5.10: Ratios and Rate (Part 1)
- Write a ratio as a fraction
- Write a rate as a fraction
- Find unit rates
- Find unit price
- Translate phrases to expressions with fractions
Before you get started, take this readiness quiz.
- Simplify: \(\dfrac{16}{24}\). If you missed this problem, review Example 4.3.1 .
- Divide: 2.76 ÷ 11.5. If you missed this problem, review Example 5.4.9 .
- Simplify: \(\dfrac{1 \dfrac{1}{2}}{2 \dfrac{3}{4}}\). If you missed this problem, review Example 4.5.7 .
Write a Ratio as a Fraction
When you apply for a mortgage, the loan officer will compare your total debt to your total income to decide if you qualify for the loan. This comparison is called the debt-to-income ratio. A ratio compares two quantities that are measured with the same unit. If we compare a and b , the ratio is written as a to b, \(\dfrac{a}{b}\), or a:b.
A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of a to b is written a to b, \(\dfrac{a}{b}\), or a:b.
In this section, we will use the fraction notation. When a ratio is written in fraction form, the fraction should be simplified. If it is an improper fraction, we do not change it to a mixed number. Because a ratio compares two quantities, we would leave a ratio as \(\dfrac{4}{1}\) instead of simplifying it to 4 so that we can see the two parts of the ratio.
Write each ratio as a fraction: (a) 15 to 27 (b) 45 to 18.
Solution
(a) 15 to 27
| Write as a fraction with the first number in the numerator and the second in the denominator. | $$\dfrac{15}{27}$$ |
| Simplify the fraction. | $$\dfrac{5}{9}$$ |
(b) 45 to 18
| Write as a fraction with the first number in the numerator and the second in the denominator. | $$\dfrac{45}{18}$$ |
| Simplify. | $$\dfrac{5}{2}$$ |
We leave the ratio in (b) as an improper fraction.
Write each ratio as a fraction: (a) 21 to 56 (b) 48 to 32.
- Answer a
-
\(\dfrac{3}{8}\)
- Answer b
-
\(\dfrac{3}{2}\)
Write each ratio as a fraction: (a) 27 to 72 (b) 51 to 34.
- Answer a
-
\(\dfrac{1}{1}\)
- Answer b
-
\(\dfrac{3}{2}\)
Ratios Involving Decimals
We will often work with ratios of decimals, especially when we have ratios involving money. In these cases, we can eliminate the decimals by using the Equivalent Fractions Property to convert the ratio to a fraction with whole numbers in the numerator and denominator.
For example, consider the ratio 0.8 to 0.05. We can write it as a fraction with decimals and then multiply the numerator and denominator by 100 to eliminate the decimals.
\[\dfrac{0.8}{0.05}\]
\[\dfrac{(0.8) \textcolor{red}{100}}{(0.05) \textcolor{red}{100}}\]
\[\dfrac{80}{5}\]
Do you see a shortcut to find the equivalent fraction? Notice that 0.8 = \(\dfrac{8}{10}\) and 0.05 = \(\dfrac{5}{100}\). The least common denominator of \(\dfrac{8}{10}\) and 5 100 is 100. By multiplying the numerator and denominator of \(\dfrac{0.8}{0.05}\) by 100, we ‘moved’ the decimal two places to the right to get the equivalent fraction with no decimals. Now that we understand the math behind the process, we can find the fraction with no decimals like this:
| "Move" the decimal 2 places. | $$\dfrac{80}{5}$$ |
| Simplify. | $$\dfrac{16}{1}$$ |
You do not have to write out every step when you multiply the numerator and denominator by powers of ten. As long as you move both decimal places the same number of places, the ratio will remain the same.
Write each ratio as a fraction of whole numbers: (a) 4.8 to 11.2 (b) 2.7 to 0.54
Solution
(a) 4.8 to 11.2
| Write as a fraction. | $$\dfrac{4.8}{11.2}$$ |
| Rewrite as an equivalent fraction without decimals, by moving both decimal points 1 place to the right. | $$\dfrac{48}{112}$$ |
| Simplify. | $$\dfrac{3}{7}$$ |
So 4.8 to 11.2 is equivalent to \(\dfrac{3}{7}\).
(b) 2.7 to 0.54
| Write as a fraction. | $$\dfrac{2.7}{0.54}$$ |
| The numerator has one decimal place and the denominator has 2. To clear both decimals we need to move the decimal 2 places to the right. | $$\dfrac{270}{54}$$ |
| Simplify. | $$\dfrac{5}{1}$$ |
So 2.7 to 0.54 is equivalent to \(\dfrac{5}{1}\).
Write each ratio as a fraction: (a) 4.6 to 11.5 (b) 2.3 to 0.69.
- Answer a
-
\(\dfrac{2}{5}\)
- Answer b
-
\(\dfrac{10}{3}\)
Write each ratio as a fraction: (a) 3.4 to 15.3 (b) 3.4 to 0.68.
- Answer a
-
\(\dfrac{2}{9}\)
- Answer b
-
\(\dfrac{5}{1}\)
Some ratios compare two mixed numbers. Remember that to divide mixed numbers, you first rewrite them as improper fractions.
Write the ratio of \(1 \dfrac{1}{4}\) to \(2 \dfrac{3}{8}\) as a fraction.
Solution
| Write as a fraction. | $$\dfrac{1 \dfrac{1}{4}}{2 \dfrac{3}{8}}$$ |
| Convert the numerator and denominator to improper fractions. | $$\dfrac{\dfrac{5}{4}}{\dfrac{19}{8}}$$ |
| Rewrite as a division of fractions. | $$\dfrac{5}{4} \div \dfrac{19}{8}$$ |
| Invert the divisor and multiply. | $$\dfrac{5}{4} \cdot \dfrac{8}{19}$$ |
| Simplify. | $$\dfrac{10}{19}$$ |
Write each ratio as a fraction: \(1 \dfrac{3}{4}\) to \(2 \dfrac{5}{8}\).
- Answer
-
\(\dfrac{2}{3}\
Write each ratio as a fraction: \(1 \dfrac{1}{8}\) to \(2 \dfrac{3}{4}\).
- Answer
-
\(\dfrac{9}{22}\)
Applications of Ratios
One real-world application of ratios that affects many people involves measuring cholesterol in blood. The ratio of total cholesterol to HDL cholesterol is one way doctors assess a person's overall health. A ratio of less than 5 to 1 is considered good.
Hector's total cholesterol is 249 mg/dl and his HDL cholesterol is 39 mg/dl. (a) Find the ratio of his total cholesterol to his HDL cholesterol. (b) Assuming that a ratio less than 5 to 1 is considered good, what would you suggest to Hector?
Solution
(a) First, write the words that express the ratio. We want to know the ratio of Hector's total cholesterol to his HDL cholesterol.
| Write as a fraction. | $$\dfrac{total\; cholesterol}{HDL\; cholesterol}$$ |
| Substitute the values. | $$\dfrac{249}{39}$$ |
| Simplify. | $$\dfrac{83}{13}$$ |
(b) Is Hector's cholesterol ratio ok? If we divide 83 by 13 we obtain approximately 6.4, so \(\dfrac{83}{13} \approx \dfrac{6.4}{1}\). Hector's cholesterol ratio is high! Hector should either lower his total cholesterol or raise his HDL cholesterol.
Find the patient's ratio of total cholesterol to HDL cholesterol using the given information. Total cholesterol is 185 mg/dL and HDL cholesterol is 40 mg/dL.
- Answer
-
\(\dfrac{37}{8}\
Find the patient’s ratio of total cholesterol to HDL cholesterol using the given information. Total cholesterol is 204 mg/dL and HDL cholesterol is 38 mg/dL.
- Answer
-
\(\dfrac{102}{19}\
Ratios of Two Measurements in Different Units
To find the ratio of two measurements, we must make sure the quantities have been measured with the same unit. If the measurements are not in the same units, we must first convert them to the same units.
We know that to simplify a fraction, we divide out common factors. Similarly in a ratio of measurements, we divide out the common unit.
The Americans with Disabilities Act (ADA) Guidelines for wheel chair ramps require a maximum vertical rise of 1 inch for every 1 foot of horizontal run. What is the ratio of the rise to the run?
Solution
In a ratio, the measurements must be in the same units. We can change feet to inches, or inches to feet. It is usually easier to convert to the smaller unit, since this avoids introducing more fractions into the problem. Write the words that express the ratio.
| Write the ratio as a fraction. | $$\dfrac{rise}{run}$$ |
| Substitute in the given values. | $$\dfrac{1\; inch}{1\; foot}$$ |
| Convert 1 foot to inches. | $$\dfrac{1\; inch}{12\; inches}$$ |
| Simplify, dividing out common factors and units. | $$\dfrac{1}{12}$$ |
So the ratio of rise to run is 1 to 12. This means that the ramp should rise 1 inch for every 12 inches of horizontal run to comply with the guidelines.
