3.4: Global Buisness Strategies
What you’ll learn to do: evaluate common strategies used to reach global markets
Globalization introduces a number of challenges that are unique to operating simultaneously in different countries and global markets. What is the best way to enter or take advantage of a global market? When should you adjust a product’s features to customize it to consumer needs in a different global market? How do you manage the costs and complexities of producing and/or promoting products in different locations, with different languages, cultural sensitivities, and consumer expectations?
While this next section doesn’t attempt to answer all of these questions, it explains common strategies and approaches used by multinational corporations to take advantage of global business opportunities.
Learning Objectives
- Explain how firms use licensing and franchising to reach global markets
- Explain how firms use joint ventures and foreign strategic alliances to reach global markets
Global Business Strategies
In today’s economy, once a nation or business has developed an advantage—either comparative or absolute—it’s likely to look beyond its own borders or storefront to seek greater economic opportunity. But how do you enter a global market? It’s certainly not as simple as loading up your products in a van, driving to the next town, and knocking on doors. Below are some of the common strategies companies and countries use to get their goods and services into global markets.
Exporting/Importing
Figure \(\PageIndex{1}\). Shipping Containers
Exporting is the easiest and most straightforward way to engage with the global market. Exporting is taking goods that were produced within a company’s home country and shipping them to another country. The party sending the good is called an exporter . It is impossible to discuss exporting without mentioning its complement, importing. Importing is the process by which a good is brought into a jurisdiction, especially across a national border, from an external source. The party bringing in the good is called an importer . Simply put, one country’s exports become another country’s imports. Examples of U.S. imports are everywhere: Take a look at the labels in your clothes or the contents of your backpack. From our vantage point, U.S. exports may be a little harder to see, but they exist all the same and are plenty visible in other countries. According to World’s Top Exports , the following export product groups represent the highest dollar value in American global shipments during 2015. In parentheses is the percentage share each export category represents in terms of overall U.S. exports:
- Machines, engines, pumps: US$205.8 billion (13.7% of total exports)
- Electronic equipment: $169.8 billion (11.3%)
- Aircraft, spacecraft: $131.1 billion (8.7%)
- Vehicles: $127.1 billion (8.4%)
- Oil: $106.1 billion (7.1%)
- Medical, technical equipment: $83.4 billion (5.5%)
- Plastics: $60.3 billion (4%)
- Gems, precious metals, coins: $58.7 billion (3.9%)
- Pharmaceuticals: $47.3 billion (3.1%)
- Organic chemicals: $38.8 billion (2.6%) [1]
Advantages and Disadvantages
Since exporting doesn’t require a company to manufacture its products in the target country, the company doesn’t have to invest in factories, equipment, or other production facilities located halfway around the globe. Most of the costs involved in exporting are associated with finding a buyer or distributor in the destination market. For these reasons, exporting is considered to be the quickest and least expensive means to enter the global market. However, there are disadvantages, too.
Once products arrive in the destination market, the business loses control of them, which can result in products being misrepresented, copied by other manufacturers, or even sold on a black market. In addition, because the business isn’t active in the new market, it can’t gain insight into or experience with local consumer preferences and demand. This lack of information can create uncertainty and potentially cost the company opportunities down the road. As you will learn later in this module, businesses operating in other countries may find themselves subject to taxes, regulations, and/or restrictions that can substantially affect the profitability of the entire export venture.
Outsourcing/Offshoring
Figure \(\PageIndex{2}\). Garment factory, Jiaxing, ChinaOutsourcing and offshoring are two additional strategies that a business can use in order to take advantage of the global market. Outsourcing contracts out a business process to another party and may include either or both foreign and domestic contracting. You may be familiar with outsourcing if your college has outsourced the bookstore to a national chain such as Barnes & Noble, or the food services are provided by a company such as Starbucks or Aramark. Although the employees work on your college campus, they are not college employees. Offshoring , on the other hand, is the actual relocation of a business process from one country to another—typically it’s an operational process, such as manufacturing, or sometimes a supporting process, such as accounting. In the case of offshoring, the employees still work for the company that’s offshoring its operations, but instead of working in a facility within the United States, they are located in a foreign country. In general, outsourcing and offshoring are strategies that companies use to try to lower their costs.
If a business chooses outsourcing as a way to engage with the global market, it might have a single component part manufactured in, say, Tibet and then shipped back to Iowa, where the factory workers in Iowa would use the outsourced part in the assembly of the final product. The business would have a contract with the company making the component part at an agreed-upon price, but it would not have an employer-employee relationship with the workers in Tibet. On the other hand, if the business wants to take advantage of offshoring, it would move the entire plant from Iowa to Tibet and hire workers in Tibet who would work directly for the business.
The following video is an example of how a small business is outsourcing its manufacturing to China. Especially for small start-up companies, using established manufacturing facilities located outside of the U.S. allows them to enter the global marketplace. Cost, logistics, finances, and speed are just some of the things that this type of arrangement can bring to businesses looking to take advantage of the growing global demand for U.S.-branded products.
Advantages and Disadvantages
Offshoring and outsourcing are both the subject of ongoing heated public debate—both in the U.S. and in other countries. Those in favor assert that these strategies benefit both sides of the arrangement: Free trade is enhanced, the destination country gains jobs, and the origin country gets cheaper goods and services. Some supporters go further and assert that outsourcing and offshoring raise the gross domestic product (GDP) and increase the total number of jobs domestically, too. This claim is based on the idea that workers who lose their jobs will move to higher-paying jobs in industries where the origin country has a comparative advantage.
On the other hand, job losses and wage erosion “at home” have sparked opposition to offshoring and outsourcing. Many argue that the jobs that are shipped overseas are not replaced by better, higher-paying ones. And it’s not just low-skilled workers who are feeling the pain. Increasingly, critics say, even highly trained workers (such as software engineers) with high-paying jobs are finding themselves replaced by cheaper workers in India and China. Some firms, while realizing financial gains from lowering their production costs, are finding that offshoring and outsourcing are very costly in terms of lack of control over product quality, working conditions, and labor relations. For example, companies like Nike and Apple have come under fire by human rights organizations and consumers over reports of worker abuse, dangerous working conditions, and ridiculously low wages. It was recently reported that apparel workers in Bangladesh are sometimes paid as little as $0.21 per hour. We will explore some of the ethical issues raised by offshoring and outsourcing later in the course in the business ethics module.
Licensing and Franchising
Figure \(\PageIndex{3}\). The Star Wars CookbookIncreasingly, businesses are getting their products and services into global markets via licensing and franchise agreements. Under a licensing agreement , the licensor agrees to let someone else (the licensee) use the property of the licensor in exchange for a fee. License agreements usually cover property that is intangible, such as trademarks, images, patents, or production techniques. Since its debut in the late 1970s, Star Wars remains the most lucrative source of licensing in the entertainment business, generating more than $42 billion from the sale of licensed merchandise.
A longer-term and more comprehensive way to access the global market is through franchising. Under the terms of a franchise agreement, a party (franchisee) acquires access to the knowledge, processes, and trademarks of a business (the franchisor) in order to sell a product or service under the business’s (franchise’s) name. In exchange for the franchise, the franchisee usually pays the franchisor both initial and annual fees. McDonald’s, Holiday Inn, Hertz Car Rental, and Dunkin’ Donuts have all expanded into foreign markets through franchising.
Advantages and Disadvantages
Licensing and franchising both offer advantages for the involved parties: The licensee and franchisee both gain a competitive advantage in the market. The licensee/franchisee gets immediate brand recognition and may quickly overtake the competition by offering a product or service for which there is existing unmet demand. For example, a local sandwich shop may have a hard time competing when a Subway franchise opens because the brand is so well known. Also, because franchises are usually “turnkey” operations in which processes, supply chains, training, and products are already in place, the new business can quickly begin efficient and profitable operations. For the franchisor, this arrangement enables them to gain inexpensive access to a new market, since the initial cost of the franchise is borne by the franchisee. Under a licensing agreement, all of the costs of production, sales, and distribution are the responsibility of the licensee. If financial capital is scarce, both approaches allow companies to have a global presence without heavy investments.
These methods do contain some risks and disadvantages, however. They are typically the least profitable way of entering a foreign market, since the profits go to the franchisee or licensee. Although the licensor or franchisor receives up-front money and/or a small percentage of future sales, the majority of the revenue remains in the destination country with the licensee or franchisee. Franchising entails a long-term commitment on the part of the franchisor to provide ongoing support in the form of training, logistics, product development, and brand marketing. Once a business begins to establish a global franchise presence, the pressure to maintain brand integrity and fiscal responsibility becomes more intense as the failure of the franchise now has global consequences. For companies selling licensing rights there is a risk that their intellectual property may be misrepresented or used in a manner that could tarnish the brand’s image. Also, once a license to use an image or other intellectual property has been granted to a company in another country, the probability that knock-off products will enter the market goes up. For both franchisors and licensors, maintaining quality standards on a global scale is a massive undertaking, and for this reason many companies are choosing to exert a higher degree of control over their products, brands, and intellectual property than they have in the past.
