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Section 3.1 One of These Things is Not Like the Other

The concept of combining like terms is not that different from the way we naturally organize information in our heads. Consider the following question: If you have one bag that contains two apples and three oranges and you had a second bag that has four apples and one orange, how many of each fruit do you have?

It should only take a moment's thought to conclude that you have six apples and four oranges. We can write this out in a mathematical notation as follows:

\begin{equation*} ( 2 \text{ apples} + 3 \text{ oranges} ) + (4 \text{ apples} + 1 \text{ orange}) = 6 \text{ apples} + 4 \text{ oranges} \end{equation*}

There's a lot to say here about the notation:

  • Notice how the parentheses serve as natural boundary markers for each bag. By looking at the mathematical notation, you know exactly what's contained in the first bag and what's contained in the second bag.

  • Notice how the addition sign corresponds to your intuition about addition. You can rearrange things so that the first set of apples are with the other set of apples, and the same thing with the oranges.

  • Notice how you have no impulse to try to combine the apples and oranges together to get "10 apple-oranges" as your final answer.

This simple word problem serves as an illustration for how we think about combining like terms. It's easier for us to see what's going on here because we have some experience handling pieces of fruit and we can easily recognize that one type of fruit is different from the other. The same concept applies to algebraic expressions, but we first need to make sure we know what we're looking at.

Definition 3.1.1. Monomial, Coefficient, Variable Part.

A monomial is any product of numbers and variables. The coefficient is the product of all of the numbers and the variable part is the product of all the variables.

Although it's possible for us to write a monomial as \(3 \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y\text{,}\) we usually take advantage of exponent notation to write it as \(3x^4 y^2\) to keep things compact. This also makes it much easier to read. In this example, the coefficient is 3 and the variable part is \(x^4 y^2\text{.}\)

There are a couple special situations to remember:

  • If you have a number by itself (such as the number 12), the term is called a constant and it is considered to have no variable part.

  • If you have a variable by itself (such as with \(x\) or \(y^2\)), the coefficient is implicitly a 1. So it would be correct to think of \(x\) as \(1x\) and \(y^2\) as \(1y^2\text{.}\)

Definition 3.1.2. Polynomial.

A polynomial is the sum of a number of monomials.

The definition says that a polynomial is a sum of monomials, not a difference of them. But we will often write polynomials with subtraction, such as with \(x^2 - 6x + 8\text{.}\) The reason for this is that we can use the idea that subtraction is addition of the opposite to rewrite it:

\begin{equation*} x^2 - 6x + 8 = x^2 + (-6x) + 8 \end{equation*}

One of the features of writing subtraction as addition is that it visually groups the negative sign with the monomial. This means that when we identify the coefficient of a monomial term, we need to account for the negative sign. Therefore, the coefficient of the \(x\) term is -6.

Activity 3.1.1. Identifying Parts of Monomials.

Understanding vocabulary is a multi-step process. It's not enough to just read the definitions and see a couple examples. In order for the definition to sink in effectively, you must take the time to work o problems yourself.

Try it!

Using the grid below, identify all of the monomials of \(x^2 + 7x - y - 8\text{.}\) Then determine the coefficient and variable part of each term.

Monomial \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\)
Coefficient
Variable Part

Solution.

Monomial \(x^2\) \(7x\) \(-y\) \(-8\)
Coefficient \(1\) \(7\) \(-1\) \(-8\)
Variable Part \(x^2\) \(x\) \(y\) NA

Definition 3.1.3. Like and Unlike Terms.

Two monomials are like terms if they have the same variable part. If their variable parts are different, then they are unlike terms.

Activity 3.1.2. Combining Like Terms.

Combining like terms is an algebraic process that uses the distributive property, but it's backwards from how it's most commonly seen. This reverse distributive property is often called "factoring out" terms. When we have like terms, we can perform the arithmetic on the coefficients to simplify the expression.

\begin{equation*} \begin{aligned} 5x + 8x \amp = (5 + 8)x \amp \eqnspacer \amp \text{Factor out the $x$} \\ \amp = 13x \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}

Notice how natural this is based on the logic of combining either the apples or the oranges. If you have 5 \(x\)-es and you add 8 more \(x\)-es, how many \(x\)-es do you have? 13 \(x\)-es.

Try it!

Simplify the expression \(7y^2 + 6y^2 - 5y^2\) using a complete presentation. Show the step where you factor out the common factor.

Solution.
\begin{equation*} \begin{aligned} 7y^2 + 6y^2 - 5y^2 \amp = (7 + 6 - 5) y^2 \amp \eqnspacer \amp \text{Factor out the $y^2$} \\ \amp = 8y^2 \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}

Activity 3.1.3. Combining Like Terms When Unlike Terms are Present.

What happens if there are unlike terms? Once again, we can think about the apples and oranges. Basically, we're just stuck with what we have. We've actually been using this already. In previous sections, if we had an expression like \(4x + 3\text{,}\) we just left it like that. And this was also true in the previous section where the different variables were all left separate.

