Example  2
 

 Multimedia



Susan and Jack have a painting business. They do not always work in the same location. When working alone, Susan takes 3 hours to paint a room, while Jack takes 2 hours to paint a room of the same size. How long would it take to paint the room if they worked together at their usual work rates?

  1. Write an equation that represents this problem.

    Step 1: Identify the variable(s) you are using.

    Let \(t\) represent the time Susan and Jack take to paint one room together.

    Step 2    : Set up a formula to solve the problem.

    \[\begin{array}{l}
     \frac{1}{{{\rm{time \thinspace taken\thinspace by\thinspace Susan}}}} + \frac{1}{{{\rm{time \thinspace taken\thinspace by \thinspace Jack }}}} = \frac{1}{{{\rm{time \thinspace Susan \thinspace and \thinspace Jack\thinspace take \thinspace together}}}} \\
     \frac{1}{3} + \frac{1}{2} = \frac{1}{t},\thinspace t \ne 0 \\
     \end{array}\]
  2. Solve and verify your solution.

    LCD: \(6t\)

    \[\begin{align}
     \left( {\frac{1}{3}} \right)\left( {6t} \right) + \left( {\frac{1}{2}} \right)\left( {6t} \right) &= \left( {\frac{1}{t}} \right)\left( {6t} \right) \\
     2t + 3t &= 6 \\
     5t &= 6 \\
     t &= \frac{6}{5} \\
     \end{align}\]

    Does this solution value make sense?

    Yes. The time of \(\frac{6}{5}h\) is positive, it is not equal to the non-permissible value of \(t = 0\), and \(\frac{6}{5}h\) is less than the time either individual would take to do the project alone.

    Verify for \(t = \frac{6}{5}\).

    Left Side Right Side
    \[\begin{array}{r}
     \frac{1}{3} + \frac{1}{2} \\
     \frac{2}{6} + \frac{3}{6} \\
     \frac{5}{6} \\
     \end{array}\]
    		\[\begin{array}{l}
     \frac{1}{t} \\
     \frac{1}{{\left( {\frac{6}{5}} \right)}} \\
     \frac{5}{6} \\
     \end{array}\]
    LS = RS

    The time it takes both Susan and Jack to paint one room is \(\frac{6}{5}\) hours.