Joint Ventures/Strategic Alliances
Figure \(\PageIndex{4}\). Honey Nut CheeriosThere are times when businesses have opportunities within the global market that are better undertaken with a partner. Sometimes these projects are extremely large and capital-intensive or are so comprehensive that it makes sense to include multiple businesses or even governments. These large-scale, global projects usually take one of two forms: strategic alliances or joint ventures.
A joint venture establishes a new business that is jointly owned by two or more otherwise independent businesses. The most common joint ventures involve two companies that are equal partners in the new firm, investing money and resources while sharing control of the newly formed firm. Often, the foreign partner provides expertise on the new market, business connections and networks, and access to other in-country aspects of business such as real estate and regulatory compliance. For example, in 2015 Fiat Chrysler entered into a joint venture with Tata Motors of India to expand the production of Jeeps in India. The company created in this joint venture is Fiat India Automobiles Private Limited.
Joint ventures require a greater commitment from firms than other global strategies, because they are riskier and less flexible. Joint ventures may afford tax advantages in many countries, particularly where foreign-owned businesses are taxed at higher rates than locally owned businesses. Some countries require all business ventures to be at least partially owned by domestic business partners.
A less permanent, but equally effective way to enter the global market is through a strategic alliance . A strategic alliance is formed between two or more corporations, each based in their home country, for a specified period of time. Unlike a joint venture, a new company is not formed. Generally, strategic alliances are pursued when businesses find that they have gained all they can from exporting and want to expand into a new geographic market or a related business. This approach can be particularly useful when a government prohibits imports in order to protect domestic industry. The cost of a strategic alliance is usually shared equitably among the corporations involved, and it’s generally the least expensive way for all concerned to form a partnership. An example of this is the alliance between General Mills and Nestlé: Honey Nut Cheerios are manufactured in bulk by General Mills in the United States and then shipped to Nestlé Europe, where they are packaged and shipped to France, Spain, and Portugal.
Advantages and Disadvantages
The greatest advantage of joint ventures and strategic alliances is the knowledge and experience of the market offered by the local partner—on everything from consumer preferences to cultural differences, language, and political/economic systems. Another advantage is that the risk of entering the market with a new product is shared by more than one firm, thereby reducing each company’s exposure to potential losses.
However, these types of partnerships also have their drawbacks. When companies share their technology and industry know-how, they run the risk that the partner firm will take that technology or innovation and use it to become a competitor in the future. This was a primary concern when Boeing collaborated with Mitsubishi (it was ultimately resolved in the legal details of the partnership agreement, which both companies signed). Conflicts over control of these partnerships can also arise if the owners of the partner firms do not agree on key business decisions.
Of all of the ways that a business can reach the global market, the most intensive approach is through foreign direct investment or FDI. Foreign direct investment is an investment in the form of a controlling ownership in a business enterprise in one country by an entity based in another country. FDI can take one of two forms: Greenfield ventures or mergers/acquisitions.
In a Greenfield venture, the company enters a foreign market and establishes a new subsidiary as a start-up business. A good example of this is the BMW US Manufacturing Company, a vehicle-assembly facility located in Greer, South Carolina, that is part of the BMW Group. Although it’s BMW’s only assembly plant in the United States, it represents a direct investment inside the United States by the German manufacturer, and it’s one of the most successful Greenfield ventures in the U.S.
Businesses that are not ready to take on the challenge of establishing a new facility or subsidiary in a foreign country will usually choose either a merger or acquisition as a means of expanding their global reach. Mergers and acquisitions represent the vast majority of FDI and range from 50 percent to 80 percent of all FDI in some industries. According to Forbes ,
"U.S. companies completed 116 emerging market acquisitions in the first half of 2013, up from 110 in the second half of 2012. . . . The most popular geographic targets for U.S. companies in the first half of 2013 were Brazil (25 deals), India (18 deals), South American countries excluding Brazil (15), South and East Asia (15), and Central America and Caribbean (14). [2] "
Mergers and acquisitions aren’t just carried out by U.S. companies, either—it’s an incredibly pervasive global business strategy, and ownership of many well-known products and brands has long been separated from the country of origin. For example, the Chinese just bought Smithfield Foods, Stolichnaya (“Stoli”) Russian vodka is actually owned by a company in the United Kingdom, Anheuser-Busch is owned by Belgian-Brazilian conglomerate InBev, and 7-Eleven is owned by the Japanese. [3]
Advantages and Disadvantages
Because the level of commitment and investment associated with FDI is so high, companies expend a great deal of time and effort scrutinizing potential opportunities. With Greenfield ventures, the amount of time it takes to build a presence in the foreign country is substantial. If a business is not already established in other global locations and lacks experience with FDI, it may be in for a series of unpleasant surprises in the form of regulations, licensing, taxes, and other “red tape”—much of which we will look at later in this module.
On the other hand, mergers and acquisitions are faster to execute than Greenfield ventures, and by merging with or acquiring an existing foreign company already in the market, outside companies can quickly take advantage of that presence. Another benefit is that a merger or acquisition involves the purchase of assets such as property, plants, and equipment that are already producing a product with a known revenue stream. The key to a successful merger or acquisition is paying the right price for the company, because, no matter how successful the business was before it was acquired (or merged), overpaying can turn a formerly profitable operation into a money pit.
- United States Top 10 Exports. (2016). Retrieved August 17, 2016, from http://www.worldstopexports.com/united-states-top-10-exports/ ↵
- Rapoza, K. (2013, September 13). U.S. Companies Buying up Foreign Competition. Retrieved August 18, 2016, from http://www.forbes.com/sites/kenrapoza/2013/09/15/u-s-companies-buying-up-foreign-competition/#7a6b71ef2177 ↵
- Frohlich, T. C., & Sauter, M. B. (n.d.). Ten Classic American Brands That Are Foreign-Owned. Retrieved August 19, 2016, from http://247wallst.com/special-report/...foreign-owned/ ↵


8.11D: Bacteroidetes and Chlorobi
- Describe the major classes of Bacteroidetes, including: Bacteroidia and Porphyromonas a well as Phlyum Chlorobi
The Phylum Bacteroidetes are characterized as rod-shaped, gram-negative bacteria that are non-spore forming and are present in anaerobic environments. Bacteroidetes are capable of thriving in numerous environments that include soil, sediments, sea water, and in the guts and on skin of animal hosts. The Bacteroidetes are classified into three large classes which include the Bacteroidia class and the Porphyromonas class. The Bacteroidia class is the most studied and is present in the gastrointestinal system of mammals which allows it to be abundant in the feces. The Porphyromonas class is characterized by their presence in the oral cavity of humans. The bacteria categorized as bacteroidetes are opportunistic and are rarely pathogenic as they constitute part of the normal flora.
The Phylum Chlorobi are characterized by bacteria that are obligately anaerobic photoautotrophic which includes green sulfur bacteria. The green sulfur bacteria are photolithotropic oxidizers of sulfur and utilize a noncyclic electron transport chain. The green sulfur bacteria are closely related to Bacteroidetes and are non-motile and can be found as sphere, rod, or spiral shaped. The most commonly studied model is Chlorobium tepidum which has had its complete genome sequences. Chlorobium species typically exist in symbiotic relationships with a colorless, nonphotosynthetic bacteria.
The Phlyum Chlorobi is often grouped with the Phlyum Bacteroidetes because their branches are very close together in the phylogenetic tree. By utilizing sequencing techniques such as comparative genomic analysis, there have been three proteins which are unique to all members of the Bacteroidetes and Chlorobi phyla but not to other bacteria indicating a conserved protein signature. Further analysis has identified additional molecular signatures that support the close relationship between these two phyla as well indicating a common ancestor.
Key Points
- The close relationship between Phylum Bacteroidetes and Phylum Chlorobi are supported by comparative genomic analysis which indicates they are derived from a common ancestor based on unique molecular signatures and common proteins.
- Phylum Bacteroidetes are composed of three large classes of gram-negative bacteria that are rod-shaped, non-spore forming, and present in anaerobic conditions. Bacteroidetes are found in numerous environments ranging from soils, sediments, sea water, and the guts and skin of animals.
- Phylum Chlorobi are composed of green sulfur bacteria that are categorized as photolithotrophic oxidizers of sulfur. Chlorobi species are commonly found in symbiotic relationships with colorless, nonphotosynthetic bacteria.
Key Terms
- photolithotropic : Obtain energy from light and use inorganic electron donors only to fuel biosynthetic reactions.


