Essential modeling

Essential modeling refers to your ability to take a scenario given by a word problem and transform it into a mathematical problem. It is a step up in difficulty from previous sections, which had you make equations from word problems. You will need to draw shapes, set up equations, and apply various different algebraic skills to solve these problems.

Words to algebra

Let’s start with a quick summary of how to turn a word problem into an algebraic equation. Look for any number, variable, and keyword like more, less, times, etc. With this information, we will develop our equations easily. The keywords describe which mathematical operators we use such as addition and subtraction, the numbers describe how much we are doing these things, and the variables show you what you are solving for. Remember that variables may be given to you in the form of letters like $a$, or they may simply be the value that the question is missing or asks you to find.

Modeling practice

Let’s go through some examples together so that we understand how to find this information and how to put it into equation or drawing form.

Example 1

Pat is a bus driver who drives to school twice a day, Monday through Friday. How many times will Pat have driven to school in $4$ weeks?

Step 1 is to identify the important information here. Pat drives twice a day, and five days a week. That means Pat drives $2\ast 5=10$ times per week. Now that we know she drives $10$ times per week, how many times did she drive in $4$ weeks?

$$10\ast 4=40$$

Pat drove to school $40$ times in $4$ weeks. In this example, we didn’t really need to write an equation. We certainly could have written an equation, but it is not necessary since it was simple enough to perform step by step.

Example 2

Ryan wants to buy a subscription to a streaming service that costs $\$9$ per month. The service allows him to stream the first month for free, but each following month will cost $\$9$. How many months can Ryan keep this subscription if he only has $\$39$?

Here we will need to make an equation because there is a bit more information. First, let’s indicate that information: It costs $\$9$ per month with the first month free, and the number of months is what we want to find. The number of months will be our variable $m$. We will have to multiply the number of months by $9$, and then we will have to subtract $9$ since that first month is free. Lastly, we must set the equation equal to his total money available, $\$39$. This is what the expression would look like before solving it.

$$9m-9=39$$

This can also come from $9(m-1)=39$ since one month is free. After you distribute the $9$ to both terms, you get the expression above.

Solving for $m$, we get:

$$\begin{array}{rl}9m& =39+9\\ m& =48/9\\ m& =5.33\end{array}$$

Our final answer cannot be $5.33$, though, because we have to find out how many entire months he can pay for. He has enough money to pay for $5$ full months, so $5$ is the final answer.

Example 3

A rectangular dining room has a length that is twice as long as its width. The tables in the room have a radius of $1$ meter. If the width of the dining room is $10$ meters and there are $30$ tables, how much of the room in square meters is not taken up by the tables?

This is a lot of information, so we will have to list it all out and then piece together where to start:

• Area of room $=$ length $\ast$ width
• Length $=2\ast$ width
• Width $=10$ meters
• Area of table $=pi\ast r^2$
• Radius ($r$) $=1$ meter
• Space not taken up by tables $=$ area of dining room $-$ (area of a table) $\ast 30$

You may be able to do this in your head, or you may prefer to write it down. Either way, let’s start piecing our information together to see what we are missing, starting with the room. We can use the width to find the length, $L=20$ meters. With that, we can find the area of the room, $A=200$ square meters.

Now, let’s do the tables. We can find the area using the radius, $A=pi\ast (1{)}^2=pi$. This is the area of one table. We need to find out how much space all of the $30$ tables take up, which is $A\ast 30=30\ast pi=94.3$ square meters.

Now we can put our information together. We want to find the space not taken up by the tables, and we know that this is equal to the area of the room minus the area taken up by all the tables.

$$200-94.3=105.7\text{square meters}$$