Mathematics Study Guide for the TEAS

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Algebra

Algebra gives us a way to write mathematical statements that can include quantities that we don’t yet know the value of. That may sound a little crazy to you. How can you write a quantity that you don’t know? The answer is at the heart of algebra. If we don’t know a quantity, we just write a letter in its place. The letters \(x\) and \(y\) are very popular choices. Algebra’s purpose is to find the value(s) of those unknown quantities using equations. These equations always have an equal sign and can be as simple as \(x=9\) or \(x+5=7\).

Solving an Equation with One Variable

To solve an equation means to find the value of a variable that makes the equation true. For example, what value, when put in place of \(x\), will make \(x+5=7\) true? You can probably see that the number is \(2\), because \(2+5=7\). That means we have solved the equation and the answer is \(x=2\).

Terms in an Equation

A variable is a symbol, typically a letter of the alphabet printed in italics, used to represent an unknown, unfixed, or nonspecific value. A variable can be contrasted with a constant, which is a quantity with an assigned value that does not change.

In the equation \(2x + 5y = 9\), \(x\) and \(y\) are variables and \(2\), \(5\), and \(9\) are constants.

Another term to know is the word coefficient. Any number that multiplies a variable is called a coefficient. In the equation just above, \(2\) is a coefficient of \(x\) and \(5\) is a coefficient of the \(y\).

Procedures to Use

The first two procedures below are usually done to accomplish the third step. These steps have the result of simplifying the equation as you do them. There’s no law that says you have to do the two steps in this order, but it’s often the best way.

1. Combining Constants

This step just means to combine all numbers without variables attached to them into one number. Suppose you had the following equation:

\[15 + 12 x -9=0\]

You could change the order of the terms and write it as \(12x+15-9=0\). (The commutative law says we can do that.) Now combine the \(15\) and \(-9\) to get \(6\). Your equation is now simpler: \(12x+6=0\).

Be aware that you cannot combine \(12x\) and \(6\) into one term.

2. Combining Variables

Algebra sometimes presents problems that require you to solve equations by simplifying expressions, using steps that involve grouping and combining like terms. This is most often accomplished by combining coefficients that share the same variable factor while keeping the variable factor the same, such as combining \(–8x^3\) and \(11x^3\) to get \(3x^3\).

You could not, however, combine \(-8x\) with \(3x^3\) because \(x\) and \(x^3\) are different terms.

3. Isolating the Variable Using Inverse Operations

Many algebra problems involve solving equations in one variable, requiring you to use elementary transformations, including inverse operations, to move all terms with the variable to one side of the equal sign, and everything else to the other. Using inverse operations just means that you perform the operation opposite to the one in the original equation on both sides of the equation.

Here is an example:

We need to get \(x\) isolated on the left side of the equation. If we add \(5\) to both sides of \(x – 5 = 13\), it becomes \(x = 18\).

\(x – 5 = 13\)
\(x – 5 + 5 = 13 + 5\)
\(x = 18\)

Proportions

There are situations in which two kinds of quantities are related to each other in such a way that when one of them changes, the other one also changes in the same way. A few are listed below:

  • time spent running and distance covered
  • hours worked and weekly pay
  • energy used and work done
  • distance on a map between two cities and actual distance between them

The sections below show how to use ratios and proportions.

Ratios

A ratio is basically a fraction that is used to compare two related quantities. Suppose in Claire’s garden she has \(10\) tomato plants and \(45\) potato plants. To show a comparison between potatoes and tomatoes, a ratio could be written comparing tomatoes to potatoes, like this:

\[\frac{10\text{ tomatoes}}{45\text{ potatoes}} = \frac{2}{9}\]

Or, we could write the ratio of potatoes to tomatoes:

\[\frac{45\text{ potatoes}}{10\text{ tomatoes}} = \frac{9}{2}\]

Ratios can also be written this way: \(\rightarrow 9:2\). You would read that as “the ratio nine to two”.