But what if there are some terms are like terms and others are unlike terms? Think back to the apples and oranges. What happened there? In that case, the apples were combined and the oranges were combined, but the fruits remained separate. We do the same thing in algebra.

\begin{equation*} \begin{aligned} (5x - 3y) + (4x + y) \amp = 5x + 4x - 3y + y \amp \eqnspacer \amp \text{Rearrange the terms} \\ \amp = (5x + 4x) + ( - 3y + y) \amp \amp \text{Group the terms} \\ \amp = (5 + 4)x + (-3 + 1)y \amp \amp \text{Factor out the $x$ and $y$} \\ \amp = 9x - 2y \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}

Students are sometimes not very careful with parentheses. They often write the wrong thing even if what's in their head is correct. Remember that the correct communication of ideas is important in math, so it's not good enough to write the wrong thing and pretend that it's correct. Here are two very common errors that students make:

\begin{equation*} \begin{aligned} 5x + 4x - 3y + y \amp \overset{\times}{=} (5x + 4x)(-3y + y) \\ 5x + 4x - 3y + y \amp \overset{\times}{=} (5x + 4x) - (3y + y) \end{aligned} \end{equation*}

The first one happens when students simply draw parentheses around symbols without thinking about how it affects the meaning. Notice that the right side is a product of terms because the parentheses are right next to each other. The second one happens when students don't pay attention to the fact that the negative sign is part of the coefficient of the \(-3y\) term. These errors are avoidable with practice. But it takes practice because you need to train you brain to see the expressions differently.

Try it!

Simplify the expression \((-4a - 3b) + (-5a + 8b)\) using a complete presentation. Show all the steps that are in the example above.

Solution.
\begin{equation*} \begin{aligned} (-4a - 3b) + (-5a + 8b) \amp = -4a - 5a - 3b + 8b \amp \eqnspacer \amp \text{Rearrange the terms} \\ \amp = (-4a - 5a) + (-3b + 8b) \amp \amp \text{Group the terms} \\ \amp = (-4 - 5)a + (-3 + 8)b \amp \amp \text{Factor out the $a$ and the $b$} \\ \amp = -9a + 5b \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}

So far, when we have combined like terms, we've been focusing on addition. But we also need to be able to do this with mathematical expressions that have subtraction and multiplication. For each of these, we need to remember to use the distributive property.

The idea behind this one comes from the concept of multiplication representing groups of things. If you have four bags that each contain two apples and three oranges, how much fruit do you have in total?

\begin{equation*} 4 \cdot (2 \text{ apples} + 3 \text{ oranges} ) = 8 \text{ apples} + 12 \text{ oranges} \end{equation*}

This is a demonstration of an important property of algebra.

Definition 3.1.4. The Distributive Property.

The distributive property states that \(a(b + c) = ab + ac\text{.}\) The full name of this the distributive property of multiplication over addition.

Activity 3.1.4. Using the Distributive Property.

You might recall that in the order of operations, multiplication comes before addition and subtraction. This means that you should use the distributive property before you combine like terms.

\begin{equation*} \begin{aligned} (2x - 5y) + 3(x - 2y) \amp = (2x - 5y) + (3x - 6y) \amp \eqnspacer \amp \text{Distributive property} \\ \amp = 2x + 3x - 5y - 6y \amp \amp \text{Rearrange the terms} \\ \amp = (2x + 3x) + (- 5y - 6y) \amp \amp \text{Group the terms} \\ \amp = (2+3)x + (-5 - 6)y \amp \amp \text{Factor out the $x$ and the $y$} \\ \amp = 5x - 11y \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}

Try it!

Simplify the expression \((3a + 2b) + 2(5a - 3b)\) using a complete presentation. Show the distributive step, the rearrangement step, the grouping step, and the arithmetic step.

Solution.
\begin{equation*} \begin{aligned} (3a + 2b) + 2(5a - 3b) \amp = (3a + 2b) + (10a - 6b) \amp \eqnspacer \amp \text{Distributive property} \\ \amp = 3a + 10a + 2b - 6b \amp \amp \text{Rearrange the terms} \\ \amp = (3a + 10a) + (2b - 6b) \amp \amp \text{Group the terms} \\ \amp = (3 + 10)a + (2 - 6)b \amp \amp \text{Factor out the $a$ and $b$} \\ \amp = 13a - 4b \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}

Activity 3.1.5. The Distributive Property With Subtraction.

Whenever we have an algebraic symbol next to a set of parentheses with multiple terms on the inside, we need to use the distributive property. When we were doing it earlier with addition, we basically just multiplied by 1, which is why there was no change. If the operation outside of the parentheses is subtraction, then we also need to distribute that negative sign.

To understand why the negative sign should be distributed, we will think about another example using apples and oranges. Suppose you have eight apples and six oranges, and someone takes four apples and five oranges away from you. How much fruit do you have left?

\begin{equation*} (8 \text{ apples} + 6 \text{ oranges}) - (4 \text{ apples} + 5 \text{ oranges}) = 4 \text{ apples} + 1 \text{ orange} \end{equation*}

Notice that in context, it is completely natural to remember to subtract off the 5 oranges from the 6 oranges. However, when manipulating symbols that context is easily lost, which leads to errors like this one:

\begin{equation*} (8x + 6y) - (4x + 5y) \overset{\times}{=} 8x + 6y - 4x + 5y \end{equation*}

Try it!

Simplify the expression \((8x + 6y) - (4x + 5y)\) using a complete presentation. Show the distributive step, the rearrangement step, the grouping step, and the arithmetic step.

Solution.
\begin{equation*} \begin{aligned} (8x + 6y) - (4x + 5y) \amp = 8x + 6y - 4x - 5y \amp \eqnspacer \amp \text{Distributive property} \\ \amp = 8x - 4x + 6y - 5y \amp \amp \text{Rearrange the terms} \\ \amp = (8x - 4x) + (6y - 5y) \amp \amp \text{Group the terms} \\ \amp = (8-4)x + (6-5)y \amp \amp \text{Factor out the $x$ and the $y$} \\ \amp = 4x + y \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}