17.6: The Partition Function of Distinguishable, Independent Molecules is the Product of the Molecular Partition Functions
A system, such as a container of gas, can consist of a large number of subsystems. How is the partition function of the system built up from those of the subsystems? This depends on whether the subsystems are distinguishable or indistinguishable. Let's start with energy. Energy is additive so that:
\[E_\text{tot}(N,V) = \epsilon_1(V) + \epsilon_2(V) + \epsilon_3(V) + \cdots \nonumber \]
Each of the molecules can have their energy distributed over their respective energy states (e.g., vibrations, rotations, etc.). This means that each \(\epsilon_i\) is already a summation over the states of the molecule. Let us assume that we can somehow distinguish all the molecules as: \(a\), \(b\), \(c\), \(d\)… and denote the energy state they are in by \(i\), \(j\), \(k\):
\[E_l(N,V) = \epsilon_i^a (V) + \epsilon_j^b (V) + \epsilon_k^c (V) + \cdots \nonumber \]
A good example would be the molecules in a molecular crystal. They only move around a fixed site and so we can distinguish by how far molecule '\(a\)' is from a given corner of the crystal. The systems partition function becomes:
\[Q(N,V,\beta) = \sum_i e^{-\beta E_l} = \sum_{i,j,k,…} e^{-\beta (\epsilon_i^a + \epsilon_j^b + \epsilon_k^c)} \nonumber \]
So far, we have done little effort to distinguish between the partition function of a molecular system \(q\) and the whole ensemble \(Q\) (e.g. the gas). If the entities that we called systems are distinguishable and independent, the whole ensemble partition function is the product of the molecular system partition functions. We get:
\[Q(N,V,\beta) = \prod_i{q_i} \nonumber \]
for \(N\) distinguishable systems. We can split up the summation into a product of molecular partition functions:
\[Q(N,V,β) = \prod_i^N q_i(V,\beta)= q_a(V,\beta) q_b(V,\beta) q_c(V,\beta) \cdots \nonumber \]
This results in each molecular system partition function being summed over independently:
\[ Q(N,V,\beta) = \sum_{i} e^{-\beta \epsilon_i^a} \sum_{j} e^{-\beta \epsilon_i^b} \sum_{k} e^{-\beta \epsilon_i^c} \cdots \nonumber \]
If the energy states of all the particles are the same, then the equation simplifies to:
\[ Q(N,V,\beta) = [q(V,\beta)]^N \nonumber \]
We can do this if, for example, the particles are embedded in a crystal where we can distinguish them by their location. We will see in the next chapter that for indistinguishable particles, such as those in a gas, we get a different result.


7.1: Exponential Growth and Decay Models
Learning Objectives
In this section, you will learn to
- Compare linear and exponential growth.
- Recognize and model exponential growth and decay.
Prerequisite Skills
Before you get started, take this prerequisite quiz.
1. Simplify \(3(2)^3\).
- Click here to check your answer
-
\(24\)
If you missed this problem, review here and scroll to Example 4. (Note that this will open a different textbook in a new window.)
2. Simplify \(-5(3)^2\).
- Click here to check your answer
-
\(-45\)
If you missed this problem, review here and scroll to Example 4. (Note that this will open a different textbook in a new window.)
3. Solve \(2^x=16\).
- Click here to check your answer
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\(x=4\)
If you missed this problem, review here . (Note that this will open a different textbook in a new window.)
4. Graph \(y=-\dfrac{1}{2}x+3\).
- Click here to check your answer
-
If you missed this problem, review Section 1.2 . (Note that this will open in a new window.)
Comparing Exponential and Linear Growth
Consider two social media sites which are expanding the number of users they have:
- Site A has 10,000 users, and expands by adding 1,500 new users each month
- Site B has 10,000 users, and expands by increasing the number of users by 10% each month.
The number of users for Site A can be modeled as linear growth. The number of users increases by a constant number, 1500, each month. If \(x\) = the number of months that have passed and \(y\) is the number of users, the number of users after \(x\) months is \(y = 10000+1500x\). For site B, the user base expands by a constant percent each month, rather than by a constant number. Growth that occurs at a constant percent each unit of time is called exponential growth.
We can look at growth for each site to understand the difference. The table compares the number of users for each site for 12 months. The table shows the calculations for the first 4 months only, but uses the same calculation process to complete the rest of the 12 months.
|
Month |
Users at Site A |
Users at Site B |
|---|---|---|
|
0 |
\(10000\) |
\(10000\) |
|
1 |
\(10000 + 1500 = 11500\) |
\(\begin{aligned}
10000+&10 \% \text { of } 10000 \\ =& 10000+0.10(10000) \\ =& 10000(1.10)=11000 \end{aligned}\) |
|
2 |
\(11500 + 1500 = 13000\) |
\(\begin{aligned}
|
|
3 |
\(13000 + 1500 = 14500\) |
\(\begin{aligned}
|
|
4 |
\(14500 + 1500 = 16000\) |
\(\begin{aligned}
|
|
5 |
\(17500\) |
\(16105\) |
|
6 |
\(19000\) |
\(17716\) |
|
7 |
\(20500\) |
\(19487\) |
|
8 |
\(22000\) |
\(21436\) |
|
9 |
\(23500\) |
\(23579\) |
|
10 |
\(25000\) |
\(25937\) |
|
11 |
\(26500\) |
\(28531\) |
|
12 |
\(28000\) |
\(31384\) |
For Site B, we can re-express the calculations to help us observe the patterns and develop a formula for the number of users after x months.
- Month 1: \(y = 10000(1.1) = 11000\)
- Month 2: \(y = 11000(1.1) = 10000(1.1)(1.1)=\mathbf{10000(1.1)^2}= 12100\)
- Month 3: \(y = 12100(1.1) = 10000(1.1)^2 (1.1)=\mathbf{10000(1.1)^3}= 13310\)
- Month 4: \(y = 13310(1.1) = 10000(1.1)^3 (1.1)=\mathbf{10000(1.1)^4}= 14641\)
By looking at the patterns in the calculations for months 2, 3, and 4, we can generalize the formula. After \(x\) months, the number of users \(y\) is given by the function \(\mathbf{y = 10000(1.1)^x}\)
Example \(\PageIndex{1}\)
Determine whether each of the following statements represent exponential growth or linear growth.
- The population of a certain town has been increasing by 2.4% over each of the last 10 years.
- The population of a certain town has been increasing by 1,000 people over each of the last 10 years.
- The number of bacterial cells increases by a factor of \(\dfrac{3}{2}\) every 24 hours.
Solution
a. Since the population has been increasing by a constant percent for each unit of time, this is an example of exponential growth.
b. Since the population has been increasing by a constant number for each unit of time, this is an example of linear growth.
c. Since the number of cells has been increasing by a constant percent \( (\dfrac{3}{2}=150\%) \) for each unit of time, this is an example of exponential growth.
Using Exponential Functions to Model Growth and Decay
In exponential growth, the value of the dependent variable \(y\) increases at a constant percentage rate as the value of the independent variable (\(x\) or \(t\)) increases. Examples of exponential growth functions include:
- the number of residents of a city or nation that grows at a constant percent rate.
- the amount of money in a bank account that earns interest if money is deposited at a single point in time and left in the bank to compound without any withdrawals.
In exponential decay, the value of the dependent variable y decreases at a constant percentage rate as the value of the independent variable (\(x\) or \(t\)) increases. Examples of exponential decay functions include:
- value of a car or equipment that depreciates at a constant percent rate over time
- the amount a drug that still remains in the body as time passes after it is ingested
- the amount of radioactive material remaining over time as a radioactive substance decays.
Exponential functions often model quantities as a function of time; thus we often use the letter \(t\) as the independent variable instead of \(x\).
The table compares exponential growth and exponential decay functions:
|
Exponential Growth |
Exponential Decay |
|---|---|
|
Quantity grows by a constant percent
|
Quantity decreases by a constant percent per unit of time |
|
\(\mathbf{y=ab^x}\)
|
\(\mathbf{y=ab^x}\)
|
In general, the domain of exponential functions is the set of all real numbers. The range of an exponential growth or decay function is the set of all positive real numbers.
In most applications, the independent variable, \(x\) or \(t\), represents time. When the independent variable represents time, we may choose to restrict the domain so that independent variable can have only non-negative values in order for the application to make sense. If we restrict the domain, then the range is also restricted as well.
- For an exponential growth function \(y=ab^x\) with \(b>1\) and \(a > 0\), if we restrict the domain so that \(x ≥ 0\), then the range is \(y ≥ a\).
- For an exponential decay function \(y=ab^x\) with \(0<b<1\) and \(a > 0\), if we restrict the domain so that \(x ≥ 0\), then the range is \(0 < y ≤ a\).
Example \(\PageIndex{2}\)
Consider the growth models for social media sites A and B, where \(x\) = number of months since the site was started and \(y\) = number of users. The number of users for Site A follows the linear growth model:
\[y = 10000+1500x \nonumber.\]
The number of users for Site B follows the exponential growth model:
\[y=10000(1.1^x) \nonumber\]
For each site, use the function to calculate the number of users at the end of the first year, to verify the values in the table. Then use the functions to predict the number of users after 30 months.
Solution
Since \(x\) is measured in months, then \(x = 12\) at the end of one year.
Linear Growth Model:
When \(x = 12\) months, then \(y = 10000 + 1500(12) = 28,000\) users
When \(x = 30\) months, then \(y = 10000 + 1500(30) = 55,000\) users
Exponential Growth Model:
When \(x = 12\) months, then \(y = 10000(1.1^{12}) = 31,384\) users
When \(x = 30\) months, then \(y = 10000(1.1^{30}) =174,494\) users
We see that as \(x\), the number of months, gets larger, the exponential growth function grows large faster than the linear function (even though in Example \(\PageIndex{2}\) the linear function initially grew faster). This is an important characteristic of exponential growth: exponential growth functions always grow faster and larger in the long run than linear growth functions.