Proportions

A proportion is a statement that shows two ratios being equal. An example is this:

\[\frac{1}{4}=\frac{3}{12}\]

In algebra, a more common proportion would have a variable like this:

\[\frac{x}{6} = \frac{15}{18}\]

We will deal with these below.

Creating a Proportion from a Word Problem

Let’s start with a sample word problem.

If \(10\) pesos equal \(\$0.55\), how many pesos are worth \(\$20\)?

Keep in mind that we need to fill in all four terms in the proportion format:

\[\frac{?}{?} = \frac{?}{?}\]

You will be given three numbers to put in the blanks. The fourth blank will be an \(x\) (or some other variable).There are four different correct ways you can set up a proportion. Let’s concentrate on one.

Step 1

Set up the proportion so that each fraction has the same units top and bottom:

\[\frac{\text{pesos}}{\text{pesos}} = \frac{\text{dollars}}{\text{dollars}}\]

Step 2

In both fractions, put the three numbers in using the pattern below. Put an \(x\) in whatever blank is left open:

\[\frac{\text{original value}}{\text{new value}} = \frac{\text{original value}}{\text{new value}}\]

For this problem, here is the setup:

\[\frac{10}{x} = \frac{0.55}{20}\]

Solving a Proportion

There are a couple of different ways to solve proportions, but we will just look at the one called cross-multiplying. We will use the example from above:

\[\frac{10}{x} = \frac{0.55}{20}\]

This is cross-multiplying:

4 Solving a Proportion.png

Each slanted line above connects a pair of numbers to be multiplied. The result is this:

\[0.55 \cdot x = 10 \cdot 20\]

Which gives us:

\[0.55 x = 200\]

Now, we can just divide both sides by \(0.55\):

\[x = \frac{200}{0.55} = 363.63\]

Remember, we were talking about pesos, so this answer is the number of pesos equal to \(\$20.00\).

Direct Proportionality

If two quantities are directly proportional, and one of them is doubled, for example, the other one will also double. If one is tripled, the other will triple. If one is cut by one third, the other will be cut by one third. In other words, they are directly related to each other such that a change in one causes a corresponding change in the other. (Be careful. This change we’re talking about does not work for addition or subtraction.)

A good example of this is speed and distance. If you drive at \(20\) mph for an hour, you will go \(20\) miles. But, if you drive at \(20\) mph for \(2\) hours, you will go \(40\) miles. Doubling your time doubled your distance (if the speed remains the same). Whatever is done to the time also happens to the distance. They are directly proportional to each other.

We can write a well-known formula showing this:

\[D = rt\]

where \(D\) is distance, \(r\) is the rate of speed, and \(t\) is the time.

Let’s see how this works:

  • At \(20\) mph for an hour, you will go \(20\) miles. In \(D=rt\) form, we can write \(20 = 20 \cdot 1\).
  • At \(20\) mph for \(2\) hours, you will go \(40\) miles. In \(D=rt\) form, we can write \(40 = 20 \cdot 2\).
  • At \(20\) mph for \(5\) hours, you will go \(100\) miles. In \(D=rt\) form, we can write \(100 = 20 \cdot 5\).

Notice how one term, \(r\), stays constant at \(20\) mph. That is called the constant of proportionality.

Any direct proportion can be written in the same form: \(y = kx\). In that form, \(k\) is the constant of proportionality, and \(y\) and \(x\) are variables. A tiny bit of algebra shows us that \(\frac{y}{x} = k\) is an equivalent form. If you can get your quantities into either of these forms, you can find the constant of proportionality.

Solving Word Problems

There are a number of problem-solving strategies available to facilitate the process of finding solutions. The most basic include reading the problem, noting the information it gives, identifying the question it asks, and expressing it as an equation.

Other strategies include acting out the problem, identifying relationships, distinguishing relevant from irrelevant facts, sequencing and prioritizing information, and finding patterns.

