This is a little bit in advance, but I wanted to let everyone know that my servers will be undergoing some maintenance on May 17 and May 18 during 8:00 AM CST until 2:00 PM CST. Hopefully the only inconvenience will be the occasional “lost/broken” connection that should be fixed by simply reloading the page. Outside of that the maintenance should (fingers crossed) be pretty much “invisible” to everyone.

Paul

May 6, 2021

*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

We now need to discuss the section that most students hate. We need to talk about applications to linear equations. Or, put in other words, we will now start looking at story problems or word problems. Throughout history students have hated these. It is my belief however that the main reason for this is that students really don’t know how to work them. Once you understand how to work them, you’ll probably find that they aren’t as bad as they may seem on occasion. So, we’ll start this section off with a process for working applications.

#### Process for Working Story/Word Problems

**READ THE PROBLEM.****READ THE PROBLEM AGAIN.**Okay, this may be a little bit of overkill here. However, the point of these first two steps is that you must read the problem. This step is the MOST important step, but it is also the step that most people don’t do properly.

You need to read the problem very carefully and as many times as it takes. You are only done with this step when you have completely understood what the problem is asking you to do. This includes identifying all the given information and identifying what you are being asked to find.

Again, it can’t be stressed enough that you’ve got to carefully read the problem. Sometimes a single word can completely change how the problem is worked. If you just skim the problem you may well miss that very important word.- Represent one of the unknown quantities with a variable and try to relate all the other unknown quantities (if there are any of course) to this variable.
- If applicable, sketch a figure illustrating the situation. This may seem like a silly step, but it can be incredibly helpful with the next step on occasion.
- Form an equation that will relate known quantities to the unknown quantities. To do this make use of known formulas and often the figure sketched in the previous step can be used to determine the equation.
- Solve the equation formed in the previous step and write down the answer to all the questions. It is important to answer all the questions that you were asked. Often you will be asked for several quantities in the answer and the equation will only give one of them.
- Check your answer. Do this by plugging into the equation, but also use intuition to make sure that the answer makes sense. Mistakes can often be identified by acknowledging that the answer just doesn’t make sense.

Let’s start things off with a couple of fairly basic examples to illustrate the process. Note as well that at this point it is assumed that you are capable of solving fairly simple linear equations and so not a lot of detail will be given for the actual solution stage. The point of this section is more on the set up of the equation than the solving of the equation.

Okay, let’s start off by defining \(p\) to be the minimum required score on the third exam.

Now, let’s recall how grades are set. Since there are no weights or anything on the grades, the grade will be set by first computing the following percentage.

\[\frac{{{\mbox{actual points}}}}{{{\mbox{total possible points}}}} = {\mbox{ grade percentage}}\]Since we are using the standard scale if the grade percentage is 0.9 or higher the student will get an A. Likewise, if the grade percentage is between 0.8 and 0.9 the student will get a B.

We know that the total possible points is 350 and the student has a total points (including the third exam) of,

\[4 + 8 + 7 + 7 + 9 + 78 + 83 + p = 196 + p\]The smallest possible percentage for an A is 0.9 and so if \(p\) is the minimum required score on the third exam for an A we will have the following equation.

\[\frac{{196 + p}}{{350}} = 0.9\]This is a linear equation that we will need to solve for \(p\).

\[196 + p = 0.9\left( {350} \right) = 315\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,p = 315 - 196 = 119\]So, the minimum required score on the third exam is 119. This is a problem since the exam is worth only 100 points. In other words, the student will not be getting an A in the Algebra class.

Now let’s check if the student will get a B. In this case the minimum percentage is 0.8. So, to find the minimum required score on the third exam for a B we will need to solve,

\[\frac{{196 + p}}{{350}} = 0.8\]Solving this for \(p\) gives,

\[196 + p = 0.8\left( {350} \right) = 280\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,p = 280 - 196 = 84\]So, it is possible for the student to get a B in the class. All that the student will need to do is get at least an 84 on the third exam.

We will first define \(x\) to be the height of the set of shelves. This means that 4\(x\) is width of the set of shelves. In this case we definitely need to sketch a figure so we can correctly set up the equation. Here it is,

Now we know that there are 72 feet of wood to be used and we will assume that all of it will be used. So, we can set up the following word equation.

\[\left( \begin{array}{c}{\mbox{length of }}\\ {\mbox{vertical pieces}}\end{array} \right) + \left( \begin{array}{c}{\mbox{length of }}\\ {\mbox{horizontal pieces}}\end{array} \right) = 72\]It is often a good idea to first put the equation in words before actually writing down the equation as we did here. At this point, we can see from the figure there are two vertical pieces; each one has a length of \(x\). Also, there are 4 horizontal pieces, each with a length of 4\(x\). So, the equation is then,

\[\begin{align*}4\left( {4x} \right) + 2\left( x \right) & = 72\\ 16x + 2x & = 72\\ 18x & = 72\\ x & = 4\end{align*}\]So, it looks like the height of the set of shelves should be 4 feet. Note however that we haven’t actually answered the question however. The problem asked us to find the dimensions. This means that we also need the width of the set of shelves. The width is 4(4)=16 feet. So the dimensions will need to be 4x16 feet.