It is helpful to use function notation, writing \(y = f(t) = ab^t\), to specify the value of \(t\) at which the function is evaluated.
Example \(\PageIndex{3}\)
A forest has a population of 2000 squirrels that is increasing at the rate of 3% per year. Let \(t\) = number of years and \(y = f(t) =\) number of squirrels at time \(t\).
- Find the exponential growth function that models the number of squirrels in the forest at the end of \(t\) years.
- Use the function to find the number of squirrels after 5 years and after 10 years
Solution
a. The exponential growth function is \(y = f(t) = ab^t\), where \(a = 2000\) because the initial population is 2000 squirrels
The annual growth rate is 3% per year, stated in the problem. We will express this in decimal form as \(r = 0.03\)
Then \(b = 1+r = 1+0.03 = 1.03\)
Answer: The exponential growth function is \(y = f(t) = 2000(1.03)^t\)
b. After 5 years, the squirrel population is \(y = f(5) = 2000(1.03)^5 \approx 2319\) squirrels
After 10 years, the squirrel population is \(y = f(10) = 2000(1.03)^{10} \approx 2688\) squirrels
Example \(\PageIndex{4}\)
A large lake has a population of 1000 frogs. Unfortunately the frog population is decreasing at the rate of 5% per year. Let \(t\) = number of years and \(y\) = \(g(t)\) = the number of frogs in the lake at time \(t\).
- Find the exponential decay function that models the population of frogs.
- Calculate the size of the frog population after 10 years.
Solution
a. The exponential decay function is \(y = g(t) = ab^t\), where \(a = 1000\) because the initial population is 1000 frogs
The annual decay rate is 5% per year, stated in the problem. The words decrease and decay indicated that \(r\) is negative. We express this as \(r = -0.05\) in decimal form.
Then, \(b = 1+ r = 1+ (-0.05) = 0.95\)
Answer: The exponential decay function is: \(y = g(t) = 1000(0.95)^t\)
b. After 10 years, the frog population is \(y = g(10) = 1000(0.95)^{10} \approx 599\) frogs
Example \(\PageIndex{5}\)
A population of bacteria is given by the function \(y = f(t) = 100(2)^t\), where \(t\) is time measured in hours and \(y\) is the number of bacteria in the population.
- What is the initial population?
- What happens to the population in the first hour?
- How long does it take for the population to reach 800 bacteria?
Solution
a. The initial population is 100 bacteria. We know this because \(a = 100\) and because at time \(t = 0\), then \(f(0) = 100(2)^0 = 100(1)=100\)
b. At the end of 1 hour, the population is \(y = f(1) = 100(2)^1 = 100(2)=200\) bacteria.
The population has doubled during the first hour.
c. We need to find the time \(t\) at which \(f(t) = 800\). Substitute 800 as the value of \(y\):
\[\begin{array}{l}
y=f(t)=100\left(2\right)^{t} \\
800=100\left(2\right)^{t}
\end{array}\nonumber\]
Divide both sides by 100 to isolate the exponential expression on the one side
\[8=1\left(2\right)^{\mathrm{t}} \nonumber\]
8 = 2 3 , so it takes \(t = 3\) hours for the population to reach 800 bacteria.
Two important notes about Example \(\PageIndex{5}\):
- In solving \(8 = 2^t\), we “knew” that \(t\) is 3. But we usually can not know the value of the variable just by looking at the equation. Later we will use logarithms to solve equations that have the variable in the exponent.
- To solve \(800 = 100(2)^t\), we divided both sides by 100 to isolate the exponential expression \(2^t\). We can not multiply 100 by 2. The exponent applies only to the quantity immediately before it, so the exponent t applies only to the base of 2.
Comparing Linear, Exponential and Polynomial Functions
To identify the type of function from its formula, we need to carefully note the position that the variable occupies in the formula.
A linear function can be written in the form \(\mathbf{y=a x+b}\)
As we studied in chapter 1, there are other forms in which linear equations can be written, but linear functions can all be rearranged to have form \(y = mx + b\).
An exponential function has form \(\mathbf{y=ab^x}\)
The variable \(\mathbf{x}\) is in the exponent. The base \(b\) is a positive number.
- If \(b>1\), the function represents exponential growth.
- If \(0 < b < 1\), the function represents exponential decay
A polynomial function has form \(\mathbf{y=ax^P+bx^{P-1}+cx^{P-2}+...}\)
The variable \(\mathbf{x}\) is in the base. The exponent \(p\) is a non-zero number, and the exponents decrease until the final exponent is zero.
We compare three functions, using increasing x values:
- linear function \(y = f(x) = 2x\)
- exponential function \(y = g(x) = 2^x\)
- polynomial function \(y = h(x) = x^2\)
|
\(\mathbf{x}\) |
\(\mathbf{y = f(x) =2x}\) |
\(\mathbf{y = g(x) =2^x}\) |
\(\mathbf{y= h(x) =x^2}\) |
|---|---|---|---|
|
0 |
0 |
1 |
0 |
|
1 |
2 |
2 |
1 |
|
2 |
4 |
4 |
4 |
|
3 |
6 |
8 |
9 |
|
4 |
8 |
16 |
16 |
|
5 |
10 |
32 |
25 |
|
6 |
12 |
64 |
36 |
|
10 |
20 |
1024 |
100 |
|
Type of function |
Linear \(y = mx+b\) |
Exponential \(y = ab^x\) |
Polynomial\(y = cx^P\) |
|
How to recognize equation for this type of function. |
all terms are first degree; \(m\) is slope; \(b\) is the \(y\) intercept |
base is a number \(b>0\); the variable is in the exponent |
variable is in the base; exponent is a number \(\mathrm{p} \neq 0\) |
|
For equal intervals of change in \(x\), \(y\) increases by a constant amount |
For equal intervals of change in \(x\), \(y\) increases by a constant ratio |
For the functions in the previous table: linear function \(y = f(x) = 2x\), exponential function \(y = g(x) = 2^x\), and polynomial function \(y = h(x) = x^2\), if we restrict the domain to \(x ≥ 0\) only, then all these functions are growth functions. When \(x ≥ 0\), the value of \(y\) increases as the value of \(x\) increases.
The exponential growth function grows large faster than the linear and polynomial functions, as \(x\) gets large. This is always true of exponential growth functions, as \(x\) gets large enough.
Example \(\PageIndex{6}\)
Classify the functions below as exponential, linear, or polynomial functions.
- \(y=10x^3\)
- \(y=-200-30x\)
- \(y=1000(1.05)^x\)
- \(y=500(0.75)^x\)
- \(y=-0.2x^4\)
- \(y=5x-1\)
- \(y=6x^2+3x\)
Solution:
The exponential functions are
c. \(y=1000(1.05)^x\) The variable is in the exponent; the base is the number \(b = 1.05\)
d. \(y=500(0.75)^x\) The variable is in the exponent; the base is the number \(b = 0.75\)
The linear functions are
b. \(y=-200-30x\)
f. \(y=5x-1\)
The polynomial functions are
a. \(y=10x^3\) The variable is the base; the exponent is a fixed number, \(p=3\).
e. \(y=-0.2x^4\) The variable is the base; the exponent is a number, \(p=4\).
g. \(y=6x^2+3x\) The variable is the base; the exponent is a number, \(p = 2\).
NATURAL BASE: e
The number e is often used as the base of an exponential function. e is called the natural base.
e is approximately 2.71828
e is an irrational number with an infinite number of decimals; the decimal pattern never repeats. Section 8.2 includes an example that shows how the value of e is developed and why this number is mathematically important.
When e is the base in an exponential growth or decay function, it is referred to as continuous growth or continuous decay . We will use e in Chapter 8 in financial calculations when we examine interest that compounds continuously.
Any exponential function can be written in the form \(\mathbf{y = ae^{kx}}\)
\(\mathbf{k}\) is called the continuous growth or decay rate.
- If \(k > 0\), the function represents exponential growth
- If \(k< 0\), the function represents exponential decay
\(\mathbf{a}\) is the initial value
We can rewrite the function in the form \(\mathbf{y = ab^x}\) , where \(\mathbf{b=e^k}\)
In general, if we know one form of the equation, we can find the other forms. For now, we have not yet covered the skills to find \(k\) when we know \(b\). After we learn about logarithms later in this chapter, we will find \(k\) using natural log: \(k = \ln b\).
The table below summarizes the forms of exponential growth and decay functions.
|
y = ab x |
y = a(1+r) x |
y = a e kx , k ≠ 0 |
|
|---|---|---|---|
|
Initial value |
a>0 |
a>0 |
a>0 |
|
Relationship between b, r, k |
b > 0 |
b=1+ r |
b = e k and k = ln b |
|
Growth |
b > 1 |
r > 0 |
k > 0 |
|
Decay |
0 < b < 1 |
r < 0 |
k < 0 |
Example \(\PageIndex{7}\)
The value of houses in a city are increasing at a continuous growth rate of 6% per year. For a house that currently costs $400,000:
- Write the exponential growth function in the form \(y=ae^{kx}\).
