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Chapter 1 Basic Algebraic Presentation

Section 1.1 How We Write Affects How We Think

Welcome to college level mathematics! It is a well-known fact that many students struggle with college mathematics. There are many people who proudly announce that they took a college math course three or four times before they passed.
But what is less known are the reasons why students struggle with college level mathematics. Some will cite math anxiety, others may blame the instructor, and others will simply say that they’re not math people. And there’s probably a hint of truth to all of those.
A more basic reason for struggling with mathematics is that most students simply have not learned to think mathematically before they get to college. Maybe they’ve memorized some formulas and some basic ways to manipulate equations, but there’s often a gap of understanding and the inability to communicate mathematics effectively.
The easiest way to spot this is when a student says, "I know what I’m doing, but I can’t explain it." This might be good enough at lower levels of mathematics, where the goal is simply to get the answer, but college level mathematics is different. We want students to be able to explain what they’re doing and why.
These materials are designed to help students bridge that gap so that they can become mathematical thinkers. But what is mathematical thinking? While it’s hard to describe it in complete detail, it includes (and is not limited to) the following:
  • The ability to think logically and analytically to solve problems
  • The proper manipulation of symbols in the process of solving problems
  • The ability to communicate and explain the ideas behind algebraic manipulations and the reasons for using them
As you work your way through these materials, you should be finding yourself being able to do these things better through thoughtful practice and repetition. Just as with a foreign language, it’s not enough to just say a phrase once to learn it. You need to do it over and over again, and you need to see it or hear in multiple contexts.

Activity 1.1. The Basic Presentation Expectations.

We’re going to start with some basic ideas for communicating mathematics. For the purposes of this book, a complete presentation of a mathematical manipulation includes the following pieces:
  • A series of equations or expressions with the equal signs lined up vertically
  • An explanation of the manipulations on the right side of each manipulation
Here is an example of solving the equation \(3x + 4 = 16\) using a complete presentation:
\begin{equation*} \begin{aligned} 3x + 4 \amp = 16 \\ 3x \amp = 12 \amp \eqnspacer \amp \text{Subtract $4$ from both sides} \\ x \amp = 4 \amp \amp \text{Divide both sides by $3$} \end{aligned} \end{equation*}
Try it!
Using Activity 1.1 as a model, solve the equation \(5x - 7 = 23\text{.}\)
Solution.
\begin{equation*} \begin{aligned} 5x - 7 \amp = 23 \\ 5x \amp = 30 \amp \eqnspacer \amp \text{Add $7$ to both sides} \\ x \amp = 6 \amp \amp \text{Divide both sides by $5$} \end{aligned} \end{equation*}

Activity 1.2. Lining up Equal Signs Vertically.

Lining up the equal signs is mostly a matter of organization and readability. Some students learned to do this because then they can use up the entire width of their paper and can cram in more problems per page. Unfortunately, as problems become more complex, this leads to all sorts of small errors that could easily be avoided by a more organized presentation.
Consider the following set of equations:
\begin{equation*} 5x + 9 = -3x - 7 \qquad 5x = - 3x + 16 \qquad 2x = 16 \qquad x = 8 \end{equation*}
There are two errors embedded into these calculations. Notice how much your eyes have to go back and forth to match up the terms. Lining up the equal signs vertically reduce that distance and make it easier to see when terms change or disappear. It also helps you to keep track of what is on the left side of the equation and what is on the right side of the equation.
Try it!
Present the calculation in Activity 1.2 using a complete presentation. Be sure to fix the errors.
Solution.
The two errors:
  1. Line 2: Wrong sign on the 16
  2. Line 3: Calculation error
\begin{equation*} \begin{aligned} 5x + 9 \amp = -3x - 7 \\ 5x \amp = -3x - 16 \amp \eqnspacer \amp \text{Subtract $9$ from both sides} \\ 8x \amp = -16 \amp \amp \text{Add $3x$ to both sides} \\ x \amp = -2 \amp \amp \text{Divide both sides by $8$} \end{aligned} \end{equation*}

Activity 1.3. A Common Presentation Problem.

There is a common way of writing equations that some people use which is less than ideal:
There are a number of problems with this. To start, there’s not really an organized way to read this. The best way to think of it is that it breaks apart into four pieces with three of them overlapping each other. Another problem is that the reader is expected to simply know what’s happening. This is sometimes at the root of students’ complaints about "skipped steps."
Try it!
Write the calculation in Activity 1.3 using a complete presentation.
Solution.
\begin{equation*} \begin{aligned} 4x + 7 \amp = 19 \\ 4x \amp = 12 \amp \eqnspacer \amp \text{Subtract $7$ from both sides} \\ x \amp = 3 \amp \amp \text{Divide both sides by $4$} \end{aligned} \end{equation*}

Activity 1.4. Describing the Steps.