#### Pricing Problems

The next couple of problems deal with some basic principles of pricing.

First, let’s define \(p\) to be the cost that the store paid for the calculator. The stores markup on the calculator is 15%. This means that 0.15\(p\) has been added on to the original price (\(p\)) to get the amount the calculator is being sold for. In other words, we have the following equation

\[p + 0.15p = 78.50\]that we need to solve for \(p\). Doing this gives,

\[1.15p = 78.50\hspace{0.5in} \Rightarrow \hspace{0.5in}p = \frac{{78.50}}{{1.15}} = 68.26087\]The store paid $68.26 for the calculator. Note that since we are dealing with money we rounded the answer down to two decimal places.

This problem is pretty much the opposite of the previous example. Let’s start with defining \(p\) to be the price of the shirt before the sale. It has been marked down by 35%. This means that 0.35\(p\) has been subtracted off from the original price. Therefore, the equation (and solution) is,

\[\begin{align*}p - 0.35p & = 15.00\\ 0.65p & = 15.00\\ p & = \frac{{15.00}}{{0.65}} = 23.0769\end{align*}\]So, with rounding it looks like the shirt was originally sold for $23.08.

#### Distance/Rate Problems

These are some of the standard problems that most people think about when they think about Algebra word problems. The standard formula that we will be using here is

\[{\mbox{Distance}} = {\mbox{Rate }} \times {\mbox{ Time}}\]All of the problems that we’ll be doing in this set of examples will use this to one degree or another and often more than once as we will see.

Let’s let \(t\) represent the amount of time that the cars are traveling before they meet. Now, we need to sketch a figure for this one. This figure will help us to write down the equation that we’ll need to solve.

From this figure we can see that the Distance Car A travels plus the Distance Car B travels must equal the total distance separating the two cars, 500 miles.

Here is the word equation for this problem in two separate forms.

\[\begin{array}{c} \left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{of Car A}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{of Car B}}\end{array} \right) = 500\\ \\ \left( \begin{array}{c}{\mbox{Rate of}}\\ {\mbox{ Car A}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time of}}\\ {\mbox{ Car A}}\end{array}\right) + \left( \begin{array}{c}{\mbox{Rate of}}\\ {\mbox{ Car B}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time of}}\\ {\mbox{ Car B}}\end{array} \right) = 500 \end{array}\]We used the standard formula here twice, once for each car. We know that the distance a car travels is the rate of the car times the time traveled by the car. In this case we know that Car A travels at 100 mph for \(t\) hours and that Car B travels at 70 mph for \(t\) hours as well. Plugging these into the word equation and solving gives us,

\[\begin{align*}100t + 70t & = 500\\ 170t & = 500\\ t & = \frac{{500}}{{170}} = 2.941176{\mbox{ hrs}}\end{align*}\]So, they will travel for approximately 2.94 hours before meeting.

For this problem we are going to need to be careful with the time traveled by each car. Let’s let \(t\) be the amount of time that the slower travel car travels. Now, since the faster car starts out 1 hour after the slower car it will only travel for \(t - 1\) hours.

Now, since we are repeating the problem from above the figure and word equation will remain identical and so we won’t bother repeating them here. The only difference is what we substitute for the time traveled for the faster car. Instead of \(t\) as we used in the previous example we will use \(t - 1\) since it travels for one hour less that the slower car.

Here is the equation and solution for this example.

\[\begin{align*}100\left( {t - 1} \right) + 70t & = 500\\ 100t - 100 + 70t & = 500\\ 170t & = 600\\ t & = \frac{{600}}{{170}} = 3.529412{\mbox{ hrs}}\end{align*}\]In this case the slower car will travel for 3.53 hours before meeting while the faster car will travel for 2.53 hrs (1 hour less than the slower car...).

Let’s start off by letting \(r\) be the speed of the boat on the right (the slower boat). This means that the boat to the left (the faster boat) is moving at a speed of 2\(r\). Here is the figure for this situation.