Find the ratio of the first length to the second length: 32 inches to 1 foot.
- Answer
-
\(\dfrac{8}{3}\
Find the ratio of the first length to the second length: 1 foot to 54 inches.
- Answer a
-
\(\dfrac{2}{9}\
Write a Rate as a Fraction
Frequently we want to compare two different types of measurements, such as miles to gallons. To make this comparison, we use a rate . Examples of rates are 120 miles in 2 hours, 160 words in 4 minutes, and $5 dollars per 64 ounces.
A rate compares two quantities of different units. A rate is usually written as a fraction.
When writing a fraction as a rate, we put the first given amount with its units in the numerator and the second amount with its units in the denominator. When rates are simplified, the units remain in the numerator and denominator.
Bob drove his car 525 miles in 9 hours. Write this rate as a fraction.
Solution
| Write as a fraction, with 525 miles in the numerator and 9 hours in the denominator. | $$\dfrac{525\; miles}{9\; hours}$$ |
| $$\dfrac{175\; miles}{3\; hours}$$ |
So 525 miles in 9 hours is equivalent to \(\dfrac{175\; miles}{3\; hours}\).
Write the rate as a fraction: 492 miles in 8 hours.
- Answer
-
\(\dfrac{123\; miles}{2\; hours}\)
Write the rate as a fraction: 242 miles in 6 hours.
- Answer
-
\(\dfrac{121\; miles}{3\; hours}\)
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


2 .
How would you respond as a nurse if your patient wanted to continue pushing efforts beyond 3 or 4 hours? Compare this to how you would respond if a patient wanted to discontinue pushing efforts and instead have a cesarean birth before 1 or 2 hours.
3 .
Think about the fourth stage of labor, with all the significant changes taking place. What would help the birthing person feel safe and well cared for during this stage?
4 .
Think about patient and staff safety in the event of threatening behavior during labor by a support person. How would you keep the patient, yourself, and other staff safe?
5 .
Should insurance coverage for a doula be required if outcomes are improved by their presence? Reflect on the implications for this.


7.5: Contact Metamorphism and Hydrothermal Processes
Contact metamorphism takes place where a body of magma intrudes into the upper part of the crust. Any type of magma body can lead to contact metamorphism, from a thin dyke to a large stock. The type and intensity of the metamorphism, and width of the metamorphic aureole will depend on a number of factors, including the type of country rock, the temperature of the intruding body and the size of the body (Figure \(\PageIndex{1}\)). A large intrusion will contain more thermal energy and will cool much more slowly than a small one, and therefore will provide a longer time and more heat for metamorphism. That will allow the heat to extend farther into the country rock, creating a larger aureole.
Contact metamorphic aureoles are typically quite small, from just a few centimeters around small dykes and sills, to several 10s of meters around a large stock. As was shown in Figure \(\PageIndex{7}\), contact metamorphism can take place over a wide range of temperatures—from around 300° to over 800°C—and of course the type of metamorphism, and new minerals formed, will vary accordingly. The nature of the country rock (or parent rock) is also important.
A hot body of magma in the upper crust can create a very dynamic situation that may have geologically interesting and economically important implications. In the simplest cases, water does not play a big role, and the main process is transfer of heat from the pluton to the surrounding rock, creating a zone of contact metamorphism (Figure \(\PageIndex{2}\)a). In that situation mudrock or volcanic rock will likely be metamorphosed to hornfels (Figure \(\PageIndex{9}\)), limestone will be metamorphosed to marble (Figure \(\PageIndex{6}\)), and sandstone to quartzite (Figure \(\PageIndex{7}\)). (But don’t forget that marble and quartzite can also form during regional metamorphism!)
In many cases, however, water is released from the magma body as crystallization takes place, and this water is dispersed along fractures in the country rock (Figure \(\PageIndex{2}\)b). The water released from a magma chamber is typically rich in dissolved minerals. As this water cools, is chemically changed by the surrounding rocks, or boils because of a drop in pressure, minerals are deposited, forming veins within the fractures in the country rock. Quartz veins are common in this situation, and they might also include pyrite, hematite, calcite, and even silver and gold.
Heat from the magma body will heat the surrounding groundwater, causing it to expand and then rise toward the surface. In some cases, this may initiate a convection system where groundwater circulates past the pluton. Such a system could operate for many thousands of years, resulting in the circulation of billions of litres of groundwater from the surrounding region past the pluton. Hot water circulating through the rocks can lead to significant changes in the mineralogy of the rock, including alteration of feldspars to clays, and deposition of quartz, calcite, and other minerals in fractures and other open spaces (Figure \(\PageIndex{3}\)). As with the magmatic fluids, the nature of this circulating groundwater can also change adjacent to, or above, the pluton, resulting in deposition of other minerals, including ore minerals. Metamorphism in which much of the change is derived from fluids passing through the rock is known as metasomatism . When hot water contributes to changes in rocks, including mineral alteration and formation of veins, it is known as hydrothermal alteration .
A special type of metasomatism can take place where a hot pluton intrudes into carbonate rock such as limestone. When magmatic fluids rich in silica, calcium, magnesium, iron, and other elements flow through the carbonate rock, their chemistry can change dramatically, resulting in the deposition of minerals that would not normally exist in either the igneous rock or limestone. These include garnet, epidote (another silicate), magnetite, pyroxene, and a variety of copper and other minerals (Figure \(\PageIndex{4}\)). This type of metamorphism is known as skarn , and again, some important types of mineral deposits can form this way.
This diagram shows a pluton that has intruded into a series of sedimentary rocks. What type of metamorphic rock would you expect to see at locations:
a) Mudstone? _____________________
b) Limestone? _____________________
c) Sandstone? _____________________
See Appendix 3 for Exercise 7.5 answers.
Media Attributions
- Figures 7.5.1, 7.5.2, 7.5.3, 7.5.5: © Steven Earle. CC BY.
- Figure \(\PageIndex{4}\): 00031 6 cm grossular calcite augite skarn © Siim. CC BY SA.


5.7B: Aerobic Hydrocarbon Oxidation
- Discuss the advantages of organisms that can undergo aerobic hydrocarbon oxidation
Microbes can utilize hydrocarbons via oxidation as an energy source.
Microbes can use many different carbon sources for energy. The best known and perhaps most common example is glucose. Microbes can utilize hydrocarbons via a stepwise oxidation of a hydrocarbon by oxygen produces water and, successively, an alcohol, an aldehyde or a ketone, a carboxylic acid, and then a peroxide. Note the presence of oxygen, thus defining this as aerobic hydrocarbon oxidation. There are examples of anaerobic hydrocarbon oxidation, which will not be discussed here. This is of special interest as many of the environment pollutants released by human industry are often hydrocarbon based. One of the best examples is oil spills. Understanding how microbes digest hydrocarbons has started the field of microbial biodegradation, a type of bioremediation. The goal of this is to find ways of using microbes to degrade hydrocarbon spills or waste into less dangerous byproducts such as alcohol.
Hydrocarbon utilizing microorganisms, mostly Cladosporium resinae and Pseudomonas aeruginosa, colloquially known as “HUM bugs,” are commonly present in jet fuel. They live in the water-fuel interface of the water droplets, form dark black/brown/green, gel-like mats, and cause microbial corrosion to plastic and rubber parts of the aircraft fuel system by consuming them, and to the metal parts by the means of their acidic metabolic products. They are also incorrectly called algae due to their appearance. FSII, which is added to the fuel, acts as a growth retardant for them. There are about 250 kinds of bacteria that can live in jet fuel, but fewer than a dozen are meaningfully harmful.
Biosurfactants are surface-active substances synthesized by living cells. Interest in microbial surfactants has been steadily increasing in recent years due to their diversity, environmentally friendly nature, possibility of large-scale production, selectivity, performance under extreme conditions, and potential applications in environmental protection. Biosurfactants enhance the emulsification of hydrocarbons, have the potential to solubilize hydrocarbon contaminants, and increase their availability for microbial degradation. The use of chemicals for the treatment of a hydrocarbon polluted site may contaminate the environment with their by-products, whereas biological treatment may efficiently destroy pollutants, while being biodegradable themselves. Therefore, biosurfactant-producing microorganisms may play an important role in the accelerated bioremediation of hydrocarbon-contaminated sites. These compounds can also be used in enhanced oil recovery and may be considered for other potential applications in environmental protection. Other applications include herbicides and pesticides formulations, detergents, healthcare and cosmetics, pulp and paper, coal, textiles, ceramic processing and food industries, uranium ore-processing, and mechanical dewatering of peat. Several microorganisms are known to synthesize surface-active agents; most of them are bacteria and yeasts. When grown on hydrocarbon substrate as the carbon source, these microorganisms synthesize a wide range of chemicals with surface activity, such as glycolipid, phospholipid, and others. These chemicals are synthesized to emulsify the hydrocarbon substrate and facilitate its transport into the cells. In some bacterial species such as Pseudomonas aeruginosa, biosurfactants are also involved in a group motility behavior called swarming motility.