One other aspect of taking the time to carefully write what you’re doing is that it causes you to think more clearly about what you’re doing. Consider the following equation: \(2x = \frac{1}{2}\text{.}\) A fair number of students see this and think they should "cancel out" the 2 from both sides. This is not correct. There are times you can cancel out numbers if one of them is in the numerator and the other is in the denominator, but this isn’t one of them.
The act of explicitly stating what mathematical operation you’re performing helps your brain to make categories of information. A more simple "error" of this type is that some students refer to all algebraic manipulations as "moving the term to the other side." Here are two examples of what that could mean to students:
Each of these is a different mathematical operation and a different algebraic step. It is not uncommon to see the following mistakes:
Usually, when I ask students to state in words what they did, they’re able to see their mistake for themselves. So the act of stating the mathematical operation in words allows students to avoid errors.
Try it!
Solve the equations \(3x + 5 = 14\) and \(3x - 7 = 14\) using a complete presentation.
Solution.
\begin{equation*} \begin{aligned} 3x + 5 \amp = 14 \\ 3x \amp = 9 \amp \eqnspacer \amp \text{Subtract $5$ from both sides} \\ x \amp = 3 \amp \amp \text{Divide both sides by $3$} \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} 3x - 7 \amp = 14 \\ 3x \amp = 21 \amp \eqnspacer \amp \text{Add $7$ to both sides} \\ x \amp = 7 \amp \amp \text{Divide both sides by $3$} \end{aligned} \end{equation*}
You will not always need to use a complete presentation. There are many times when it is tedious and unnecessary to state what is happening at every single algebraic step. This becomes more and more true as your basic algebra gets stronger and stronger. What ends up happening is that the steps that need to be described become less about the small details and more about the big picture. When solving for \(x\text{,}\) it may be enough to say, "Apply the quadratic formula" and not have to explain all of the individual arithmetic steps.
That being said, it is very important to develop the habit of keeping your equal signs lined up when writing math. If there is any one writing habit you develop from these worksheets, it should be that one. This one change by itself does a lot of things for you that you may not be consciously aware of.

Section 1.2 Worksheets

PDF Version of these Worksheets
 1 
external/worksheets/01-Worksheets.pdf

Worksheet Worksheet 1

1.
Circle and identify the two features that make the equations below make a complete presentation of the mathematical manipulation.
\begin{equation*} \begin{aligned} -6x + 16 \amp = -8 \\ -6x \amp = -24 \amp \eqnspacer \amp \text{Subtract $16$ from both sides} \\ x \amp = 4 \amp \amp \text{Divide both sides by $6$} \end{aligned} \end{equation*}
2.
Solve the equation \(5x + 9 = 34\) using a complete presentation.
3.
Solve the equation \(4x - 13 = 8\) using a complete presentation.
4.
Work backwards from the given information to derive the original presentation.
\begin{equation*} \begin{aligned} \\ \\ \amp \amp \eqnspacer \amp \text{Add $8$ to both sides} \\ \\ x \amp = -3 \amp \amp \text{Divide both sides by $5$} \end{aligned} \end{equation*}

Worksheet Worksheet 2

1.
Solve the equation \(6x + 9 = -21\) using a complete presentation.
2.
Solve the equation \(4x + 7 = 7\) using a complete presentation.
3.
Solve the equation \(-5t - 27 = -6t\) using a complete presentation.
4.
Work backwards from the given information to derive the original presentation.
\begin{equation*} \begin{aligned} \\ \amp \amp \eqnspacer \amp \text{Add $3y$ to both sides} \\ \\ \amp \amp \amp \text{Subtract $6$ from both sides} \\ \\ 7 \amp = y \amp \amp \text{Divide both sides by $-2$} \\ \\ y \amp = 7 \amp \amp \text{Rewrite in the conventional order} \end{aligned} \end{equation*}

Worksheet Worksheet 3

1.
Solve the equation \(-9x + 3 = 66\) using a complete presentation.
2.
Solve the equation \(3x + 9 = -9\) using a complete presentation.
3.
Solve the equation \(-8x + 17 = -4x + 41\) using a complete presentation.
4.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*} \begin{aligned} -9x + 16 \amp = 7x - 8 \\ -9x \amp = 7x - 24 \amp \eqnspacer \amp \text{Subtract $16$ from both sides} \\ -2x \amp = -24 \amp \amp \text{Subtract $7x$ from both sides} \\ x \amp = 12 \amp \amp \text{Divide both sides by $-2$} \end{aligned} \end{equation*}

Worksheet Worksheet 4

1.
Solve the equation \(-3x + 12 = 5x - 8\) using a complete presentation.
2.
Solve the equation \(4x + 7 = -3x + 7\) using a complete presentation.
3.
Perform the indicated algebraic manipulations.
\begin{equation*} \begin{aligned} x^2 \amp = 6x + 7 \\ \\ \amp \amp \eqnspacer \amp \text{Subtract $6x$ from both sides} \\ \\ \amp \amp \amp \text{Add $9$ to both sides} \end{aligned} \end{equation*}
4.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*} \begin{aligned} 5x + 8 \amp = -2x - 8 \\ 5x \amp = -2x \amp \eqnspacer \amp \text{Subtract $8$ from both sides} \\ 7x \amp = 0 \amp \amp \text{Add $2x$ to both sides} \\ x \amp = 0 \amp \amp \text{Divide both sides by $7$} \end{aligned} \end{equation*}