From the figure it looks like we’ve got the following word equation.

\[100 + \left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{of Boat B}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Distance}}\\ {\mbox{of Boat A}}\end{array} \right)\]Upon plugging in the standard formula for the distance gives,

\[100 + \left( \begin{array}{c}{\mbox{Rate of}}\\ {\mbox{ Boat B}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time of}}\\ {\mbox{ Boat B}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Rate of}}\\ {\mbox{ Boat A}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time of}}\\ {\mbox{ Boat A}}\end{array} \right)\]For this problem we know that the time each is 5 hours and we know that the rate of Boat A is 2\(r\) and the rate of Boat B is \(r\). Plugging these into the work equation and solving gives,

\[\begin{align*}100 + \left( r \right)\left( 5 \right) & = \left( {2r} \right)\left( 5 \right)\\ 100 + 5r & = 10r\\ 100 & = 5r\\ 20 & = r\end{align*}\]So, the slower boat is moving at 20 mph and the faster boat is moving at 40 mph (twice as fast).

#### Work/Rate Problems

These problems are actually variants of the Distance/Rate problems that we just got done working. The standard equation that will be needed for these problems is,

\[\left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{done in given time}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Work}}\\ {\mbox{Rate}}\end{array} \right) \times \left( \begin{array}{c}{\mbox{Time Spent}}\\ {\mbox{Working}}\end{array} \right)\]As you can see this formula is very similar to the formula we used above.

Let \(t\) be the time that it takes both machines, working together, to stuff a batch of envelopes. The word equation for this problem is,

\[\begin{array}{c}\left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{done by Machine A}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{done by Machine B}}\end{array} \right) = 1{\mbox{ Job}}\\ \\ \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Machine A}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time Spent}}\\ {\mbox{Working}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Machine B}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time Spent}}\\ {\mbox{Working}}\end{array} \right) = 1\end{array}\]We know that the time spent working is \(t\) however we don’t know the work rate of each machine. To get these we’ll need to use the initial information given about how long it takes each machine to do the job individually. We can use the following equation to get these rates.

\[1{\mbox{ Job}} = \left( \begin{array}{c}{\mbox{Work}}\\ {\mbox{Rate}}\end{array} \right) \times \left( \begin{array}{c}{\mbox{Time Spent}}\\ {\mbox{Working}}\end{array} \right)\]Let’s start with Machine A.

\[1{\mbox{ Job}} = \left( {{\mbox{Work Rate of A}}} \right) \times \left( 5 \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\mbox{Work Rate of A}} = \frac{1}{5}\]Now, Machine B.

\[1{\mbox{ Job}} = \left( {{\mbox{Work Rate of B}}} \right) \times \left( 3 \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\mbox{Work Rate of B}} = \frac{1}{3}\]Plugging these quantities into the main equation above gives the following equation that we need to solve.

\[\begin{align*}\frac{1}{5}t + \frac{1}{3}t & = 1\hspace{0.5in}{\mbox{Multiplying both sides by 15}}\\ 3t + 5t & = 15\\ 8t & = 15\\ t & = \frac{{15}}{8} = 1.875{\mbox{ hours}}\end{align*}\]So, it looks like it will take the two machines, working together, 1.875 hours to stuff a batch of envelopes.

Let \(t\) be the amount of time it would take John to clean the office complex by himself. The basic word equation for this problem is,

\[\begin{array}{c}\left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{done by Mary}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{done by John}}\end{array} \right) = 1{\mbox{ Job}}\\ \\ \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Mary}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time Spent}}\\ {\mbox{Working}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of John}}\end{array} \right)\left( \begin{array}{c}{\mbox{Time Spent}}\\ {\mbox{Working}}\end{array} \right) = 1\end{array}\]This time we know that the time spent working together is 3.5 hours. We now need to find the work rates for each person. We’ll start with Mary.

\[1{\mbox{ Job}} = \left( {{\mbox{Work Rate of Mary}}} \right) \times \left( 5 \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\mbox{Work Rate of Mary}} = \frac{1}{5}\]Now we’ll find the work rate of John. Notice however, that since we don’t know how long it will take him to do the job by himself we aren’t going to be able to get a single number for this. That is not a problem as we’ll see in a second.

\[1{\mbox{ Job}} = \left( {{\mbox{Work Rate of John}}} \right) \times \left( t \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\mbox{Work Rate of John}} = \frac{1}{t}\]Notice that we’ve managed to get the work rate of John in terms of the time it would take him to do the job himself. This means that once we solve the equation above we’ll have the answer that we want. So, let’s plug into the work equation and solve for the time it would take John to do the job by himself.

\[\begin{align*}\frac{1}{5}\left( {3.5} \right) + \frac{1}{t}\left( {3.5} \right) & = 1\hspace{0.5in}{\mbox{Multiplying both sides by }}5t\\ 3.5t + \left( {3.5} \right)\left( 5 \right) & = 5t\\ 17.5 & = 1.5t\\ \frac{{17.5}}{{1.5}} & = t\hspace{0.5in} \Rightarrow \hspace{0.5in}t = 11.67{\mbox{ hrs}}\end{align*}\]So, it looks like it would take John 11.67 hours to clean the complex by himself.