Key Points
- Microbes in aerobic conditions can use hydrocarbons via oxidation of the hydrocarbon. This leads to byproducts such as water, alcohol, and peroxide.
- Many hydrocarbons are environmentally damaging, thus the break down of hydrocarbons by microbes is of special interest.
- HUM bugs can function as biosurfactants to facilitate the emulsification of hydrocarbons.
Key Terms
- hydrocarbon : A compound consisting only of carbon and hydrogen atoms.
- biosurfactant : Surface-active substances synthesized by living cells.
- bioremediation : The use of biological organisms, usually microorganisms, to remove contaminants, especially from polluted water.


2.4: Rational Equations
- Solve rational equations
- Use rational functions
- Solve a rational equation for a specific variable
Before you get started, take this readiness quiz.
1. Solve \(\dfrac{1}{6} x+\dfrac{1}{2}=\dfrac{1}{3}\).
2. Solve
3. Solve the formula \(5x+2y=10\) for \(y\)
After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve rational equations .
A rational equation is an equation that contains a rational expression.
Let's recall the difference between rational expressions and rational equations. The equation contains an equal sign.
| Rational Expression | Rational Equation |
| \(\dfrac{1}{8} x+\dfrac{1}{2}\) |
\(\dfrac{1}{8} x+\dfrac{1}{2}=\dfrac{1}{4}\) |
| \(\dfrac{y+6}{y^{2}-36}\) | \(\dfrac{y+6}{y^{2}-36}=y+1\) |
| \(\dfrac{1}{n-3}+\dfrac{1}{n+4} \) | \(\dfrac{1}{n-3}+\dfrac{1}{n+4}=\dfrac{15}{n^{2}+n-12} \) |
As we explained in the introduction to this unit, in the previous unit we have used equal signs during a simplification of an expression. These equal signs implicitly mean that equation holds for all values of the variables except those that leave an expression undefined. In the case of the equations dealt with here, these equalities say something about the variable, and for 'most' values of the variable the equation is false. We will seek the values which make the equality true, i.e., we seek to solve these equations.
Solve Rational Equations
We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.
We will use the same strategy to solve rational equations. As is often the case when facing a new kind of problem, we aim to somehow reformulate it in a way that is more familiar. So, we will aim at rewriting the equation as a polynomial equation. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve since we already know how to solve linear and quadratic equations. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.
So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard. Alternatively, we can check to make sure your possible solutions make sense in our equation.
A solution to an an equation that is equivalent (except for a few values) to a rational equation for which the rational expressions are undefined is called an extraneous solution to a rational equation .
An extraneous solution to a rational equation is a solution to an equation which is that is equivalent to the original except for a certain finite number of values that would cause any of the expressions in the original equation to be undefined.
We note any possible extraneous solutions, \(c\), by writing \(x\neq c\) next to the equation.
Solve \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}\).
Solution
| \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}\) | |
| Note any value of the variable that would make any denominator zero. |
If \(x=0\), then \(\dfrac{1}{x}\) is undefined. So we'll write \(x \neq 0\) next to the equation. \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}, x \neq 0 \nonumber \) |
| Find the least common denominator of all denominators in the equation. | Find the LCD of \(\dfrac{1}{x}\), \(\dfrac{1}{3}\), and \(\dfrac{5}{6}\). The LCD is \(6x\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. |
Multiply both sides of the equation by the LCD, \(6x\). \({\color{red}6 x} \cdot\left(\dfrac{1}{x}+\dfrac{1}{3}\right)={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \) Use the Distributive Property. \({\color{red}6 x} \cdot \dfrac{1}{x}+{\color{red}6 x} \cdot \dfrac{1}{3}={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \) Simplify - and notice, no more fractions! \(6+2 x=5 x \nonumber \) |
| Solve the resulting equation. |
Simplify. \(\begin{aligned} &6=3 x\\ &2=x \end{aligned} \nonumber \) |
| Check. If any values found in Step 1 are extraneous solutions, discard them. Check any remaining solutions in the original equation. (This is not strictly necessary, but is helpful in case an error was made). |
Our solution is not an extraneous solution. \(\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6} \nonumber \) We substitute \(x=2\) into the original equation. \(\begin{aligned} \frac{1}{2}+\frac{1}{3}&\overset{?}{=}\frac{5}{6} \\ \frac{3}{6}+\frac{2}{6}&\overset{?}{=}\frac{5}{6} \\ \frac{5}{6}&=\frac{5}{6} \surd \end{aligned} \nonumber \) |
| Conclude. | The solution is \(x=2\). |
Solve \(\dfrac{1}{y}+\dfrac{2}{3}=\dfrac{1}{5}. \nonumber \)
- Answer
-
\(y=-\dfrac{15}{7}\)
Solve \(\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}. \nonumber \)
- Answer
-
\(x=\dfrac{15}{13}\)
The steps of this method are shown.
- Step 1. Note any value of the variable that would make any denominator zero.
- Step 2. Find the least common denominator of all denominators in the equation.
- Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
- Step 4. Solve the resulting equation.
-
Step 5. Check:
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.
We always start by noting the values that would cause any denominators to be zero.
Solve \(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}}. \nonumber \)
Solution
| \(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber\) | |
| Note any value of the variable that would make any denominator zero. | \(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}}, y \neq 0 \nonumber \) |
|
Find the least common denominator of all denominators in the equation. |
The LCD is \(y^2\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. | \(y^{2}\left(1-\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber\) |
| Distribute. | \(y^{2} \cdot 1-y^{2}\left(\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber\) |
| Multiply. | \(y^{2}-5 y=-6 \nonumber \) |
| Solve the resulting equation. First write the quadratic equation in standard form. | \(y^{2}-5 y+6=0 \nonumber \) |
| Factor. | \((y-2)(y-3)=0 \nonumber \) |
| Use the Zero Product Property. |
\(y-2=0 \text { or } y-3=0 \nonumber \) |
| Solve. | \(y=2 \text { or } y=3 \nonumber \) |
|
Check. There are no extraneous solutions. Check \(y=2\) and \(y=3\) in the original equation. |
\(1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \quad \quad \quad 1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber \) \(1-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{2^{2}} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{3^{2}} \nonumber \) \(1-\dfrac{5}{2} \overset{?}{=}-\dfrac{6}{4} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \) \(\dfrac{2}{2}-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad \dfrac{3}{3}-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \) \(-\dfrac{3}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad -\dfrac{2}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \) \(-\dfrac{3}{2}=-\dfrac{3}{2} \surd \quad \quad \quad -\dfrac{2}{3}=-\dfrac{2}{3} \surd \nonumber \) |
| Conclude. | The solutions are \(y=2\) and \(y=3\). |
Solve \(1-\dfrac{2}{x}=\dfrac{15}{x^{2}}. \nonumber \)
- Answer
-
\(x=-3, x=5\)
Solve \(1-\dfrac{4}{y}=\dfrac{12}{y^{2}}. \nonumber \)
- Answer
-
\(y=-2, y=6\)
In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.
Solve \(\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{x^{2}-4}. \nonumber \)
Solution
| \(\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{x^{2}-4} \nonumber \) | |
| Note any value of the variable that would make any denominator zero. | \(\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{(x+2)(x-2)}, x \neq-2, x \neq 2 \nonumber \) |
| Find the least common denominator of all denominators in the equation. | The LCD is \((x+2)(x-2)\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. | \((x+2)(x-2)\left(\dfrac{2}{x+2}+\dfrac{4}{x-2}\right)=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \) |
| Distribute. |
\((x+2)(x-2) \dfrac{2}{x+2}+(x+2)(x-2) \dfrac{4}{x-2}=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \) |
| Remove common factors. | \(\cancel {(x+2)}(x-2) \dfrac{2}{\cancel {x+2}}+(x+2){\cancel {(x-2)}} \dfrac{4}{\cancel {x-2}}=\cancel {(x+2)(x-2)}\left(\dfrac{x-1}{\cancel {x^{2}-4}}\right) \nonumber \) |
| Simplify. | \(2(x-2)+4(x+2)=x-1 \nonumber \) |
|
Distribute. |
\(2 x-4+4 x+8=x-1 \nonumber \) |
| Solve. | \(\begin{aligned} 6 x+4&=x-1\\ 5 x&=-5 \\ x&=-1 \end{aligned}\) |
|
Check: We have no extraneous solutions. Check \(x=-1\) in the original equation. |
\(\begin{aligned} \dfrac{2}{x+2}+\dfrac{4}{x-2} &=\dfrac{x-1}{x^{2}-4} \\ \dfrac{2}{(-1)+2}+\dfrac{4}{(-1)-2} &\overset{?}{=} \dfrac{(-1)-1}{(-1)^{2}-4} \\ \dfrac{2}{1}+\dfrac{4}{-3} &\overset{?}{=} \dfrac{-2}{-3} \\ \dfrac{6}{3}-\dfrac{4}{3} &\overset{?}{=} \dfrac{2}{3} \\ \dfrac{2}{3} &=\dfrac{2}{3} \surd \end{aligned} \nonumber \) |
| Conclude. | The solution is \(x=-1\). |
Solve \(\dfrac{2}{x+1}+\dfrac{1}{x-1}=\dfrac{1}{x^{2}-1}. \nonumber \)
- Answer
-
\(x=\dfrac{2}{3}\)
Solve \(\dfrac{5}{y+3}+\dfrac{2}{y-3}=\dfrac{5}{y^{2}-9}. \nonumber \)
- Answer
-
\(y=2\)
In the next example, the first denominator is a trinomial. Remember to factor it first to find the LCD.