- What would be the value of this house 4 years from now?
- Rewrite the exponential growth function in the form \(y=ab^x\).
- Find and interpret \(r\).
Solution
a. The initial value of the house is \(a\) = $400000
The problem states that the continuous growth rate is 6% per year, so \(k\) = 0.06
The growth function is : \(y=400000e^{0.06x}\)
b. After 4 years, the value of the house is \(y=400000e^{0.06 (4)}\) = $508,500.
c. To rewrite \(y=400000e^{0.06x}\) in the form \(y = ab^x\), we use the fact that \(b=e^k\).
\[\begin{array}{l}
\mathrm{b}=e^{0.06} \\
\mathrm{b}=1.06183657 \approx 1.0618 \\
\mathrm{y}=400000(1.0618)^{\mathrm{x}}
\end{array}\nonumber\]
d. To find \(r\), we use the fact that \(b=1+ r\)
\[\begin{array}{l}
\mathrm{b}=1.0618 \\
1+\mathrm{r}=1.0618 \\
\mathrm{r}=0.0618
\end{array} \nonumber \]
The value of the house is increasing at an annual rate of 6.18%.
Example \(\PageIndex{8}\)
Suppose that the value of a certain model of new car decreases at a continuous decay rate of 8% per year. For a car that costs $20,000 when new:
- Write the exponential decay function in the form \(y=ae^{kx}\).
- What would be the value of this car 5 years from now?
- Rewrite the exponential decay function in the form \(y=ab^x\).
- Find and interpret \(r\).
Solution
a. The initial value of the car is \(a\) = $20000
The problem states that the continuous decay rate is 8% per year, so \(k\) = -0.08
The growth function is : \(y=20000e^{-0.08x}\)
b. After 5 years, the value of the car is \(y=20000 e^{-0.08 (5)}\) = $13,406.40.
c. To rewrite \(y=20000e^{-0.08x}\) in the form \(y=ab^x\), we use the fact that \(b=e^k\).
\[\begin{array}{l}
\mathrm{b}=e^{-0.08} \\
\mathrm{b}=0.9231163464 \approx 0.9231 \\
\mathrm{y}=20000(0.9231)^{\mathrm{x}}
\end{array} \nonumber \]
d. To find \(r\), we use the fact that \(b=1+ r\)
\[\begin{array}{l}
\mathrm{b}=0.9231 \\
\mathrm{l}+\mathrm{r}=0.9231 \\
\mathrm{r}=0.9231-1=-0.0769
\end{array} \nonumber\]
The value of the car is decreasing at an annual rate of 7.69%.


21.8: Introduction
This handbook is a brief yet comprehensive reference for you to consult as you write papers and other assignments for a college course. You can refer to it as you draft paragraphs and polish sentences for clarity, conciseness, and point of view. You can read it to learn how to identify and revise common sentence errors and confused words. You can use it to help you edit your writing and fine-tune your use of verbs, pronouns, punctuation, and mechanics. And you can have it open as you integrate and cite quotations as well as other source material in your papers in MLA or APA style.
Designed as a reference tool, the handbook is organized to help you get answers to your questions. You do not need to read the entire handbook to get helpful information from it. For example, if your instructor has noted that you need to work on comma splices, you can refer to Sentence Errors, before you turn in a final draft of your writing. If you know you frequently misuse commas, refer to Punctuation, and check your sentences against the advice there. And if you, like many writers, can’t remember which punctuation marks go inside and outside quotation marks, refer to Quotations. Becoming familiar with the handbook and the various topics will allow you to use it efficiently.


12.13.2: Social and Personal
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Social media growth is another trend that continues. Facebook now has over one billion users! In 2013, 80% of Facebook users were outside of the US and Canada. [1] Countries where Facebook is growing rapidly include Indonesia, Mexico, and the Philippines. [2]
Besides Facebook, other social media sites are also seeing tremendous growth. Over 70% of YouTube’s users are outside the US, with the UK, India, Germany, Canada, France, South Korea, and Russia leading the way. [3] Pinterest gets over 50% of its users from outside the US, with over 9% from India. [4] Twitter now has over 230 million active users. [5] Social media sites not based in the US are also growing. China’s QQ instant-messaging service is the eighth most-visited site in the world. [6]
Ever since the advent of Web 2.0 and e-commerce, users of information systems have expected to be able to modify their experiences to meet their personal tastes. From custom backgrounds on computer desktops to unique ringtones on mobile phones, makers of digital devices provide the ability to personalize how we use them. More recently, companies such as Netflix have begun assisting their users with personalizations by making suggestions. In the future, we will begin seeing devices perfectly matched to our personal preferences, based upon information collected about us in the past.


18: Crossings and Candles
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‘It is better to light a candle than curse the darkness’ W. L. Watkinson 1838–1925
An old lesson teaches that endings are more difficult to write than beginnings. This may be so, but I have found it difficult to even begin writing about the world International Relations (IR) makes without reflecting on a near-fortyyear career in both the theory and practice of IR. This is because my intellectual engagement in IR is indivisible from who I am. To make the same point in a slightly more elevated mode, although trained in the tradition that a scholar’s gaze is objective, my academic pilgrimage has been one of continuous crossings between the personal, the political and the professional. My early professional life was conducted during a particularly nasty period of apartheid in South Africa. Not only was the minority-white-ruled government cracking down on all forms of political dissent, it was also wedded to a fierce anti-communism. In these circumstances it was difficult to exercise academic objectivity when it came to thinking about the world. Those years taught me a valuable lesson in life and learning: to believe that there is a totally objective or value-free view in IR is to call up the old Russian saying that ‘he lied like an eye witness!’ We all come to understand the world through our own experiences. Because of this, even the most objective person has predetermined understandings about the world.
A standard dictionary definition of international relations runs that the term ‘is used to identify all interactions between state-based actors across state boundaries’ (Evans and about the future or fearful for one’s family. So we ought to require, perhaps, that a definition does something more than simply demarcate boundaries. A more reflective gaze points to what it is that we, the prospective student or emeritus professor, actually do when we ‘do’ academic IR and why it matters to us.Newnham 1998, 274). This is certainly suggestive of the scholarly field of IR but unhelpful in explaining the international relationships that fall between the cracks of the discipline’s many boundaries and the personal anxiety and fear around these issues. After all, at the height of the Cold War there was real fear that the entire planet would be destroyed by nuclear warfare. In these circumstances, it was difficult not to be anxious about the future or fearful for one’s family. So we ought to require, perhaps, that a definition does something more than simply demarcate boundaries. A more reflective gaze points to what it is that we, the prospective student or emeritus professor, actually do when we ‘do’ academic IR and why it matters to us.


1.1: Identifying Word Parts in Medical Terms
Word Parts
Medical terms are built from word parts. Those word parts are prefix , word root , suffix , and combining form vowel . When a word root is combined with a combining form vowel the word part is referred to as a combining form .
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://ecampusontario.pressbooks.pub/medicalterminology/?p=5
Identifying Word Parts in Medical Terms
By the end of this resource, you will have identified hundreds of word parts within medical terms. Let’s start with some common medical terms that many non-medically trained people may be familiar with.
Osteoarthritis
Oste/o/arthr/itis – Inflammation of bone and joint.
Oste/o is a
combining form
that means bone
arthr/o is a
combining form
that means joint
-itis is a
suffix
that means inflammation
Intravenous
Intra/ven/ous – Pertaining to within a vein.
Intra- is a
prefix
that means within
ven/o – is a
combining form
that means vein
-ous is a
suffix
that means pertaining to
Notice, when breaking down words that you place slashes between word parts and a slash on each side of a combining form vowel .
Language Rules
Language rules are a good place to start when building a medical terminology foundation. Many medical terms are built from word parts and can be translated literally . At first, literal translations sound awkward. Once you build a medical vocabulary and become proficient at using it, the awkwardness will slip away. For example, suffixes will no longer be stated and will be assumed. The definition of intravenous then becomes within the vein .
Since you are at the beginning of building your medical terminology foundation stay literal when applicable. It should be noted that as with all language rules there are always exceptions and we refer to those as rebels . So let’s begin by analyzing the language rules for medical terminology.


4.2: Solve Systems of Linear Equations with Two Variables
By the end of this section, you will be able to:
- Determine whether an ordered pair is a solution of a system of equations
- Solve a system of linear equations by graphing
- Solve a system of equations by substitution
- Solve a system of equations by elimination
- Choose the most convenient method to solve a system of linear equations
Before you get started, take this readiness quiz.
-
For the equation \(y=\frac{2}{3}x−4\),
ⓐ Is \((6,0)\) a solution? ⓑ Is \((−3,−2)\) a solution?
If you missed this problem, review [link] . -
Find the slope and
y
-intercept of the line \(3x−y=12\).
If you missed this problem, review [link] . -
Find the
x-
and
y
-intercepts of the line \(2x−3y=12\).
If you missed this problem, review [link] .
Determine Whether an Ordered Pair is a Solution of a System of Equations
In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .
When two or more linear equations are grouped together, they form a system of linear equations .
In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.