#### Mixing Problems

This is the final type of problems that we’ll be looking at in this section. We are going to be looking at mixing solutions of different percentages to get a new percentage. The solution will consist of a secondary liquid mixed in with water. The secondary liquid can be alcohol or acid for instance.

The standard equation that we’ll use here will be the following.

\[\left( \begin{array}{c}{\mbox{Amount of secondary}}\\ {\mbox{liquid in the water}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Percentage of}}\\ {\mbox{Solution}}\end{array} \right) \times \left( \begin{array}{c}{\mbox{Volume of}}\\ {\mbox{Solution}}\end{array} \right)\]Note as well that the percentage needs to be a decimal. So if we have an 80% solution we will need to use 0.80.

Okay, let \(x\) be the amount of 50% solution that we need. This means that there will be \(x + 10\) gallons of the 40% solution once we’re done mixing the two.

Here is the basic work equation for this problem.

\[\begin{array}{c}\left( \begin{array}{c}{\mbox{Amount of alcohol}}\\ {\mbox{in 50% Solution}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Amount of alcohol}}\\ {\mbox{in 35% Solution}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Amount of alcohol}}\\ {\mbox{in 40% Solution}}\end{array} \right)\\ \\ \left( {0.5} \right)\left( \begin{array}{c}{\mbox{Volume of }}\\ {\mbox{50% Solution}}\end{array} \right) + \left( {0.35} \right)\left( \begin{array}{c}{\mbox{Volume of }}\\ {\mbox{35% Solution}}\end{array} \right) = \left( {0.4} \right)\left( \begin{array}{c}{\mbox{Volume of }}\\ {\mbox{40% Solution}}\end{array} \right)\end{array}\]Now, plug in the volumes and solve for \(x\).

\[\begin{align*}0.5x + 0.35\left( {10} \right) & = 0.4\left( {x + 10} \right)\\ 0.5x + 3.5 & = 0.4x + 4\\ 0.1x & = 0.5\\ x & = \frac{{0.5}}{{0.1}} = 5\,{\mbox{gallons}}\end{align*}\]So, we need 5 gallons of the 50% solution to get a 40% solution.

Let \(x\) be the amount of water we need to add to the 40% solution. Now, we also don’t how much of the 40% solution we’ll need. However, since we know the final volume (75 liters) we will know that we will need \(75 - x\) liters of the 40% solution.

Here is the word equation for this problem.

\[\left( \begin{array}{c}{\mbox{Amount of acid}}\\ {\mbox{in the water}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Amount of acid}}\\ {\mbox{in 40% Solution}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Amount of acid}}\\ {\mbox{in 15% Solution}}\end{array} \right)\]Notice that in the first term we used the “Amount of acid in the water”. This might look a little weird to you because there shouldn’t be any acid in the water. However, this is exactly what we want. The basic equation tells us to look at how much of the secondary liquid is in the water. So, this is the correct wording. When we plug in the percentages and volumes we will think of the water as a 0% percent solution since that is in fact what it is. So, the new word equation is,

\[\left( 0 \right)\left( \begin{array}{c}{\mbox{Volume }}\\ {\mbox{of Water}}\end{array} \right) + \left( {0.4} \right)\left( \begin{array}{c}{\mbox{Volume of }}\\ {\mbox{40% Solution}}\end{array} \right) = \left( {0.15} \right)\left( \begin{array}{c}{\mbox{Volume of }}\\ {\mbox{15% Solution}}\end{array} \right)\]Do not get excited about the zero in the first term. This is okay and will not be a problem. Let’s now plug in the volumes and solve for \(x\).

\[\begin{align*}\left( 0 \right)\left( x \right) + \left( {0.4} \right)\left( {75 - x} \right) & = \left( {0.15} \right)\left( {75} \right)\\ 30 - 0.4x & = 11.25\\ 18.75 & = 0.4x\\ x & = \frac{{18.75}}{{0.4}} = 46.875{\mbox{ liters}}\end{align*}\]So, we need to add in 46.875 liters of water to 28.125 liters of a 40% solution to get 75 liters of a 15% solution.