Solve \(\dfrac{m+11}{m^{2}-5 m+4}=\dfrac{5}{m-4}-\dfrac{3}{m-1}. \nonumber \)
Solution
| \(\dfrac{m+11}{m^{2}-5 m+4}=\dfrac{5}{m-4}-\dfrac{3}{m-1} \nonumber \) | |
| Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator. | \(\dfrac{m+11}{(m-4)(m-1)}=\dfrac{5}{m-4}-\dfrac{3}{m-1}, m \neq 4, m \neq 1 \nonumber \) |
| Find the least common denominator of all denominators in the equation. | The LCD is \((m-4)(m-1)\). |
| Clear the fractions by multiplying both sides of the equation by the LCD. | \((m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1)\left(\dfrac{5}{m-4}-\dfrac{3}{m-1}\right) \nonumber \) |
| Distribute. | \((m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1) \dfrac{5}{m-4}-(m-4)(m-1) \dfrac{3}{m-1} \nonumber \) |
| Remove common factors. | \(\cancel {(m-4)(m-1)}\left(\dfrac{m+11}{\cancel {(m-4)(m-1)}}\right)=\cancel {(m-4)}(m-1) \dfrac{5}{\cancel{m-4}}-(m-4)\cancel {(m-1)} \dfrac{3}{\cancel {m-1}} \nonumber \) |
| Simplify. | \(m+11=5(m-1)-3(m-4) \nonumber \) |
| Solve the resulting equation. | \(\begin{aligned} m+11&=5 m-5-3 m+12 \\ 4&=m \end{aligned} \nonumber \) |
| Check. | We see that substituting 4 for \(m\) in the original equation would make a denominator equal to zero. The only solution to the nearly equivalent linear equation is an extraneous solution. |
| Conclude. | There is no solution to the original equation. |
Solve \(\dfrac{x+13}{x^{2}-7 x+10}=\dfrac{6}{x-5}-\dfrac{4}{x-2}. \nonumber \)
- Answer
-
There is no solution.
Solve \(\dfrac{y-6}{y^{2}+3 y-4}=\dfrac{2}{y+4}+\dfrac{7}{y-1}. \nonumber \)
- Answer
-
There is no solution.
The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.
Solve \(\dfrac{y}{y+6}=\dfrac{72}{y^{2}-36}+4. \nonumber \)
Solution
| \(\dfrac{y}{y+6}=\dfrac{72}{y^{2}-36}+4 \nonumber \) | |
|
Factor all the denominators, so we can note any value of the variable that would make any denominator zero. |
\(\dfrac{y}{y+6}=\dfrac{72}{(y-6)(y+6)}+4, y \neq 6, y \neq-6 \nonumber \) |
| Find the least common denominator. | The LCD is \((y-6)(y+6)\). |
| Clear the fractions. |
\((y-6)(y+6)\left(\dfrac{y}{y+6}\right)=(y-6)(y+6)\left(\dfrac{72}{(y-6)(y+6)}+4\right) \nonumber \) Note that the above equation has exactly the same solutions as the original except for, perhaps, \(y=-6\) and \(y=6\). |
| Simplify. | \((y-6) \cdot y=72+(y-6)(y+6) \cdot 4 \nonumber \) |
| Simplify. | \(y(y-6)=72+4\left(y^{2}-36\right) \nonumber \) |
| Solve the resulting equation. | \(\begin{aligned} y^{2}-6 y&=72+4 y^{2}-144\\ 0&=3 y^{2}+6 y-72 \\ 0&=3\left(y^{2}+2 y-24\right) \\ 0&=3(y+6)(y-4) \\ y&=-6, y=4 \end{aligned} \nonumber \) |
|
Check. \(y=-6\) is an extraneous solution. Check \(y=4\) in the original equation (not strictly necessary). |
\(\begin{aligned} \dfrac{y}{y+6} &=\dfrac{72}{y^{2}-36}+4 \\ \dfrac{4}{4+6} &\overset{?}{=}\dfrac{72}{4^{2}-36}+4 \\ \dfrac{4}{10} &\overset{?}{=} \dfrac{72}{-20}+4 \\ \dfrac{4}{10} &\overset{?}{=} -\dfrac{36}{10}+\dfrac{40}{10} \\ \dfrac{4}{10} &=\dfrac{4}{10} \surd \end{aligned} \nonumber \) |
| Conclude. | The solution is \(y=4\). |
Solve \(\dfrac{x}{x+4}=\dfrac{32}{x^{2}-16}+5. \nonumber \)
- Answer
-
\(x=3\)
Solve \(\dfrac{y}{y+8}=\dfrac{128}{y^{2}-64}+9. \nonumber \)
- Answer
-
\(y=7\)
In some cases, all the algebraic solutions are extraneous.
Solve \(\dfrac{x}{2 x-2}-\dfrac{2}{3 x+3}=\dfrac{5 x^{2}-2 x+9}{12 x^{2}-12}. \nonumber \)
Solution
| \(\dfrac{x}{2 x-2}-\dfrac{2}{3 x+3}=\dfrac{5 x^{2}-2 x+9}{12 x^{2}-12} \nonumber \) | |
| We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD. | \(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}=\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)} \nonumber \) |
| Note any value of the variable that would make any denominator zero. | \(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}=\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}, x \neq 1, x \neq-1 \nonumber \) |
| Find the least common denominator. | The LCD is \(12(x-1)(x+1)\). |
| Clear the fractions. | \(12(x-1)(x+1)\left(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}\right)=12(x-1)(x+1)\left(\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}\right) \nonumber \) |
| Simplify. | \(6(x+1) \cdot x-4(x-1) \cdot 2=5 x^{2}-2 x+9 \nonumber \) |
| Simplify. | \(6 x(x+1)-4 \cdot 2(x-1)=5 x^{2}-2 x+9 \nonumber \) |
| Solve the resulting equation. | \(\begin{aligned} 6 x^{2}+6 x-8 x+8&=5 x^{2}-2 x+9\\ x^{2}-1&=0 \\ (x-1)(x+1)&=0 \\ x&=1 \text { or } x=-1 \end{aligned} \nonumber \) |
| Check. | \(x=1\) and \(x=-1\) are extraneous solutions. |
| Conclude. | The equation has no solution. |
Solve \(\dfrac{y}{5 y-10}-\dfrac{5}{3 y+6}=\dfrac{2 y^{2}-19 y+54}{15 y^{2}-60}. \nonumber \)
- Answer
-
There is no solution.
Solve \(\dfrac{z}{2 z+8}-\dfrac{3}{4 z-8}=\dfrac{3 z^{2}-16 z-16}{8 z^{2}+2 z-64}. \nonumber \)
- Answer
-
There is no solution.
Solve \(\dfrac{4}{3 x^{2}-10 x+3}+\dfrac{3}{3 x^{2}+2 x-1}=\dfrac{2}{x^{2}-2 x-3}. \nonumber \)
Solution
| \(\dfrac{4}{3 x^{2}-10 x+3}+\dfrac{3}{3 x^{2}+2 x-1}=\dfrac{2}{x^{2}-2 x-3} \nonumber \) | |
| Factor all the denominators, so we can note any value of the variable that would make any denominator zero. | \(\dfrac{4}{(3 x-1)(x-3)}+\dfrac{3}{(3 x-1)(x+1)}=\dfrac{2}{(x-3)(x+1)}, x \neq-1, x \neq \dfrac{1}{3}, x \neq 3\nonumber \) |
| Find the least common denominator. | The LCD is \((3 x-1)(x+1)(x-3)\). |
| Clear the fractions. |
\(\quad (3 x-1)(x+1)(x-3)\left(\dfrac{4}{(3 x-1)(x-3)}+\dfrac{3}{(3 x-1)(x+1)}\right)\) \(=(3 x-1)(x+1)(x-3)\left(\dfrac{2}{(x-3)(x+1)}\right) \nonumber \) |
| Simplify. | \(4(x+1)+3(x-3)=2(3 x-1) \nonumber \) |
| Distribute. | \(4 x+4+3 x-9=6 x-2 \nonumber \) |
| Simplify. |
\(7 x-5=6 x-2 \nonumber \) \(x=3 \nonumber \) But this is an extraneous solution. |
| Conclude. | There is no solution to this equation. |
Solve \(\dfrac{15}{x^{2}+x-6}-\dfrac{3}{x-2}=\dfrac{2}{x+3}. \nonumber \)
- Answer
-
There is no solution.
