Paul's Online Notes
Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Applications of Linear Equations
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.3 : Applications of Linear Equations

6. John can paint a house in 28 hours. John and Dave can paint the house in 17 hours working together. How long would it take Dave to paint the house by himself?

Show All Steps Hide All Steps

Start Solution

So, if we consider painting the house to be a single job we have the following word equation if both John and Dave work together to paint the house.

\[\left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{ done by John}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Portion of job}}\\ {\mbox{done by Dave}}\end{array} \right) = 1{\mbox{ Job}}\]

We know that Portion of Job = Work Rate X Work Time so this gives the following word equation.

\[\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{ of John}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{ of John}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{ of Dave}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{ of Dave}}\end{array} \right) = 1\] Show Step 2

We know that John can paint the house in 28 hours so we can use the following equation to determine the John’s work rate.

\[\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{ of John}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{ of John}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{ of John}}\end{array} \right)\left( {28} \right) = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}{\mbox{Work Rate of John = }}\frac{1}{{28}}\]

Similarly, if we let \(t\) be the amount of time it takes Dave to paint the house by himself we have the following relationship between the time and work rate of Dave.

\[\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{ of Dave}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{ of Dave}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{ of Dave}}\end{array} \right)\left( t \right) = 1 \hspace{0.25in} \Rightarrow \hspace{0.25in}{\mbox{Work Rate of Dave = }}\frac{1}{t}\] Show Step 3

We can now plug in the information from the second step as well as the fact that it takes John and Dave 17 hours to paint the house by themselves into the word equation from the first step to get,

\[\begin{align*}\left( {\frac{1}{{28}}} \right)\left( {17} \right) + \left( {\frac{1}{t}} \right)\left( {17} \right) & = 1\\ \frac{{17}}{{28}} + \frac{{17}}{t} & = 1\end{align*}\] Show Step 4

Now we can solve this for \(t\).

\[\begin{align*}\frac{{17}}{t} & = 1 - \frac{{17}}{{28}}\\ \frac{{17}}{t} & = \frac{{11}}{{28}}\\ 17 & = \frac{{11}}{{28}}t\\ \frac{{476}}{{11}} & = t\hspace{0.25in} \Rightarrow \hspace{0.25in}t = 43.2727\end{align*}\]

So, it will take Dave approximately 43.2727 hours to paint the house by himself.