An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.
\[ \left\{ \begin{aligned} 2x+y & = 7 \\ x−2y & = 6 \end{aligned} \right. \nonumber \]
A linear equation in two variables, such as \(2x+y=7\), has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.
To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs \((x,y)\) that make both equations true. These are called the solutions of a system of equations .
The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair \((x,y)\).
To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.
Determine whether the ordered pair is a solution to the system \(\left \{ \begin{array} {l} x−y = −1 \\ 2x−y = −5 \end{array} \right. \).
ⓐ \((−2,−1)\) ⓑ \((−4,−3)\)
- Answer
-
ⓐ
ⓑ
Determine whether the ordered pair is a solution to the system \(\left \{ \begin{array} 3x+y = 0 \\ x+2y = −5 \end{array} \right. \).
ⓐ \((1,−3)\) ⓑ \((0,0)\)
- Answer
-
ⓐ yes ⓑ no
Determine whether the ordered pair is a solution to the system \(\left \{ \begin{array} x−3y = −8 \\ −3x−y = 4 \end{array} \right. \).
ⓐ \((2,−2)\) ⓑ \((−2,2)\)
- Answer
-
ⓐ no ⓑ yes
Solve a System of Linear Equations by Graphing
In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.
The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.
Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.
Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.
Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.
Solve the system by graphing \( \left\{ \begin{array} {l} 2x+y = 7 \\ x−2y = 6 \end{array} \right. \).
- Answer
-
Solve the system by graphing: \( \left\{ \begin{array} {l} x−3y = −3 \\ x+y = 5 \end{array} \right. \).
- Answer
-
\((3,2)\)
Solve the system by graphing: \( \left\{ \begin{array} {l} −x+y = 1 \\ 3x+2y = 12 \end{array} \right.\)
- Answer
-
\((2,3)\)
The steps to use to solve a system of linear equations by graphing are shown here.
- Graph the first equation.
- Graph the second equation on the same rectangular coordinate system.
- Determine whether the lines intersect, are parallel, or are the same line.
-
Identify the solution to the system.
- If the lines intersect, identify the point of intersection. This is the solution to the system.
- If the lines are parallel, the system has no solution.
- If the lines are the same, the system has an infinite number of solutions.
- Check the solution in both equations.
In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.
Solve the system by graphing: \(\left\{ \begin{array} {l} 3x+y = −1 \\ 2x+y = 0 \end{array}\right.\)
- Answer
-
We’ll solve both of these equations for \(y\) so that we can easily graph them using their slopes and \(y\)-intercepts.
Solve the first equation for y . Find the slope and y -intercept. Solve the second equation for y . Find the slope and y -intercept. Graph the lines. Determine the point of intersection. The lines intersect at \((−1,2)\). Check the solution in both equations.
The solution is \((−1,2)\).
Solve the system by graphing: \(\left\{ \begin{array} {l} −x+y = 1 \\2x+y = 10 \end{array}\right. \).
- Answer
-
\((3,4)\)
Solve the system by graphing: \(\left\{ \begin{array} {l} 2x+y = 6 \\x+y = 1 \end{array}\right. \).
- Answer
-
\((5,−4)\)
In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.
Solve the system by graphing: \(\left\{ \begin{array} {l} y = \tfrac{1}{2}x-3 \\ x-2y = 4 \end{array}\right. \).
- Answer
-
To graph the first equation, we will use its
slope and y -intercept.To graph the second equation, we will use
the intercepts.Graph the lines. Determine the points of intersection. The lines are parallel.
Since no point is on both lines, there is no
ordered pair that makes both equations
true. There is no solution to this system.
Solve the system by graphing: \(\left\{ \begin{array} {l} y = -\tfrac{1}{4}x+2 \\ x+4y = 4 \end{array}\right. \).
- Answer
-
no solution
Solve the system by graphing: \(\left\{ \begin{array} {l} y = 3x-1 \\ 6x-2y = 6 \end{array}\right. \).
- Answer
-
no solution
Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.
Solve the system by graphing: \(\left\{ \begin{array} {l} y = 2x-3 \\ -6x+3y = 9 \end{array}\right. \).
- Answer
-
Find the slope and y -intercept of the first equation. Find the intercepts of the second equation. Graph the lines. The lines are the same!
Since every point on the line makes both
equations true, there are infinitely many
ordered pairs that make both equations true.
There are infinitely many solutions to this system.If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.
Solve the system by graphing: \(\left\{ \begin{array} {l} y = -3x-6 \\ 6x+2y = -12 \end{array}\right. \).
- Answer
-
infinitely many solutions
Solve the system by graphing: \(\left\{ \begin{array} {l} y = \tfrac{1}{2}x-4 \\ 2x-4y = 16 \end{array}\right. \).
- Answer
-
infinitely many solutions
When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.
Coincident lines have the same slope and same y- intercept.
The systems of equations in Example and Example each had two intersecting lines. Each system had one solution.
In Example , the equations gave coincident lines, and so the system had infinitely many solutions.
The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.
A system with parallel lines, like Example , has no solution. We call a system of equations like this inconsistent. It has no solution.
A consistent system of equations is a system of equations with at least one solution.
An inconsistent system of equations is a system of equations with no solution.
We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.
If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.
Let’s sum this up by looking at the graphs of the three types of systems. See below and Table .
| Lines | Intersecting | Parallel | Coincident |
|---|---|---|---|
| Number of solutions | 1 point | No solution | Infinitely many |
| Consistent/inconsistent | Consistent | Inconsistent | Consistent |
| Dependent/ independent | Independent | Independent | Dependent |
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ \( \left\{ \begin{array} {l} y = 3x−1 \\ 6x−2y = 12 \end{array}\right. \) ⓑ \( \left\{ \begin{array} {l} 2x+y=−3 \\ x−5y=5 \end{array} \right. \)
- Answer
-
ⓐ We will compare the slopes and intercepts of the two lines.
\(\begin{array} {lll} {} &{} &{ \left\{ \begin{array} {l} {y=3x-1} \\ {6x−2y=12} \end{array} \right. } \\ {} &{} &{y = 3x-1} \\ {\text{The first equation is already in slope-intercept form.}} &{} &{} \\ {\text{Write the second equation in slope-intercept form.}} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{6x-2y=12} \\ {} &{} &{-2y=-6x+12} \\ {} &{} &{\frac{-2y}{-2}=\frac{-6x+12}{-2}} \\ {} &{} &{y=3x-6} \\ {} &{y=3x-1} &{y=3x-6} \\ {} &{m=3} &{m=3} \\ {} &{b=-1} &{b=-6} \\ {\text{Find the slope and intercept of each line.}} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{\text{Since the slopes are the same andy-intercepts are}} &{} \\ {} &{\text{different, the lines are parallel.}} &{} \\ \end{array}\)
ⓑ We will compare the slope and intercepts of the two lines.
\(\begin{array} {lll} {} &{} &{} \\ {} &{ \left\{ \begin{array} {l} 2x+y=-3 \\ x-5y=5 \\ \end{array} \right. } &{} \\ {\text{Write both equations in slope–intercept form.}} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{2x+y=-3} &{x-5y=5} \\ {} &{y=-2x-3} &{-5y=-x+5} \\ {} &{} &{\frac{-5y}{-5}=\frac{-x+5}{-5}} \\ {} &{} &{y=\frac{1}{5}-1} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {} &{} &{} \\ {\text{Find the slope and intercept of each line.}} &{} &{} \\ {} &{} &{} \\ {} &{y=-2x-3} &{y=\frac{1}{5}-1} \\ {} &{m=-2} &{m=\frac{1}{5}} \\ {} &{b=-3} &{b=-1} \\ {} &{} &{} \\ {} &{\text{Since the slopes are different, the lines intersect.}} &{} \\ \end{array}\)
A system of equations whose graphs are intersect has 1 solution and is consistent and independent.
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ \(\left\{ \begin{array} {l} y=−2x−4 \\ 4x+2y=9 \end{array} \right. \) ⓑ \(\left\{ \begin{array} {l} 3x+2y=2 \\ 2x+y=1 \end{array} \right. \)
- Answer
-
ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ \(\left\{ \begin{array} {l} y=\frac{1}{3}x−5 \\ x−3y=6 \end{array} \right. \) ⓑ \(\left\{ \begin{array} {l} x+4y=12 \\ −x+y=3 \end{array} \right. \)
- Answer
-
ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between \(−10\) and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.
Solve a System of Equations by Substitution
We will now solve systems of linear equations by the substitution method.
We will use the same system we used first for graphing.
\[ \left\{ \begin{array} {l} 2x+y=7 \\ x−2y=6 \end{array} \right. \nonumber \]
We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.
Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!
After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.
Solve the system by substitution: \( \left\{ \begin{array} {l} 2x+y=7 \\ x−2y=6 \end{array} \right. \)
- Answer
-
Solve the system by substitution: \( \left\{ \begin{array} {l} −2x+y=−11 \\ x+3y=9 \end{array} \right. \)
- Answer
-
\((6,1)\)
Solve the system by substitution: \( \left\{ \begin{array} {l} 2x+y=−1 \\ 4x+3y=3 \end{array} \right. \)
- Answer
-
\((−3,5)\)
- Solve one of the equations for either variable.
