Mathematics Study Guide for the TABE Test

Algebraic Concepts: Operations and Algebraic Thinking

Algebra is one of the core categories covered on the TABE. To ease into these algebraic processes, we will learn about some of the key concepts and terms that you’ll need to know to navigate this subject matter.

Equations and Expressions

To start, let’s dive deeper into two concepts that you’ve learned previously, equations and expressions, because they are at the heart of everything we do in algebra. Recall that an expression is any group of numbers, variables, and math operations, while an equation combines numbers, variables, and operations, as well, but also includes an equal sign. Equations have a value or expression on both sides of the equal sign. In other words, you can have an equation that is made up of two expressions.

Vocabulary

Expressions and equations are made up various terms, each with a specific name:

constant—Any number you can think of, say $7$ or $44$, is a constant. Constants never change in value; $7$ always means the same thing.
variable—Variables are letters that take the place of numbers that are not yet known. The variable $x$ can mean $18$ in one problem and $5$ in another one.
coefficients—Coefficients are numbers that are paired with variables, such as the $5$ in $5x$.

In the expression $12 + 5x$, there is one constant, $12$, and one coefficient, $5$, paired with the variable $x$.

You should also know like terms, which are terms that share a variable to the same power. For instance, in the expression $2x^2 + 3x + 7+ 2x + 9$, $3x$ and $2x$ are like terms and so are $7$ and $9$. The term $2x^2$ stands alone.

Note: If you’re confused about why $7$ and $9$ are like terms even though they aren’t variables, just imagine that each number is paired with a variable raised to the power of $0$, which just makes it $1$.

Independent and Dependent Variables

Sometimes an equation will have more than one variable. In an equation with two variables, it can be useful to think of one variable as the independent variable and the other one as the dependent variable. This example should help show how that works.

Suppose a man is sawing a tree trunk into sections. Out of curiosity, he weighs a few of them. He finds that a foot-long piece weighs $30$ pounds, a three-foot piece weighs $90$ pounds, and a six-foot piece weighs $180$ pounds. Letting $w$ be weight and $l$ be length, we can make a table:

\[\begin {array}{|c|c|} \hline l & w \\ \hline 1 & 30\\ \hline 3 & 90\\ \hline 6 & 180\\ \hline \end{array}\]

Is there a simple equation with two variables that shows the relationship between weight and length? Yes. We can see that the weight is always $30$ times the length, so we can write this:

\[w = 30\ l\]

This formula tells us how changing the length makes the weight change. The weight depends on the length in a specific way. For that reason, $w$, the weight, is called the dependent variable. That makes the length, $l$, the independent variable.

A good way to think of this is to think of cause and effect. Changing the value of $l$, the cause, produces a change in $w$, the effect. The value that is deliberately changed is the independent variable, and the value produced is the dependent variable.

The Equal Sign and Unknown Numbers

When you first learned about the equal sign, you basically learned that it means, “Here’s the answer” (e.g., $3 + 5 - 2 = 6$). However, in algebraic thinking an equal sign can also be thought of as a sign of balance. The total value on the left side of the equal sign must always be in balance with the total value on the right side:

\[9 = 9\] \[9 = 10 - 1\] \[4 + 3 + 11 = 9 + 9\]

This idea is fundamental for when you’re finding unknown numbers, or variables. For instance, consider this example:

\[\text{?} + 7 = 15\]

On one hand, this is simply asking “What plus $7$ equals $15$?” However, you can also think of it as, “What added to $7$ will make the left side balanced with the right side?”

The ability to balance sides will be very important when you start seeing equations with variables on both sides of the equal sign.

Relating Word Problems and Equations

For every word problem, there is an equation that can be solved to answer it. Your job will be to use key words, logical reasoning, and experience to translate the words into equations. You’ve learned about key words when first learning about operations. For instance, the key word “sum” indicates addition and the key word “per” will usually indicate division. The more familiar you are with those key words, the easier it will be to translate a word problem into an equation, or vice versa.

Word Problem to Equation

Translating basic word problems into equations often requires using one or more variables to represent the unknown information. It could be distance, weight, or something else, but the important thing is to understand how it fits into the equation you are building. We’ll illustrate this with an example problem.

Susie Q. is on her lunch break from work. She went down to the local deli and bought a sandwich for $\$3.50$ and a root beer for $\$1.75$. How much did she spend in total on her lunch?

Solution

We see the key word total, which means add. What do we add? The cost of the sandwich and the root beer. And what is the unknown? The total, so let’s use the letter $t$ as our variable and write an equation. We can skip the dollar signs for now:

\[3.50 + 1.75 = t\]

Now it’s clear that, once we stripped away all the extra details, the initial word problem is a basic addition problem. Our answer ($t$ in our equation) is $\$5.25$.

Here’s another example.

Maria works at a bookstore and earns $\$12$ per hour. If she works for five hours, how much money will she earn in total?

Solution

In this problem, we are asked to find the total amount earned, but unlike the previous example, we are not combining two separate amounts. Instead, we are dealing with equal groups. Maria earns the same amount each hour. This tells us we should use multiplication.

