Question

two mechanics, bonita rich and pamela pearson, take 9 hours to rebuild an engine when they work together. if each worked alone, bonita, the more experienced mechanic, could complete the job 1 hour faster than pamela. how long would it take each of them to rebuild the engine working alone?
the time it takes bonita is about □ hours.
(round to two decimal places as needed.)
the time it takes pamela is about □ hours.
(round to two decimal places as needed.)

Explanation:

Step1: Set up variables

Let \( t\) be the time it takes Pamela to rebuild the engine alone. Then the time it takes Bonita is \( t - 1\) hours. The rate of work is the reciprocal of the time. The combined rate of work is \(\frac{1}{9}\) (since they take 9 hours together). So, \(\frac{1}{t}+\frac{1}{t - 1}=\frac{1}{9}\).

Step2: Simplify the equation

Multiply through by \(9t(t - 1)\) to clear the fractions: \(9(t - 1)+9t=t(t - 1)\). Expand: \(9t-9 + 9t=t^{2}-t\). Rearrange to get a quadratic equation: \(t^{2}-19t + 9=0\).

Step3: Solve the quadratic equation

Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=-19\), \(c = 9\). \(t=\frac{19\pm\sqrt{361-36}}{2}=\frac{19\pm\sqrt{325}}{2}=\frac{19\pm5\sqrt{13}}{2}\). Calculate \(\sqrt{13}\approx3.606\). \(t=\frac{19\pm18.03}{2}\). We get two solutions: \(t_1=\frac{19 + 18.03}{2}\approx18.52\) and \(t_2=\frac{19-18.03}{2}\approx0.49\). But \(t>1\) (since Bonita's time \(t - 1>0\)), so \(t\approx18.52\) (Pamela's time). Bonita's time is \(t-1\approx17.52\).

Answer:

Bonita: \(17.52\) hours, Pamela: \(18.52\) hours