- Substitute the expression from Step 1 into the other equation.
- Solve the resulting equation.
- Substitute the solution in Step 3 into either of the original equations to find the other variable.
- Write the solution as an ordered pair.
- Check that the ordered pair is a solution to both original equations.
Be very careful with the signs in the next example.
Solve the system by substitution: \( \left\{ \begin{array} {l} 4x+2y=4 \\ 6x−y=8 \end{array} \right. \)
- Answer
-
We need to solve one equation for one variable. We will solve the first equation for y .
Solve the first equation for y .
Substitute \(−2x+2\) for y in the second equation.Replace the y with \(−2x+2\). Solve the equation for x . Substitute \(x=54\) into \(4x+2y=4\) to find y . The ordered pair is \((54,−12)\). Check the ordered pair in both equations.
The solution is \((54,−12)\).
Solve the system by substitution: \( \left\{ \begin{array} {l} x−4y=−4 \\ −3x+4y=0 \end{array} \right. \)
- Answer
-
\((2,32)\)
Solve the system by substitution: \( \left\{ \begin{array} {l} 4x−y=0 \\ 2x−3y=5 \end{array} \right. \)
- Answer
-
\((−12,−2)\)
Solve a System of Equations by Elimination
We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.
The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.
The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.
For any expressions a, b, c, and d .
\[\begin{array} {ll} {\text{if}} &{a=b} \\ {\text{and}} &{c=d} \\ {\text{then}} &{a+c=b+d.} \\ \nonumber \end{array}\]
To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.
Notice how that works when we add these two equations together:
\[\left\{ \begin{array} {l} 3x+y=5 \\ \underline{2x−y=0} \end{array} \right. \nonumber\]
\[5x=5 \nonumber\]
The y ’s add to zero and we have one equation with one variable.
Let’s try another one:
\[ \left\{ \begin{array} x+4y=2 \\ 2x+5y=−2 \end{array} \right. \nonumber\]
This time we don’t see a variable that can be immediately eliminated if we add the equations.
But if we multiply the first equation by \(−2\), we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by \(−2\).
Then rewrite the system of equations.
Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.
Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.
Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.
Solve the system by elimination: \(\left\{ \begin{array} {l} 2x+y=7 \\ x−2y=6 \end{array} \right. \)
- Answer
-
Solve the system by elimination: \(\left\{ \begin{array} {l} 3x+y=5 \\ 2x−3y=7 \end{array} \right.\)
- Answer
-
\((2,−1)\)
Solve the system by elimination: \(\left\{ \begin{array} {l} 4x+y=−5 \\ −2x−2y=−2 \end{array} \right.\)
- Answer
-
\((−2,3)\)
The steps are listed here for easy reference.
- Write both equations in standard form. If any coefficients are fractions, clear them.
-
Make the coefficients of one variable opposites.
- Decide which variable you will eliminate.
- Multiply one or both equations so that the coefficients of that variable are opposites.
- Add the equations resulting from Step 2 to eliminate one variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Write the solution as an ordered pair.
- Check that the ordered pair is a solution to both original equations.
Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.
Solve the system by elimination: \(\left\{ \begin{array} {l} 4x−3y=9 \\ 7x+2y=−6 \end{array} \right. \)
- Answer
-
In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.
Both equations are in standard form.
To get opposite coefficients of y , we will
multiply the first equation by 2 and the
second equation by 3.Simplify. Add the two equations to eliminate y . Solve for x . Substitute x=0x=0 into one of the original equations. Solve for y . Write the solution as an ordered pair. The ordered pair is \((0,−3)\).
Check that the ordered pair is a solution to
both original equations.
The solution is \((0,−3)\).
Solve the system by elimination: \(\left\{ \begin{array} {l} 3x−4y=−9 \\ 5x+3y=14\end{array} \right. \)
- Answer
-
\((1,3)\)
Solve each system by elimination: \(\left\{ \begin{array} {l} 7x+8y=4 \\ 3x−5y=27 \end{array} \right.\)
- Answer
-
\((4,−3)\)
When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.
Solve the system by elimination: \(\left\{ \begin{array} {l} x+\tfrac{1}{2}y=6 \\ \tfrac{3}{2}x+\tfrac{2}{3}y=\tfrac{17}{2} \end{array} \right.\)
- Answer
-
In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.
To clear the fractions, multiply each
equation by its LCD.Simplify. Now we are ready to eliminate one
of the variables. Notice that both equations are in
standard form.We can eliminate y by multiplying the top equation by \(−4\). Simplify and add.
Substitute \(x=3\) into one of the original equations.Solve for y.
Write the solution as an ordered pair. The ordered pair is \((3,6)\). Check that the ordered pair is a solution to
both original equations.
The solution is \((3,6)\).
Solve each system by elimination: \(\left\{ \begin{array} {l} \tfrac{1}{3}x−\tfrac{1}{2}y=1 \\ \tfrac{3}{4}x−y=\tfrac{5}{2} \end{array} \right.\)
- Answer
-
\((6,2)\)
Solve each system by elimination: \(\left\{ \begin{array} {l} x+\tfrac{3}{5}y=−\tfrac{1}{5} \\ −\tfrac{1}{2}x−\tfrac{2}{3}y=\tfrac{5}{6} \end{array} \right.\)
- Answer
-
\((1,−2)\)
When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.
The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.
Solve the system by elimination: \(\left\{ \begin{array} {l} 3x+4y=12 \\ y=3−\tfrac{3}{4}x \end{array} \right. \)
- Answer
-
\(\begin{array} {ll} {} &{ \left\{ \begin{array} {l} 3x+4y=12 \\ y=3−\frac{3}{4}x \end{array} \right.} \\ {} &{} \\ {\text{Write the second equation in standard form.}} &{\left\{ \begin{array} {l} 3x+4y=12 \\ \frac{3}{4}x+y=3 \end{array} \right. } \\ {} &{} \\ {\text{Clear the fractions by multiplying the } \\ \text{second equation by 4.}} &{\left\{ \begin{array} {l} 3x+4y=12 \\ 4(\frac{3}{4}x+y)=4(3) \end{array} \right. } \\ {} &{} \\ {\text{Simplify.}} &{\left\{ \begin{array} {l} 3x+4y=12 \\ 3x+4y=12 \end{array} \right. } \\ {} &{} \\ {\text{To eliminate a variable, we multiply the} \\ \text{second equation by−1. Simplify and add.}} &{\begin{array} {l} {\left\{ \begin{array} {l} 3x+4y=12 \\ \underline{-3x-4y=-12 } \end{array} \right.} \\ {\hspace{16mm} 0=0} \end{array}} \\ \end{array} \)
This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.
After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.
Solve the system by elimination: \(\left\{ \begin{array} {l} 5x−3y=15 \\ 5y=−5+\tfrac{5}{3}x \end{array} \right. \)
- Answer
-
infinitely many solutions
Solve the system by elimination: \(\left\{ \begin{array} {l} x+2y=6 \\ y=−\tfrac{1}{2}x+3\end{array} \right. \)
- Answer
-
infinitely many solutions
Choose the Most Convenient Method to Solve a System of Linear Equations
When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.
\[ \textbf{Choose the Most Convenient Method to Solve a System of Linear Equations} \\ \begin{array} {lll} {\underline{\textbf{Graphing}}} &{\underline{\textbf{Substitution}}} &{\underline{\textbf{Elimination}}} \\ {\text{Use when you need a}} &{\text{Use when one equation is}} &{\text{Use when the equations a}} \\ {\text{picture of the situation.}} &{\text{already solved or can be}} &{\text{rein standard form.}} \\ {\text{}} &{\text{easily solved for one}} &{\text{}} \\ {\text{}} &{\text{variable.}} &{\text{}} \\ \end{array} \nonumber \]
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ \(\left\{ \begin{array} {l} 3x+8y=40 \\ 7x−4y=−32 \end{array} \right.\) ⓑ \(\left\{ \begin{array} {l} 5x+6y=12 \\ y=\tfrac{2}{3}x−1 \end{array} \right.\)
- Answer
-
ⓐ
\[\left\{ \begin{array} {l} 3x+8y=40 \\ 7x−4y=−32 \end{array} \right.\nonumber\]
Since both equations are in standard form, using elimination will be most convenient.
ⓑ
\[\left\{ \begin{array} {l} 5x+6y=12 \\ y=\tfrac{2}{3}x−1 \end{array} \right.\nonumber \]
Since one equation is already solved for y , using substitution will be most convenient.
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ \(\left\{ \begin{array} {l} 4x−5y=−32 \\ 3x+2y=−1 \end{array} \right.\) ⓑ \(\left\{ \begin{array} {l} x=2y−1 \\ 3x−5y=−7 \end{array} \right.\)
- Answer
-
ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x , using substitution will be most convenient.
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ \(\left\{ \begin{array} {l} y=2x−1 \\ 3x−4y=−6 \end{array} \right.\) ⓑ \(\left\{ \begin{array} {l} 6x−2y=12 \\ 3x+7y=−13 \end{array} \right.\)
- Answer
-
ⓐ Since one equation is already solved for y , using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.