First, let’s identify the values:

dollar amount earned per hour: $12$
number of hours worked: $5$
unknown total earnings: let’s call it $t$

Now, we can write the equation:

\[t = 12 \times 5\]

Solving, we get:

\[t = 60\]

Thus, Maria earns a total of $\$60$.

Equation to Word Problem

One way to strengthen your ability to create equations from word problems is reading an equation and imagining a situation in which the equation could fit. In other words, you are reversing the process. Think about this equation with multiplication:

\[t = 3 \times 12\]

It could be used for a lot of different situations. In general, it means you have $12$ groups of three things or three groups of $12$ things. Here are a few examples of what this equation could mean in a real-world situation:

“Find the number of cookies in three dozen cookies.”
“If a yard is three feet long, how many feet are in $12$ yards?”
“A bike moves at $12$ miles per hour. How far will it go in three hours?”

Now, think about this equation with division:

\[n = 66 \div 22\]

Here we are talking about a large group being divided into smaller groups. In the real world, this could present the following questions:

“If there are $66$ cookies and $22$ kids, how many cookies will each kid get?”
“If a chain is $66$ inches long and has $22$ links, how long is each link?”
“If a barrel is $22$ inches around, how many times can you wrap a $66$-inch string around it?”

Addition and Subtraction

In algebra, you need to be able to use addition and subtraction to solve one- or two-step word problems where a value is unknown. As you’ve seen, this is done using a variable to represent the unknown value (most commonly the letter $x$).

You saw how this works with a one-step problem above. Let’s try an example with a two-step problem.

Johnny Applepicker picked $27$ apples on Saturday and $41$ apples on Sunday. If he used nine apples to make pies on Monday, how many apples does he have left?

Solution

There are two steps to this problem because there are two unknowns. The first unknown is how many total apples Johnny picked over the two days. That’s found with simple addition:

\[27+41=68\]

The second unknown is how many of those total apples are left over after he made pies on Monday. That’s found with simple subtraction:

\[68 - 9 = 59\]

So, the answer to the question is Johnny Applepicker has $59$ apples left.

Though it wasn’t necessary to do here, both of those operations could have been set up as equations with variables to represent the unknown values, as such:

\[27+41=x\] \[x-9=y\]

Note: Notice that because there were two unknowns, there were two variables in the equations we set up. That will be important later on.

Tips for Multi-Step Problems

Word problems with multiple steps can be tricky. They require that you are clear on the information that you are starting with and what you have to do to get your answer. One simple strategy for keeping things straight iscircling the given numbers and underlining what is wanted. Then, read the problem carefully and determine exactly what you’ll be trying to find.

Another strategy that can help is to start at the end and see if you can work out the step before it. So, think about what information you would need to know to get the answer, and determine how much of that information the problem has already given you. Your thinking can be something like this: “I could find the answer if only I knew the ____.” After that figuring out the missing information is your first step. Let’s look at an example.

A farmer has $34$ chickens, and each one eats one-fourth of a pound of commercial chicken feed every day. If he increases his flock to $50$ chickens, how much more chicken feed will the farmer need to buy each week?

Solution

First off, let’s circle and underline the important information:

A farmer has $\require{enclose} \enclose{circle}{34}$ chickens, and each one eats $\enclose{circle}{\text{one-fourth}}$ of a pound of commercial chicken feed every day. If he increases his flock to $\enclose{circle}{50}$ chickens, how $\underline{\text{much more chicken feed}}$ will the farmer need to buy each week?

Now that we have that information marked, the next thing we need to determine is how many more chickens the farmer has now compared to before. Knowing that will allow us to calculate the answer to the problem. So, if he now has $50$ chickens, that’s $50 - 34 = 16$ more than he had.

Now, how much will those $16$ chickens eat in a day? Each chicken will each eat one-fourth of a pound of feed a day, so in total that gives us:

\[16 \times \frac{1}{4} = \frac{16}{4} = 4\]

All the chickens will eat four pounds of chicken feed each day, so for a week the farmer will need to buy $7 \times 4 = 28$ more pounds. That’s the answer.

Additive and Multiplicative Comparisons

You will be presented with problems that say things like, “The height of a box is four inches more than a foot,” or “A chicken weighs $15$ pounds less than a turkey.” These two phrases, or phrases with similar constructions, indicate problems involving numeric comparisons.

For instance, say you are given this question: “A grapefruit weighs one pound and a watermelon weighs $11$ pounds more. How much does the watermelon weigh?”

The math itself is quite simple once you know what the question is asking. We are told the watermelon weighs “more,” so this is an additive comparison:

\[1 + 11 = 12\]

The watermelon weighs $12$ pounds.

Numeric comparisons can also be given in terms of multiples. In such cases, there are other phrases or terms that will let you know that you need to do multiplication. Consider this example:

A squirrel is $14$ inches long, and a desk is $4$ times longer than that squirrel. How long is the desk?

Solution

The desk is longer than the squirrel, but we aren’t told it is a certain amount of inches longer. Instead, we are told it is a certain amount “times” longer. That times is a key word that lets you know to multiply:

\[4 \times 14 = 56\]

The desk is $56$ inches long.