Solving a simpler problem and then generalizing the same principles to more complex problems is another problem-solving technique, as are guessing and checking and sketching drawings, diagrams, or other visual representations to clarify the information given.

1. Study the Information Given

Read the word problem thoroughly, maybe several times, noting the question being asked. What are you being asked to find? Be careful, though. Sometimes problems contain information that is not required to find the solution but is nonetheless framed as if it is part of the problem. It is, therefore, important that you read an entire problem before beginning to work it out. This will eliminate such distractions and help you decide what information is important and what information is not.

2. Determine Operations to Use

There are always keys in word problems telling us which operations to use when solving. Here are some words and phrases with which you need to be familiar:

  1. The words add, plus, combine, total, sum, together, more, increase, older by, and similar words imply addition.

  2. The words less, minus, difference, reduce, decrease, fewer, and similar words imply subtraction.

  3. The words product, times, twice, thrice, and of indicate multiplication.

  4. The words quotient, ratio, out of, and per indicate division.

3. Create an Equation

Solving word problems first requires us to translate the words into numerical sentences, or equations, inequalities, or expressions. We use variables to represent unknown quantities.

For example, “The number of visitors that came to the museum Saturday was twice the number that came Friday,” can be represented by \(n = 2v\), where \(n\) represents the number of visitors on Saturday and \(v\) represents the number of visitors on Friday.

4. Perform the Operations

The hard part of doing a word problem is setting it up as an equation to solve. That’s what the three sections above can help with. Once we have an equation, we are on more solid ground and can use the algebraic methods for solving equations we have used many times before.

5. Review for Accuracy

Once you have an answer, it’s good to check it out and see if it makes sense. Check for things that don’t seem to fit.

For example, if the problem is finding the dimensions of something and your answer is negative, there’s a mistake somewhere. Lengths can’t be negative.

Another case may be an answer that is surprisingly big or small. Maybe you have calculated that a car has a gas tank that holds something like \(110\) gallons or maybe \(1.2\) gallons. One is too big and the other is too small.

It’s not always easy, but trying to do this kind of thing can often help you catch mistakes.

Remember that estimating an answer is a useful tool to help judge how reasonable the answer may be.

Problems with Expressions, Equations, and Inequalities

You need to be able to compose expressions, equations, and inequalities from word problems. As you work through the problem, you will first write the expressions and then fit them into an equation. As an example, suppose you read this statement:

Five times a number plus six is \(36\).

“Five times a number” translates to the expression \(5x\).
“Plus six” translates to the expression \(5x+6\).
“Is” means equals here, so now you have the equation \(5x + 6 = 36\).

More examples are below.

When to Use an Expression

A mathematical expression is some combination of constants and/or variables linked by symbols, such as \(+, -,\times,\) and \(\div\).

A few examples are \(7+x,\ \ y^2-3,\ \ x^2+y^2, \ \ 2xy + 9x - 32\).

Often the multiplication symbol is understood and not written, giving us expressions like these: \(5x, \,2(a+b), \,(x+y)(x-y)\).

It’s probably fair to say that almost anything you write in mathematics will have at least one expression in it.

If you have an equation or inequality, you will very likely have expressions on both sides, but keep in mind that an expression does not include any equal or inequality signs.

When to Use an Equation

A very common form of math problem goes like this:

The sum of a number and twice the number is \(27\). What is the number?

What word in the first sentence tells you that this problem needs an equation? That is the word is. Is generally means equals. The phrase that comes before the is goes on one side of the equal sign and what comes after the is goes on the other side.

In this case, we would write \(x + 2x = 27\).

Watch for the word is and similar words such as has, have, are, yields, had, produces…. There are a lot of them that could mean equals, but be careful, because they can be used in other ways. The main thing to look for is the structure of the wording. It should be similar to a sentence, with the verb being the equal sign. In fact, equations are sometimes referred to as number sentences. Look for this structure:

Given Fact \(1\) and Fact \(2\), what is their result?