Solve \(\dfrac{5}{x^{2}+2 x-3}-\dfrac{3}{x^{2}+x-2}=\dfrac{1}{x^{2}+5 x+6}. \nonumber \)
- Answer
-
There is no solution.
Solve a Rational Equation for a Specific Variable (Optional)
When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.
When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.
\(\begin{aligned} m &=\frac{y-y_{1}}{x-x_{1}} \\ m\left(x-x_{1}\right) &=\left(\frac{y-y_{1}}{x-x_{1}}\right)\left(x-x_{1}\right) \quad \text{Multiply both sides of the equation by } x-x_1.\\ m\left(x-x_{1}\right) &=y-y_{1} \quad \text {Simplify.}\\ y-y_{1} &=m\left(x-x_{1}\right) \quad \text {Rewrite the equation with the y terms on the left.} \end{aligned} \nonumber \)
In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point \((2,3)\). We will add one more step to solve for \(y\).
Solve \(m=\dfrac{y-2}{x-3}\) for \(y\).
Solution
| \(m=\dfrac{y-2}{x-3} \nonumber \) | |
| Note any value of the variable that would make any denominator zero. | \(m=\dfrac{y-2}{x-3}, x \neq 3 \nonumber \) |
| Clear the fractions by multiplying both sides of the equation by the LCD, \(x-3\). | \((x-3) m=(x-3)\left(\dfrac{y-2}{x-3}\right) \nonumber \) |
| Simplify. | \(x m-3 m=y-2 \nonumber \) |
| Isolate the term with \(y\). | \(x m-3 m+2=y \nonumber \) |
| Conclude. | \(y=x m-3 m+2\) |
Solve \(m=\dfrac{y-5}{x-4}\) for \(y\).
- Answer
-
\(y=m x-4 m+5\)
Solve \(m=\dfrac{y-1}{x+5}\) for \(y\).
- Answer
-
\(y=m x+5 m+1\)
Remember to multiply both sides by the LCD in the next example.
Solve \(\dfrac{1}{c}+\dfrac{1}{m}=1\) for \(c\).
Solution
| \(\dfrac{1}{c}+\dfrac{1}{m}=1 \text { for } c \nonumber \) | |
| Note any value of the variable that would make any denominator zero. | \(\dfrac{1}{c}+\dfrac{1}{m}=1, c \neq 0, m \neq 0 \nonumber \) |
| Clear the fractions by multiplying both sides of the equations by the LCD, \(cm\). | \(cm\left(\dfrac{1}{c}+\dfrac{1}{m}\right)=cm(1) \nonumber \) |
| Distribute. | \(cm\left(\frac{1}{c}\right)+cm \frac{1}{m}=cm(1) \nonumber \) |
| Simplify. | \(m+c=cm \nonumber \) |
| Collect the terms with \(c\) to the right. | \(m=cm-c \nonumber \) |
| Factor the expression on the right. | \(m=c(m-1) \nonumber \) |
| To isolate \(c\), divide both sides by \(m-1\). | \(\dfrac{m}{m-1}=\dfrac{c(m-1)}{m-1} \nonumber \) |
| Simplify by removing common factors. | \(\dfrac{m}{m-1}=c \nonumber \) |
| Conclude. | \(c=\dfrac{m}{m-1}\) |
Notice that even though we excluded \(c=0\) and \(m=0\) from the original equation, we must also now state that \(m \neq 1\).
Solve \(\dfrac{1}{a}+\dfrac{1}{b}=c\) for \(a\).
- Answer
-
\(a=\dfrac{b}{c b-1}\)
Solve \(\dfrac{2}{x}+\dfrac{1}{3}=\dfrac{1}{y}\) for \(y\).
- Answer
-
\(y=\dfrac{3 x}{x+6}\)
- What is the difference between an expression and an equation?
- What does it mean to solve an equation?
- Is it necessary to check your answer if you know you have not made a mistake? Explain.
Solve \(\dfrac{y}{y+3}+\dfrac{3}{y-3}=\dfrac{18}{y^2-9}\).
Key Concepts
Rational equation
Solultion to a rational equation
Extraneous solution to a rational equation
Solving a rational equation


8.1.4: Carbon
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- Tom Buckley
- University of California, Davis
Skills to Develop
- Explain why carbon is important for life
- Describe the role of functional groups in biological molecules
Cells are made of many complex molecules called macromolecules, such as proteins, nucleic acids (RNA and DNA), carbohydrates, and lipids.  The fundamental component for all of these macromolecules is carbon. The carbon atom has unique properties that allow it to form covalent bonds to as many as four different atoms, making this versatile element ideal to serve as the basic structural component, or “backbone,” of the macromolecules.
Individual carbon atoms have an incomplete outermost electron shell. With an atomic number of 6 (six electrons and six protons), the first two electrons fill the inner shell, leaving four in the second shell. Therefore, carbon atoms can form up to four covalent bonds with other atoms to satisfy the octet rule. The methane molecule provides an example: it has the chemical formula CH 4 . Each of its four hydrogen atoms forms a single covalent bond with the carbon atom by sharing a pair of electrons. This results in a filled outermost shell.
Hydrocarbons
Hydrocarbons are organic molecules consisting entirely of carbon and hydrogen, such as methane (CH 4 ) described above. We often use hydrocarbons in our daily lives as fuels—like the propane in a gas grill or the butane in a lighter. The many covalent bonds between the atoms in hydrocarbons store a great amount of energy, which is released when these molecules are burned (oxidized). Methane, an excellent fuel, is the simplest hydrocarbon molecule, with a central carbon atom bonded to four different hydrogen atoms, as illustrated in Figure \(\PageIndex{1}\). The geometry of the methane molecule, where the atoms reside in three dimensions, is determined by the shape of its electron orbitals. The carbons and the four hydrogen atoms form a shape known as a tetrahedron, with four triangular faces; for this reason, methane is described as having tetrahedral geometry.
As the backbone of the large molecules of living things, hydrocarbons may exist as linear carbon chains, carbon rings, or combinations of both. Furthermore, individual carbon-to-carbon bonds may be single, double, or triple covalent bonds, and each type of bond affects the geometry of the molecule in a specific way. This three-dimensional shape or conformation of the large molecules of life (macromolecules) is critical to how they function.
Hydrocarbon Chains
Hydrocarbon chains are formed by successive bonds between carbon atoms and may be branched or unbranched. Furthermore, the overall geometry of the molecule is altered by the different geometries of single, double, and triple covalent bonds, illustrated in Figure \(\PageIndex{2}\). The hydrocarbons ethane, ethene, and ethyne serve as examples of how different carbon-to-carbon bonds affect the geometry of the molecule. The names of all three molecules start with the prefix “eth-,” which is the prefix for two carbon hydrocarbons. The suffixes “-ane,” “-ene,” and “-yne” refer to the presence of single, double, or triple carbon-carbon bonds, respectively. Thus, propane, propene, and propyne follow the same pattern with three carbon molecules, butane, butane, and butyne for four carbon molecules, and so on. Double and triple bonds change the geometry of the molecule: single bonds allow rotation along the axis of the bond, whereas double bonds lead to a planar configuration and triple bonds to a linear one. These geometries have a significant impact on the shape a particular molecule can assume.
Hydrocarbon Rings
So far, the hydrocarbons we have discussed have been aliphatic hydrocarbons , which consist of linear chains of carbon atoms. Another type of hydrocarbon, aromatic hydrocarbons , consists of closed rings of carbon atoms. Ring structures are found in hydrocarbons, sometimes with the presence of double bonds, which can be seen by comparing the structure of cyclohexane to benzene in Figure \(\PageIndex{3}\). Examples of biological molecules that incorporate the benzene ring include some amino acids and cholesterol and its derivatives, including the hormones estrogen and testosterone. The benzene ring is also found in the herbicide 2,4-D. Benzene is a natural component of crude oil and has been classified as a carcinogen. Some hydrocarbons have both aliphatic and aromatic portions; beta-carotene is an example of such a hydrocarbon.
Isomers
The three-dimensional placement of atoms and chemical bonds within organic molecules is central to understanding their chemistry. Molecules that share the same chemical formula but differ in the placement (structure) of their atoms and/or chemical bonds are known as isomers . Structural isomers (like butane and isobutene shown in Figure \(\PageIndex{4}\)a differ in the placement of their covalent bonds: both molecules have four carbons and ten hydrogens (C 4 H 10 ), but the different arrangement of the atoms within the molecules leads to differences in their chemical properties. For example, due to their different chemical properties, butane is suited for use as a fuel for cigarette lighters and torches, whereas isobutene is suited for use as a refrigerant and a propellant in spray cans.
