{D}{T}\)

   Rearrange the formula to highlight time.

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:modeling/x2ec2f6f830c9fb89:manipulating-formulas/e/manipulating-formulas
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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:

   \(T = \frac{D}{S}\)