An equal sign would replace the what is:

Fact \(1\) and Fact \(2\) = Result

When to Use an Inequality

The clues that tell you to use an inequality in a problem are the phrases like less than, greater than, no more than, no less than, no longer than, no shorter than, at least, at most, or other similar phrases. Here’s an example:

Max bought a pen for \(\$3\) and four pencils. If he spent less than \(\$8\), what is the most a pencil could cost?

Let \(x\) = the cost of a pencil. Write an expression for the total cost using \(x\):

\[3 + 4x\]

This must be less than \(8\), so write an inequality showing that, and solve it:

\[3+4x<8\] \[4x<5\] \[x < 5 \div 4\] \[x < 1.25\]

This shows that \(x\), the cost of a pencil, cannot be any larger than \(\$1.25\).

Additional Word Problem Solving Procedures

Many many word problems require multiple steps to solve. We will also take another look at percentages, ratios, and proportions, as you may see them in word problems.

Multi-Step Problems

Some problems, including multiple-step problems, can be solved more easily by breaking them into simpler parts. It might be helpful to create a brief outline of the steps before actually solving the problem. After solving each part, take a few moments to analyze how each relates to the whole and double check your answer by asking yourself if it seems reasonable.

Example:

How old is my sister if \(240\) reduced by three times her age is \(165\)?

Pick a variable to represent my sister’s age. How about \(a\)? Now zero in on the phrase three times her age and write an expression for that: \(3a\).

Now look at the phrase “\(240\) reduced by three times her age.” An expression for that is \(240-3a\).

Finally, we have the phrase “is \(165\).”
Is means “equal” here so now we have the equation \(240-3a=165\).

Getting the equation is the hardest part. Now all we have to do is solve it, and this is where the multiple steps come in.

First we subtract \(240\) from both sides:

\[240-3a=165\] \[-3a= 165-240\] \[-3a=-75\]

Now divide both sides by \(-3\) and get:

\[a=25\]

Problems Using Basic Percentage

Solving (word) problems involving percentages requires you to be familiar with the basic percent equation: percent x base = amount, where percents are rewritten in their equivalent decimal forms, and the base is usually the value that follows the word of. You also need to remember that the word of translates to the multiplication sign and the word is translates to the equal sign.

Example:

Marlo has been promised \(5\%\) of the company profits. If the company makes \(\$43\text{,}000\) this year, how much can Marlo expect to be paid?

\[$43\text{,}000\; \times .05 = $2\text{,}150\]

Finding Percentage Increase or Decrease

There are two ways to calculate percent increase or decrease. We’ll show them below.

Percent Increase

In Michigan there is a \(6\%\) sales tax. What would the final price be if you bought a lawn chair priced at \(\$11.50\)?

Method 1

  • Change \(6\%\) to \(0.06\) and multiply it by \(\$11.50\). You will get a tax of \(\$0.69\).
  • Add \(\$0.69\) to \(\$11.50\). You will get \(\$12.19\).

Method 2

  • Change \(6\%\) to \(0.06\) and add it to \(1\). You will get \(1.06\).
  • Multiply \(1.06\) by \(\$11.50\). You will get \(\$12.19\).

Method 2 is a bit faster because you can probably do the first step in your head.

Percent Decrease

A T-shirt originally priced at \(\$11.50\) is reduced by \(6\%\). What is the new price? Ignore any tax on the T-shirt.

Method 1

  • Change \(6\%\) to \(0.06\) and multiply it by \(\$11.50\). You will get \(\$0.69\).
  • Subtract \(\$0.69\) from \(\$11.50\). You will get \(\$10.81\).

Method 2

  • Change \(6\%\) to \(0.06\) and subtract it from \(1\). You will get \(0.94\).
  • Multiply \(0.94\) by \(\$11.50\). You will get \(\$10.81\).