Geometric isomers , on the other hand, have similar placements of their covalent bonds but differ in how these bonds are made to the surrounding atoms, especially in carbon-to-carbon double bonds. In the simple molecule butene (C 4 H 8 ), the two methyl groups (CH 3 ) can be on either side of the double covalent bond central to the molecule, as illustrated in Figure \(\PageIndex{4}\)b. When the carbons are bound on the same side of the double bond, this is the cis configuration; if they are on opposite sides of the double bond, it is a trans configuration. In the trans configuration, the carbons form a more or less linear structure, whereas the carbons in the cis configuration make a bend (change in direction) of the carbon backbone.
Art Connection
Which of the following statements is false?
- Molecules with the formulas CH 3 CH 2 COOH and C 3 H 6 O 2 could be structural isomers.
- Molecules must have a double bond to be cis-trans isomers.
- To be enantiomers, a molecule must have at least three different atoms or groups connected to a central carbon.
- To be enantiomers, a molecule must have at least four different atoms or groups connected to a central carbon.
In triglycerides (fats and oils), long carbon chains known as fatty acids may contain double bonds, which can be in either the cis or trans configuration, illustrated in Figure \(\PageIndex{5}\). Fats with at least one double bond between carbon atoms are unsaturated fats. When some of these bonds are in the cis configuration, the resulting bend in the carbon backbone of the chain means that triglyceride molecules cannot pack tightly, so they remain liquid (oil) at room temperature. On the other hand, triglycerides with trans double bonds (popularly called trans fats), have relatively linear fatty acids that are able to pack tightly together at room temperature and form solid fats. In the human diet, trans fats are linked to an increased risk of cardiovascular disease, so many food manufacturers have reduced or eliminated their use in recent years. In contrast to unsaturated fats, triglycerides without double bonds between carbon atoms are called saturated fats, meaning that they contain all the hydrogen atoms available. Saturated fats are a solid at room temperature and usually of animal origin.
Enantiomers
Enantiomers are molecules that share the same chemical structure and chemical bonds but differ in the three-dimensional placement of atoms so that they are mirror images. As shown in Figure \(\PageIndex{6}\), an amino acid alanine example, the two structures are non-superimposable. In nature, only the L-forms of amino acids are used to make proteins. Some D forms of amino acids are seen in the cell walls of bacteria, but never in their proteins. Similarly, the D-form of glucose is the main product of photosynthesis and the L-form of the molecule is rarely seen in nature.
Functional Groups
Functional groups are groups of atoms that occur within molecules and confer specific chemical properties to those molecules. They are found along the “carbon backbone” of macromolecules. This carbon backbone is formed by chains and/or rings of carbon atoms with the occasional substitution of an element such as nitrogen or oxygen. Molecules with other elements in their carbon backbone are substituted hydrocarbons .
The functional groups in a macromolecule are usually attached to the carbon backbone at one or several different places along its chain and/or ring structure. Each of the four types of macromolecules—proteins, lipids, carbohydrates, and nucleic acids—has its own characteristic set of functional groups that contributes greatly to its differing chemical properties and its function in living organisms.
A functional group can participate in specific chemical reactions. Some of the important functional groups in biological molecules are shown in Figure \(\PageIndex{7}\); they include: hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl. These groups play an important role in the formation of molecules like DNA, proteins, carbohydrates, and lipids. Functional groups are usually classified as hydrophobic or hydrophilic depending on their charge or polarity characteristics. An example of a hydrophobic group is the non-polar methane molecule. Among the hydrophilic functional groups is the carboxyl group found in amino acids, some amino acid side chains, and the fatty acids that form triglycerides and phospholipids. This carboxyl group ionizes to release hydrogen ions (H + ) from the COOH group resulting in the negatively charged COO - group; this contributes to the hydrophilic nature of whatever molecule it is found on. Other functional groups, such as the carbonyl group, have a partially negatively charged oxygen atom that may form hydrogen bonds with water molecules, again making the molecule more hydrophilic.
Hydrogen bonds between functional groups (within the same molecule or between different molecules) are important to the function of many macromolecules and help them to fold properly into and maintain the appropriate shape for functioning. Hydrogen bonds are also involved in various recognition processes, such as DNA complementary base pairing and the binding of an enzyme to its substrate, as illustrated in Figure \(\PageIndex{8}\).
Summary
The unique properties of carbon make it a central part of biological molecules. Carbon binds to oxygen, hydrogen, and nitrogen covalently to form the many molecules important for cellular function. Carbon has four electrons in its outermost shell and can form four bonds. Carbon and hydrogen can form hydrocarbon chains or rings. Functional groups are groups of atoms that confer specific properties to hydrocarbon (or substituted hydrocarbon) chains or rings that define their overall chemical characteristics and function.
Art Connections
Figure \(\PageIndex{4}\): Which of the following statements is false?
- Molecules with the formulas CH 3 CH 2 COOH and C 3 H 6 O 2 could be structural isomers.
- Molecules must have a double bond to be cis - trans isomers.
- To be enantiomers, a molecule must have at least three different atoms or groups connected to a central carbon.
- To be enantiomers, a molecule must have at least four different atoms or groups connected to a central carbon.
- Answer
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C
Glossary
- aliphatic hydrocarbon
- hydrocarbon consisting of a linear chain of carbon atoms
- aromatic hydrocarbon
- hydrocarbon consisting of closed rings of carbon atoms
- enantiomers
- molecules that share overall structure and bonding patterns, but differ in how the atoms are three dimensionally placed such that they are mirror images of each other
- functional group
- group of atoms that provides or imparts a specific function to a carbon skeleton
- geometric isomer
- isomer with similar bonding patterns differing in the placement of atoms alongside a double covalent bond
- hydrocarbon
- molecule that consists only of carbon and hydrogen
- isomers
- molecules that differ from one another even though they share the same chemical formula
- organic molecule
- any molecule containing carbon (except carbon dioxide)
- structural isomers
- molecules that share a chemical formula but differ in the placement of their chemical bonds
- substituted hydrocarbon
- hydrocarbon chain or ring containing an atom of another element in place of one of the backbone carbons


7.3: California’s Political Geography
California has an idiosyncratic political culture compared to the rest of the country and is sometimes regarded as a trendsetter. In socio-cultural mores and national politics, Californians are perceived as more liberal than other Americans, especially those who live in the inland states. In the 2016 United States presidential election, California had the third highest percentage of Democratic votes behind the District of Columbia and Hawaii.
Among the political idiosyncrasies, California was the second state to recall their state governor (the first state being North Dakota in 1921), the second state to legalize abortion, and the only state to ban marriage for gay couples twice by vote (including Proposition 8 in 2008). Voters also passed Proposition 71 in 2004 to fund stem cell research, making California the second state to legalize stem cell research after New Jersey, and Proposition 14 in 2010 to completely change the state's primary election process. California has also experienced disputes over water rights; and a tax revolt, culminating with the passage of Proposition 13 in 1978, limiting state property taxes. California voters have rejected affirmative action on multiple occasions, most recently in November 2020.
The state's trend towards the Democratic Party and away from the Republican Party can be seen in state elections. From 1899 to 1939, California had Republican governors. Since 1990, California has generally elected Democratic candidates to federal, state, and local offices, including current Governor Gavin Newsom; however, the state has elected Republican Governors, though many of its Republican Governors, such as Arnold Schwarzenegger, tend to be considered moderate Republicans and more centrist than the national party.


29.7: The Inflationary Universe
By the end of this section, you will be able to:
- Describe two important properties of the universe that the simple Big Bang model cannot explain
- Explain why these two characteristics of the universe can be accounted for if there was a period of rapid expansion (inflation) of the universe just after the Big Bang
- Name the four forces that control all physical processes in the universe
The hot Big Bang model that we have been describing is remarkably successful. It accounts for the expansion of the universe, explains the observations of the CMB, and correctly predicts the abundances of the light elements. As it turns out, this model also predicts that there should be exactly three types of neutrinos in nature, and this prediction has been confirmed by experiments with high-energy accelerators. We can’t relax just yet, however. This standard model of the universe doesn’t explain all the observations we have made about the universe as a whole.
Problems with the Standard Big Bang Model
There are a number of characteristics of the universe that can only be explained by considering further what might have happened before the emission of the CMB. One problem with the standard Big Bang model is that it does not explain why the density of the universe is equal to the critical density. The mass density could have been, after all, so low and the effects of dark energy so high that the expansion would have been too rapid to form any galaxies at all. Alternatively, there could have been so much matter that the universe would have already begun to contract long before now. Why is the universe balanced so precisely on the knife edge of the critical density?
Another puzzle is the remarkable uniformity of the universe. The temperature of the CMB is the same to about 1 part in 100,000 everywhere we look. This sameness might be expected if all the parts of the visible universe were in contact at some point in time and had the time to come to the same temperature. In the same way, if we put some ice into a glass of lukewarm water and wait a while, the ice will melt and the water will cool down until they are the same temperature.