Key Concepts
-
How to solve a system of linear equations by graphing.
- Graph the first equation.
- Graph the second equation on the same rectangular coordinate system.
- Determine whether the lines intersect, are parallel, or are the same line.
-
Identify the solution to the system.
If the lines intersect, identify the point of intersection. This is the solution to the system.
If the lines are parallel, the system has no solution.
If the lines are the same, the system has an infinite number of solutions. - Check the solution in both equations.
-
How to solve a system of equations by substitution.
- Solve one of the equations for either variable.
- Substitute the expression from Step 1 into the other equation.
- Solve the resulting equation.
- Substitute the solution in Step 3 into either of the original equations to find the other variable.
- Write the solution as an ordered pair.
- Check that the ordered pair is a solution to both original equations.
-
How to solve a system of equations by elimination.
- Write both equations in standard form. If any coefficients are fractions, clear them.
-
Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites. - Add the equations resulting from Step 2 to eliminate one variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Write the solution as an ordered pair.
- Check that the ordered pair is a solution to both original equations. \[ \textbf{Choose the Most Convenient Method to Solve a System of Linear Equations} \\ \begin{array} {lll} {\underline{\textbf{Graphing}}} &{\underline{\textbf{Substitution}}} &{\underline{\textbf{Elimination}}} \\ {\text{}} &{\text{Use when one equation is}} &{\text{}} \\ {\text{Use when you need a}} &{\text{already solved or can be}} &{\text{Use when the equations a}} \\ {\text{picture of the situation.}} &{\text{easily solved for one}} &{\text{rein standard form.}} \\ {\text{}} &{\text{variable.}} &{\text{}} \\ \end{array} \nonumber \]
Glossary
- coincident lines
- Coincident lines have the same slope and same y -intercept.
- consistent and inconsistent systems
- Consistent system of equations is a system of equations with at least one solution; inconsistent system of equations is a system of equations with no solution.
- solutions of a system of equations
- Solutions of a system of equations are the values of the variables that make all the equations true; solution is represented by an ordered pair (x,y).(x,y).
- system of linear equations
- When two or more linear equations are grouped together, they form a system of linear equations.


6: Trade-Area Analysis and Site Selection Last updated Save as PDF Page ID 44964 6.1: Introduction to Available Retail Locations 6.2: Comparing Location Benefits 6.3: Introduction to Site Selection Decisions 6.4: Evaluating Site Characteristics 6.5: Influence of Societal Views 6.6: Huff Gravity Model 6.7: Common Lease Types 6.8: Putting It Together- Trade-Area Analysis and Site Selection 6.9: Discussion- Trade-Area Analysis and Site Selection 6.10: Why It Matters- Trade-Area Analysis and Site Selection 6.11: Introduction to Selecting a Store Location 6.12: Environmental Analysis 6.13: Data Analysis for Location Selection 6.14: Location Effects on Planning 6.15: Introduction to Retail Trading Area 6.16: U.S. Census Bureau and Demographic Data 6.17: GIS and Location Decisions 6.18: Factors in Analyzing Trading Area


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adaptive model
-
method of entrepreneurial problem solving that seeks solutions in ways that are tested and known to be effective
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assessment
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lean problem-solving process phase in which the entrepreneur experiments and analyzes the potential process and its capabilities
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brainstorming
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generation of ideas in an environment free of judgment or dissension with the goal of creating solutions
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clarify
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first step of the creative process, which involves recognizing the existence of a gap between the current state and a desired state
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communication skills
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skills entrepreneurs use to pool resources for the purposes of investigating solutions leading to innovative problem solving and competitive advantage
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creativity
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development of original ideas
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critical thinking
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complex analysis of a problem or issue with the goal of solving the problem or making a decision
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crowdsourcing
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teams of amateurs and non-experts working together to form a solution to a problem
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data analysis
-
process of analyzing data and modeling the data into a structure that leads to innovative conclusions
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decisiveness
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ability to make a quick, effective decision, not letting too much time go by in the process
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design thinking
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method to focus the design and development decisions of a product on the needs of the customer, typically involving an empathy-driven process to define complex problems and create solutions that address those problems
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develop
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stage in which the entrepreneur takes the list of ideas generated and tests each solution for feasibility
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entrepreneurial problem solving
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process of using innovation and creative solutions to close the opportunity gap by resolving complex societal, business, or technological problems
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evaluate
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stage in which the final solution is assessed
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evaluation
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lean problem-solving process phase in which behaviors are analyzed to assess success
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external environment
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outside the enterprise and encompasses customers, industry trends, and competition
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hackathon
-
event, usually hosted by a tech company or organization, which brings together programmers and workers with other degrees of specialization within the company, community, or organization to collaborate on a project over a short period of time
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human-centered design
-
focuses on people during design and development
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ideate
-
stage of the creative problem-solving process leading to the detailing and formation of idea generation by the entrepreneur
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implement
-
means to enact the plan and test for success
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innovative model
-
method of entrepreneurial problem solving that uses techniques unknown to the market and that bring competitive advantage to an organization
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internal environment
-
comprises the factors inside the enterprise, such as employees, and internal practices and processes
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lean process
-
systematic process for the maximizing of continuous improvement and the minimization of surplus or unused material in the production of a process
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need awareness
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occurs when the entrepreneur notes a gap between societal or customer needs and actual circumstances
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observation
-
lean problem-solving process phase in which the entrepreneur studies the challenge and notes all facets of the challenge requiring solution
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petitioner problem solvers
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problem solvers who see a problem and ask others for solution ideas
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resourcefulness
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ability to discover clever solutions to obstacles
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self-regulating problem solvers
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problem solvers who are autonomous and work on their own, without external influence; they have the ability to see a problem, immediately visualize a possible solution to the problem, and seek to devise a solution
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storyboarding
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process of presenting an idea in a step-by-step graphic format
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team creativity
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process of a team creating an unexpected solution for an issue or challenge
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theorist problem solvers
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problem solvers who see a problem and begin to consider a path toward solving the problem using a theory
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whiteboarding
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type of graphing that permits the entrepreneur to plot each step in a process to build comprehension and detailing of the process


 Are you considered a sedentary person?
Activity: Your daily movement patterns
Keep a journal and record your movement throughout the day. It is best to record a regular work day and also a weekend day.
Track the time you spend sitting, laying down, doing low intensity activities of daily living, and purposeful exercise.
Writing down a diary can help you to clearly see your active and inactive patterns in your life to help you identify opportunities for increasing physical activity.
A person who is described as sedentary means that they are sitting, reclining, or lying down. Most desk-based office work, driving a car, and watching television are examples of sedentary behaviors. It is important to limit the amount of time spent being sedentary. Replacing sedentary time with physical activity of any intensity (including light intensity) provides health benefits. If you are a sedentary person, the first thing you should do is identify opportunities throughout your day to move more and sit less.
Remember:
- Some physical activity is better than none.
- Everything counts! Move more!
Examples of strategies to increase activity throughout your day [5] :
- Walk instead of drive, whenever you can
- Walk your children to school
- Take the stairs instead of the escalator or elevator
- Take a family walk after dinner
- Replace a Sunday drive with a Sunday walk
- Go for a half-hour walk instead of watching TV
- Get off the bus a stop early, and walk
- Park farther from the store and walk
- Make a Saturday morning walk a family habit
- Walk briskly in the mall
- Take the dog on longer walks
- Go up hills instead of around them
- Garden, or make home repairs
- Do yard work. Get your children to help rake, weed, or plant
- Work around the house. Ask your children to help with active chores
- Wash the car by hand
- Use a rake instead of a leaf blower
- Avoid labor-saving devices, such as a remote control or electric mixers
- Do sit-ups in front of the TV. Have a sit-up competition with your children/friends
Some possible ways that fitness and health outcomes may relate to physical activity are:
- Physical activity leads to improvements in physical fitness, and physical fitness causes improvements in health outcomes;
- Physical fitness may modify the amount of the effect that physical activity has on health outcomes; or
- Physical activity can lead to improved physical fitness as a health outcome.


19.6: Putting It Together- Human Resources in Small and Entrepreneurial Businesses
As we discussed earlier, small businesses account for 99% of the businesses in the United States, and almost half of the country works for a small business. As an HR professional, you could find yourself happily employed for a small business. There are many reasons to work for a small business, such as:
- You are likely to have some autonomy to work independently
- It can be easier to put ideas into action
- There may be greater job satisfaction and opportunity for job growth
- You may have an increased sense of ownership
- The company is able to react quickly to change
…and so many more. The reasons why you might find yourself happy at a small business are the same reasons others are attracted to them. Leverage those great qualities to bring like-minded individuals into the company with you. You’re armed with the information to make small business HR a success.
Contributors and Attributions
- Putting It Together: Human Resources in Small and Entrepreneurial Businesses. Authored by : Freedom Learning Group. Provided by : Lumen Learning. License : CC BY: Attribution
- Authored by : WOCinTech Chat. Provided by : Flickr. Located at : https://www.flickr.com/photos/wocintechchat/25900820842/ . License : CC BY: Attribution