Operation Properties

The common operations of addition and multiplication have certain patterns known as properties that will ensure you always get the right answer. As the problems you see get more complex and start involving variables, knowing these properties will also allow you to take shortcuts that will make things easier.

Addition

There are two main properties for addition:

commutative property of addition—This property states that when adding, the order of numbers makes no difference:

\[15 + 9 = 9 + 15\]

Note: There is no commutative property of subtraction because the order does matter: $13-8$ doesn’t equal $8-13$.

associative property of addition—This property states that when adding three or more numbers, how they are grouped makes no difference:

\[(19 + 9) +11 = 19 + (9 + 11)\] \[28 + 11 = 19 + 20\] \[39 = 39\]

Note: Again, there is no associative property of subtraction because in subtraction the order always matters.

#### Multiplication

There are two main properties for multiplication:

commutative property of multiplication—This property states that with multiplication, the order of numbers doesn’t matter:

\[12 \times 15 = 15 \times 12\]

Note: As with subtraction, there is no commutative property for division because the order matters: $10 \div 2$ doesn’t equal $2 \div 10$.

associative property of multiplication—This property states that with multiplication, how you group numbers doesn’t matter:

\[(7 \times 3) \times 4 = 7 \times (3 \times 4)\] \[21 \times 4 = 7 \times 12\] \[84 = 84\]

There is one other property that relates to both operations:

distributive property of multiplication over addition—This property states that multiplying a number by a specific sum is the same as multiplying it by each individual part:

\[7 \times (3 + 5) = 7 \times 3 + 7 \times 5\] \[7 \times 8 = 21 + 35\] \[56=56\]

Factoring with Properties

Suppose you have the sum of two numbers that have a common factor; for instance, $21+49$, which both have $7$ for a factor. You could factor out the common factor ($7$) and write the result like this:

\[7(3+7)\]

You can then show that these two expressions are equivalent:

\[21+49=7(3+7)\]

Using the distributive property on the right side and adding the left, we get $70 = 7(10)$, which is of course simply $70=70$.

While this is an overly complicated process for a simple addition problem, the ability to factor out common factors (including variables and expressions) will be very important in algebra. It helps to understand how it works with basic numbers first.

Parentheses, Brackets, and Braces

At times, you will have multiple expressions or operations next to each other or grouped together on one side of an equation. To differentiate these different expressions or determine which operations are performed first, we use three types of grouping symbols:

$()$ are called parentheses.
$[\ ]$ are called brackets.
$\{\}$ are called braces.

Parentheses are the first level, meaning they are the first we use for grouping. Parentheses basically say, “Do inside here first.” When we have groups within groups, the next level is brackets; if there are brackets, parentheses are always used inside them. You deal with everything in brackets only after you address the parentheses. The third level is braces, with brackets always used inside braces, if there are any. Only deal with the braces after you’re done with the brackets.

This order looks like this:

\[\{[(\;)]\}\]

parentheses alone: $31 +(6 -2) +13 -(4+4)$
parentheses and brackets: $31 +[(6-2) + 13] - (4+4)$
parentheses, brackets, and braces: $\{31 + [(6-2)+13] -(4+4)\}$

Consider this expression:

\[12 +[3-5 +(8-3)]\]

The first step is inside the parentheses:

\[8-3 = 5\]

So, let’s write $5$ in place of $(8-3)$ to get:

\[12 +[3-5+5]\]

The second step is inside the brackets:

\[3-5+5 = 3\]

So, let’s write that in place of $[3-5+5]$ to get:

\[12 + 3\]

After that, there’s just one final step:

\[12 + 3 = 15\]

The Order of Operations

Those grouping symbols introduced an important concept in math: the order of operations. Lengthy and complicated mathematical operations can be simplified by following the PEMDAS rule, which states the order in which operations must be done:

Do everything in parentheses (P), left to right.
Evaluate any exponents (E), left to right.
Do all multiplication and division (MD), left to right.
Then do all addition and subtraction (AS), left to right.
A good way to remember: Please Excuse My Dear Aunt Sally.

If you do a problem in the wrong order, you will get the wrong answer. On a TABE multiple-choice question, the incorrect answers may be what you get if you do operations out of order, so it’s important you always follow PEMDAS. Here’s an example problem that illustrates the process.

\[5 - (4 + 1)^2 \div (5^2 \cdot \frac{1}{5})\]

Solution

Start with performing the operations inside the parentheses:

\[5 - (4 + 1)^2 \div (5^2 \cdot \frac{1}{5})\] \[5-(5)^2 \div (25 \cdot \frac{1}{5})\] \[5-(5)^2 \div (5)\]

Note: Within the second set of parentheses, we followed PEMDAS by doing the exponent before the multiplication.

The next step in the order of operations is exponents, so the expression becomes:

\[5-25 \div 5\]

There is no multiplication at this point, but there is division, so the expression is now:

\[5 - 5\]

Finally, we do any addition or subtraction that’s left; in this case, there is just subtraction:

\[0\]

Note: If we had incorrectly done the subtraction before the division, we would have gotten $-4$.