However, if we accept the standard Big Bang model, all parts of the visible universe were not in contact at any time. The fastest that information can go from one point to another is the speed of light. There is a maximum distance that light can have traveled from any point since the time the universe began—that’s the distance light could have covered since then. This distance is called that point’s horizon distance because anything farther away is “below its horizon”—unable to make contact with it. One region of space separated by more than the horizon distance from another has been completely isolated from it through the entire history of the universe.
If we measure the CMB in two opposite directions in the sky, we are observing regions that were significantly beyond each other’s horizon distance at the time the CMB was emitted. We can see both regions, but they can never have seen each other. Why, then, are their temperatures so precisely the same? According to the standard Big Bang model, they have never been able to exchange information, and there is no reason they should have identical temperatures. (It’s a little like seeing the clothes that all the students wear at two schools in different parts of the world become identical, without the students ever having been in contact.) The only explanation we could suggest was simply that the universe somehow started out being absolutely uniform (which is like saying all students were born liking the same clothes). Scientists are always uncomfortable when they must appeal to a special set of initial conditions to account for what they see.
The Inflationary Hypothesis
Some physicists suggested that these fundamental characteristics of the cosmos—its flatness and uniformity—can be explained if shortly after the Big Bang (and before the emission of the CMB), the universe experienced a sudden increase in size. A model universe in which this rapid, early expansion occurs is called an inflationary universe . The inflationary universe is identical to the Big Bang universe for all time after the first 10 –30 second. Prior to that, the model suggests that there was a brief period of extraordinarily rapid expansion or inflation, during which the scale of the universe increased by a factor of about 10 50 times more than predicted by standard Big Bang models (Figure 29.25).
Prior to (and during) inflation, all the parts of the universe that we can now see were so small and close to each other that they could exchange information, that is, the horizon distance included all of the universe that we can now observe. Before (and during) inflation, there was adequate time for the observable universe to homogenize itself and come to the same temperature. Then, inflation expanded those regions tremendously, so that many parts of the universe are now beyond each other’s horizon.
Another appeal of the inflationary model is its prediction that the density of the universe should be exactly equal to the critical density. To see why this is so, remember that curvature of spacetime is intimately linked to the density of matter. If the universe began with some curvature of its spacetime, one analogy for it might be the skin of a balloon. The period of inflation was equivalent to blowing up the balloon to a tremendous size. The universe became so big that from our vantage point, no curvature should be visible (Figure 29.26). In the same way, Earth’s surface is so big that it looks flat to us no matter where we are. Calculations show that a universe with no curvature is one that is at critical density. Universes with densities either higher or lower than the critical density would show marked curvature. But we saw that the observations of the CMB in Figure 29.18, which show that the universe has critical density, rule out the possibility that space is significantly curved.
Grand Unified Theories
While inflation is an intriguing idea and widely accepted by researchers, we cannot directly observe events so early in the universe. The conditions at the time of inflation were so extreme that we cannot reproduce them in our laboratories or high-energy accelerators, but scientists have some ideas about what the universe might have been like. These ideas are called grand unified theories or GUTs.
In GUT models, the forces that we are familiar with here on Earth, including gravity and electromagnetism, behaved very differently in the extreme conditions of the early universe than they do today. In physical science, the term force is used to describe anything that can change the motion of a particle or body. One of the remarkable discoveries of modern science is that all known physical processes can be described through the action of just four forces: gravity, electromagnetism, the strong nuclear force, and the weak nuclear force (Table 29.3).
| Force | Relative Strength Today | Range of Action | Important Applications |
|---|---|---|---|
| Gravity | 1 | Whole universe | Motions of planets, stars, galaxies |
| Electromagnetism | 10 36 | Whole universe | Atoms, molecules, electricity, magnetic fields |
| Weak nuclear force | 10 33 | 10 –17 meters | Radioactive decay |
| Strong nuclear force | 10 38 | 10 –15 meters | The existence of atomic nuclei |
Gravity is perhaps the most familiar force, and certainly appears strong if you jump off a tall building. However, the force of gravity between two elementary particles—say two protons—is by far the weakest of the four forces. Electromagnetism—which includes both magnetic and electrical forces, holds atoms together, and produces the electromagnetic radiation that we use to study the universe—is much stronger, as you can see in Table 29.3. The weak nuclear force is only weak in comparison to its strong “cousin,” but it is in fact much stronger than gravity.
Both the weak and strong nuclear forces differ from the first two forces in that they act only over very small distances—those comparable to the size of an atomic nucleus or less. The weak force is involved in radioactive decay and in reactions that result in the production of neutrinos. The strong force holds protons and neutrons together in an atomic nucleus.
Physicists have wondered why there are four forces in the universe—why not 300 or, preferably, just one? An important hint comes from the name electromagnetic force . For a long time, scientists thought that the forces of electricity and magnetism were separate, but James Clerk Maxwell (see the chapter on Radiation and Spectra) was able to unify these forces—to show that they are aspects of the same phenomenon. In the same way, many scientists (including Einstein) have wondered if the four forces we now know could also be unified. Physicists have actually developed GUTs that unify three of the four forces (but not gravity).
In these theories, the strong, weak, and electromagnetic forces are not three independent forces but instead are different manifestations or aspects of what is, in fact, a single force. The theories predict that at high enough temperatures, there would be only one force. At lower temperatures (like the ones in the universe today), however, this single force has changed into three different forces (Figure 29.27). Just as different gases or liquids freeze at different temperatures, we can say that the different forces “froze out” of the unified force at different temperatures. Unfortunately, the temperatures at which the three forces acted as one force are so high that they cannot be reached in any laboratory on Earth. Only the early universe, at times prior to 10 –35 second, was hot enough to unify these forces.
Many physicists think that gravity was also unified with the three other forces at still higher temperatures, and scientists have tried to develop a theory that combines all four forces. For example, in string theory , the point-like particles of matter that we have discussed in this book are replaced by one-dimensional objects called strings. In this theory, infinitesimal strings, which have length but not height or width, are the building blocks used to construct all the forms of matter and energy in the universe. These strings exist in 11-dimensional space (not the 4-dimensional spacetime with which we are familiar). The strings vibrate in the various dimensions, and depending on how they vibrate, they are seen in our world as matter or gravity or light. As you can imagine, the mathematics of string theory is very complex, and the theory remains untested by experiments. Even the largest particle accelerators on Earth do not achieve high enough energy to show whether string theory applies to the real world.
String theory is interesting to scientists because it is currently the only approach that seems to have the potential of combining all four forces to produce what physicists have termed the Theory of Everything. 3 Theories of the earliest phases of the universe must take both quantum mechanics and gravity into account, but at the simplest level, gravity and quantum mechanics are incompatible. General relativity, our best theory of gravity, says that the motions of objects can be predicted exactly. Quantum mechanics says you can only calculate the probability (chance) that an object will do something. String theory is an attempt to resolve this paradox. The mathematics that underpins string theory is elegant and beautiful, but it remains to be seen whether it will make predictions that can be tested by observations in yet-to-be-developed, high-energy accelerators on Earth or by observations of the early universe.
The earliest period in the history of the universe from time zero to 10 –43 second is called the Planck time . The universe was unimaginably hot and dense, and theorists believe that at this time, quantum effects of gravity dominated physical interactions—and, as we have just discussed, we have no tested theory of quantum gravity. Inflation is hypothesized to have occurred somewhat later, when the universe was between perhaps 10 –35 and 10 –33 second old and the temperature was 10 27 to 10 28 K. This rapid expansion took place when three forces (electromagnetic, strong, and weak) are thought to have been unified, and this is when GUTs are applicable.
After inflation, the universe continued to expand (but more slowly) and to cool. An important milestone was reached when the temperature was down to 10 15 K and the universe was 10 –10 second old. Under these conditions, all four forces were separate and distinct. High-energy particle accelerators can achieve similar conditions, and so theories of the history of the universe from this point on have a sound basis in experiments.
As yet, we have no direct evidence of what the conditions were during the inflationary epoch, and the ideas presented here are speculative. Researchers are trying to devise some experimental tests. For example, the quantum fluctuations in the very early universe would have caused variations in density and produced gravitational waves that may have left a detectable imprint on the CMB. Detection of such an imprint will require observations with equipment whose sensitivity is improved from what we have today. Ultimately, however, it may provide confirmation that we live in a universe that once experienced an epoch of rapid inflation.
If you are typical of the students who read this book, you may have found this brief discussion of dark matter, inflation, and cosmology a bit frustrating. We have offered glimpses of theories and observations, but have raised more questions than we have answered. What is dark matter? What is dark energy? Inflation explains the observations of flatness and uniformity of the university, but did it actually happen? These ideas are at the forefront of modern science, where progress almost always leads to new puzzles, and much more work is needed before we can see clearly. Bear in mind that less than a century has passed since Hubble demonstrated the existence of other galaxies. The quest to understand just how the universe of galaxies came to be will keep astronomers busy for a long time to come.