Problems Involving Ratio

A ratio is a relationship between two quantities that stipulates how much of one accompanies a given amount of the other. For instance, if every boy had three pieces of candy, the ratio of boys to candy would be \(1\) to \(3\), written as \(1 / 3\). The ratio of candy to boys, however, would be \(3\) to \(1\), written as \(3 / 1\). In healthcare professions, ratios are often used with medications, such as \(150 \,\text{mg}\) per \(15 \, \text{mL}\), written as \(150 \,\text{mg}\) / \(15 \,\text{mL}\). Ratios can also be written with a colon, such as \(3:1\) or \(1:3\).

Example word problem: When mixing concrete, a common recipe to use is \(100\) pounds of cement, \(300\) pounds of sand, and \(200\) pounds of aggregate. What is the ratio of aggregate to sand?

When writing a ratio, the quantity named first will go on top and the other quantity will go on the bottom. In this case we will write the ratio as you see below. It’s good practice to always simplify any ratio.

\[\frac{200 \text{ pounds of aggregate}}{300 \text{ pounds of sand}} = \frac{2}{3}\]

Problems Involving Proportion

A proportion is a comparison in which the relationships between the corresponding parts (ratios) are equal. So, a proportion compares two ratios. For example, the ratios \(1 / 3\) and \(3 / 9\) are in proportion because each one’s denominator is three times larger than its numerator. We can write a proportion that looks like this:

\[\frac{1}{3}=\frac{3}{9}\]

Example:

A doctor orders \(3\text{,}500\) units of heparin and the heparin bottle states that it contains \(5\text{,}000\) units of heparin per mL. How many mL of heparin did the doctor order?

We can solve this with a proportion, keeping units on the top of each ratio:

\[\frac{5\text{,}000 \text{ units}} {1\text{ mL}} = \frac{3\text{,}500\text{ units}}{ x}\]

Cross-multiply to get:

\[5\text{,}000 \times x = 3\text{,}500 \times 1\]

Dividing both sides by \(5\text{,}000\) gives us:

\[x = \frac{3\text{,}500}{5\text{,}000}\] \[x = 0.7 \text{ mL}\]

Problems Involving Rates

Certain ratios are considered “rates.” You will notice that the word per shows up in them.

Rates of speed

  • miles per hour
  • meters per second
  • snowfall: \(8\) inches per day

Others

  • hours of work per week
  • calories per candy bar

Finding the Unit Rate

The five rates above are examples of unit rates. This is because in each case the second term in the rate is an understood \(1\). Miles per hour means miles traveled in one hour. Calories per candy bar means calories in one candy bar, and so on.

Often, problems with rates will give you information and ask you to find the unit rate. Here’s an example:

If an auto factory can build \(336\) cars in one day, what is their building rate per hour? Tip: Write a fraction so that whatever quantity follows the word per goes on the bottom (denominator) of the ratio. Since we want cars per hour, we need the number of hours on the bottom. Then we simplify the ratio.

\[\frac{336 \text{ cars}}{24 \text{ hours}} = \frac{14 \text{ cars}}{\text{ hour}}\]

What if the question asks for the unit rate of hours per car?

\[\frac{24 \text{ hours}}{336 \text{ cars}} = \frac{0.071 \text{ hour}}{\text{ car}}\]

What if the question asks for the unit rate of minutes per car? Convert hours to minutes.

\[\require{cancel}\] \[\frac{0.071\cancel{\text{ hour}}}{\text{ car}} \times \frac{60 \text{ minutes}}{1 \cancel{\text{ hour}}} = \frac{4.26 \text{ minutes}}{ \text{ car}}\]

Finding the Rate of Change

Closely related to ratio is the concept of rate, which is a special kind of ratio that compares two different types of measurements, such as time and distance. Rate of change—often the focus of problems involving dependent and independent variables—is found by determining the amount of change in one type of unit given a specified amount of change in another.

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