“One important key to success is self-confidence. An important key to self-confidence is preparation.”
– Arthur Ashe
If your résumé and cover letter have served their purposes well, you will be invited to participate in an interview with the company or organization you’re interested in. Congratulations! It’s an exciting time, and your prospects for employment are very strong if you put in the time to be well prepared.
In this section we look at how to get ready for an interview, what types of interviews you might need to engage in, and what kinds of questions you might be asked.
Preparing Effectively for a Job Interview
Review the Job Description
When you prepare for an interview, your first step will be to carefully read and reread the job posting or job description. This will help you develop a clearer idea of how you meet the skills and attributes the company seeks.
Research the Company or Organization
Researching the company will give you a wider view of what the company is looking for and how well you might fit in. Your prospective employer may ask you what you know about the company. Being prepared to answer this question shows that you took time and effort to prepare for the interview and that you have a genuine interest in the organization. It shows good care and good planning—soft skills you will surely need on the job.
Practice Answering Common Questions
Most interviewees find that practicing for the interview in advance with a family member, a friend, or a colleague eases possible nerves during the actual interview. It also creates greater confidence when you walk through the interview door. In the “Interview Questions” section below, you’ll learn more about specific questions you will likely be asked and corresponding strategies for answering them.
Plan to Dress Appropriately
Interviewees are generally most properly dressed for an interview in business attire, with the goal of looking highly professional in the eyes of the interviewer.
Come Prepared
Plan to bring your résumé, cover letter, and a list of references to the interview. You may also want to bring a portfolio of representative work. Leave behind coffee, chewing gum, and any other items that could be distractions.
Be Confident
Above all, interviewees should be confident and “courageous.” By doing so you make a strong first impression. As the saying goes, “There is never a second chance to make a first impression.”
Job Interview Types and Techniques
Every interview you participate in will be unique: The people you meet with, the interview setting, and the questions you’ll be asked will all be different from interview to interview.
The various factors that characterize any given interview can contribute to the sense of adventure and excitement you feel. But it’s also normal to feel a little nervous about what lies ahead. With so many unknowns, how can you plan to “nail the interview” no matter what comes up?
A good strategy for planning is to anticipate the type of interview you may find yourself in. There are common formats for job interviews, described in detail, below. By knowing a bit more about each type and being aware of techniques that work for each, you can plan to be on your game no matter what form your interview takes.
Screening Interviews
Screening interviews might best be characterized as “weeding-out” interviews. They ordinarily take place over the phone or in another low-stakes environment in which the interviewer has maximum control over the amount of time the interview takes. Screening interviews are generally short because they glean only basic information about you. If you are scheduled to participate in a screening interview, you might safely assume that you have some competition for the job and that the company is using this strategy to whittle down the applicant pool. With this kind of interview, your goal is to win a face-to-face interview. For this first shot, though, prepare well and challenge yourself to shine. This type of interview should be treated like a real interview. This may mean dressing for the interview and having a resume in front of you so that it can be referred to. Another suggestion is to use a land line phone if possible and/or make sure a cell phone is fully charged and that the screening interview takes place in a location that is free of distractions. Try to stand out from the competition and be sure to follow up with a thank-you note.
Phone or Web Conference Interviews
If you are geographically separated from your prospective employer, you may be invited to participate in a phone interview or online interview, instead of meeting face-to-face. Technology, of course, is a good way to bridge distances. The fact that you’re not there in person doesn’t make it any less important to be fully prepared, though. In fact, you may wish to be all the more “on your toes” to compensate for the distance barrier. Make sure your equipment (phone, computer, Internet connection, etc.) is fully charged and works. If you’re at home for the interview, make sure the environment is quiet and distraction-free. If the meeting is online, make sure your video background is pleasing and neutral, like a wall hanging or even a white wall.
One-on-One Interviews
The majority of job interviews are conducted in this format—just you and a single interviewer—likely with the manager you would report to and work with. The one-on-one format gives you both a chance to see how well you connect and how well your talents, skills, and personalities mesh. You can expect to be asked questions like “Why would you be good for this job?” and “Tell me about yourself.” Many interviewees prefer the one-on-one format because it allows them to spend in-depth time with the interviewer. Rapport can be built. As always, be very courteous and professional. Have handy a portfolio of your best work.
Panel Interviews
An efficient format for meeting a candidate is a panel interview, in which perhaps four to five coworkers meet at the same time with a single interviewee. The coworkers comprise the “search committee” or “search panel,” which may consist of different company representatives such as human resources, management, and staff. One advantage of this format for the committee is that meeting together gives them a common experience to reflect on afterward. In a panel interview, listen carefully to questions from each panelist, and try to connect fully with each questioner. Be sure to write down names and titles, so you can send individual thank-you notes after the interview.
Serial Interviews
Serial interviews are a combination of one-on-one meetings with a group of interviewers, typically conducted as a series of meetings staggered throughout the day. Ordinarily this type of interview is for higher-level jobs, when it’s important to meet at length with major stakeholders. If your interview process is designed this way, you will need to be ultraprepared, as you will be answering many in-depth questions. Be prepared.
Lunch Interviews
In some higher-level positions, candidates are taken to lunch or dinner, especially if this is a second interview (a “call back” interview). If this is you, count yourself lucky and be on your best behavior, because even if the lunch meeting is unstructured and informal, it’s still an official interview. Do not order an alcoholic beverage, and use your best table manners. You are not expected to pay or even to offer to pay. But, as always, you must send a thank-you note.
Group Interviews
Group interviews are comprised of several interviewees and perhaps only one or two interviewers who may make a presentation to the assembled group. This format allows an organization to quickly prescreen candidates. It also gives candidates a chance to quickly learn about the company. As with all interview formats, you are being observed. How do you behave with your group? Do you assume a leadership role? Are you quiet but attentive? What kind of personality is the company looking for? A group interview may reveal this.
For a summary of the interview formats we’ve just covered (and a few additional ones), take a look at the following video,
Job Interview Guide—10 Different Types of Interviews in Today’s Modern World.


9.3: Examples of Cardiovascular Terms Easily Defined By Their Word Components
Here are examples of common medical terms related to the cardiovascular system that can be easily defined by breaking the terms into their word components.
Cardiologist
- Break down the medical term into word components: Cardi / o / logist
- Label the word components: Cardi = WR; o = CV; logist = S
- Define the word components: Cardi = heart; logist = specialist who studies and treats
- Create a final definition of the medical term: A specialist who studies and treats diseases of the heart
Bradycardia
- Break down the medical term into word components: Brady / cardi / a
- Label the word components: Brady = P; card = WR; ia = S
- Define the word components: Brady = slow; card = heart; ia =condition of
- Create a final definition of the medical term: Condition of slow heart (rate)
Electrocardiogram
- Break down the medical term into word components: Electr / o / cardi / o / gram
- Label the word components: Electr = WR; o = CV; cardi = WR; o = CV; gram = S
- Define the word components: Electr = electricity; cardi = heart; gram = record
- Create a final definition of the medical term: A record of electrical (activity) of the heart
Interactive Learning Activity: Practice defining and pronouncing cardiovascular system medical terms by breaking them down into word parts.
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Interactive Learning Activity: Practice defining and pronouncing cardiovascular system medical terms by breaking them down into word parts.
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You can also print these flashcard activities as a Chapter 9 Student Companion Worksheet and check your answers with this Answer Key .


1.2.9: Emotional and Mental Health- Optional Continued Learning
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The videos below provide an opportunity to expand on your learning from the chapter. As you watch the video, make a list of what you notice and what you wonder. This is a great strategy to listen purposefully, engage with the content, and look for deeper opportunities for self growth, self-reflection, and learning.
What do you notice ?
- How does this topic support, expand, or challenge the content in the chapter?
- How does this topic connect to health topics from other chapters?
- How does this topic connect with your prior learning, your experiences, your work, your family, or your life in general?
- How does this topic help you to more fully understand health and wellness?
What do you wonder ?
- How has this topic sparked your curiosity?
- What would you like to know more about?
- What questions did you have as you reviewed the topic?
- What other popular and scholarly sources support or refute this topic?
Videos for continued learning and application
Note: These videos are intended to more fully reflect on health. You might agree or disagree with the videos and that is ok! Utilize these videos to critically think through the topics and identify other sources, both scholarly and popular, to convey your learning.
A new way to help young people with their mental health
Break the silence for suicide attempt survivors
Debunking the myths of OCD
How do antidepressants work?
How gratitude rewires your brain
How to discover your “why” in difficult times
How to protect your mental well-being online — from a Gen-Zer
The gift and power of emotional courage
The Housing First approach to homelessness
The psychology of post-traumatic stress disorder
The voices in my head
What causes panic attacks, and how can you prevent them?
What is bipolar disorder?
What makes a good life? Lessons from the longest study on happiness
What’s normal anxiety — and what’s an anxiety disorder?
What’s your happiness score?
Why we get mad — and why it’s healthy