Worksheet Worksheet 5

1.
Solve the equation \(7x - 15 = -2x + 7\) using a complete presentation.
2.
Solve the equation \(-5x - 3 = -3x + 3\) using a complete presentation.
3.
Work backwards from the given information to derive the original presentation.
\begin{equation*} \begin{aligned} \amp \amp \eqnspacer \amp \\ \\ \amp \amp \amp \text{Subtract $3x$ from both sides} \\ \\ 2x + 1 \amp = 5x -4 \amp \amp \text{Add $9$ to both sides} \end{aligned} \end{equation*}
4.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*} \begin{aligned} 6x + 4 \amp = -3x - 10 \\ 6x \amp = -3x - 14 \amp \eqnspacer \amp \text{Subtract $4$ from both sides} \\ 9x \amp = -14 \amp \amp \text{Add $3x$ to both sides} \\ x \amp = \frac{14}{9} \amp \amp \text{Divide both sides by $9$} \end{aligned} \end{equation*}

Section 1.3 Deliberate Practice: Solving Equations for a Variable

Algebra is a skill, which means it requires practice to become proficient. But it will take more than rote repetition to get there. Deliberate practice is the thoughtful repetition of a task. For each of these sections, you will be given a list of specific skills or ideas to focus on as you practice thinking through the problems.
Focus on these skills:
  • Write the original equation.
  • Line up your equal signs.
  • Correctly state the algebraic steps using the correct phrasing:
    • Add (expression) to both sides.
    • Subtract (expression) from both sides.
    • Multiply both sides by (expression).
    • Divide both sides by (expression).
  • Execute the algebra and arithmetic correctly.
  • Present your work legibly.

Worksheet Worksheet

Instructions: Solve the equations for the variable using a complete presentation.
1.
\(7x + 34 = 13\)
2.
\(2y + 24 = -32\)
3.
\(5t - 7 = 7\)
4.
\(6u - 18 = -31\)
5.
\(3a - 44 = -38\)
6.
\(2z + 24 = 24\)
7.
\(5b - 1 = 40\)
8.
\(11x + 40 = 13\)
9.
\(6t - 19 = -35\)
10.
\(9y - 40 = 32\)
11.
\(-4u - 31 = -25u - 10\)
12.
\(4v - 40 = -14v + 14\)
13.
\(7r + 38 = 8r + 32\)
14.
\(-9b + 3 = -8b + 14\)
15.
\(6c + 7 = 7c + 13\)
16.
\(2t - 10 = -3t + 10\)
17.
\(17u - 5 = 3u + 23\)
18.
\(-x - 1 = 2x + 83\)
19.
\(2m + 6 = 4m - 2\)
20.
\(s + 9 = -2s + 18\)

Section 1.4 Closing Ideas

We have been implicitly using some very core mathematical ideas throughout this section. Specifically, we have been working with the axioms of equality. All of the axioms are built on the idea that if you start with two quantities that are equal, and then perform the same manipulation to both of them, that the results should be equal. Here’s what this formally looks like:

Definition 1.1. Axioms of Equality.

Let \(a\text{,}\) \(b\text{,}\) and \(c\) be real numbers. The axioms of equality state that
  1. If \(a = b\text{,}\) then \(a + c = b + c\text{.}\)
  2. If \(a = b\text{,}\) then \(a - c = b - c\text{.}\)
  3. If \(a = b\text{,}\) then \(ac = bc\text{.}\)
  4. If \(a = b\text{,}\) then \(\frac{a}{c} = \frac{b}{c}\text{,}\) provided that \(c \neq 0\text{.}\)
This may look intimidating at first, but it’s much less so once you recognize that this is what you’ve been doing the entire section. It just says that if we add to, subtract from, multiply, or divide (as long as we don’t divide by zero) both sides by the same value, then we maintain the equality. And this is what allows all of our algebraic manipulations to make sense.
It should also be fairly intuitive that if we started with an equality but treated the two sides differently, we would break the equality. For example, if we start off with \(5 = 5\text{,}\) but then add 3 to the left side and subtract 2 from the right side, the two sides will simply be different values.
The important takeaway from this is that these axioms simply make sense. It’s hard to imagine a world where we each start with five apples, we both eat one of our own apples, but we somehow end up with two different numbers of apples. And that’s supposed to be true about math in general. Math is supposed to make sense. In some ways, mathematics represents the epitome of strict logical reasoning.
Unfortunately, for many students, math is anything but that. There are so many symbols, so many rules, and so many manipulations that they have been forced to memorize that the whole thing is an incomprehensible mess. This book is our attempt to change that. The focus here is not going to be just on manipulating symbols, but really understanding what’s going on. The emphasis will be placed on making sure you are clear in your thinking and clear in your communication. These things are far more important than just running you through the gamut of algebraic manipulations for the second, third, or fourth time in your life.
The hope is that as you start to think about math differently, you will find more and more of it simply makes sense.

Section 1.5 Going Deeper: Clearing the Denominator

For the problems in this section, it was possible that you ended up with a fraction in your final answer, but you were not given any fractions in the original problems. Students tend to have a negative relationship with fractions. Later in the book, we’re going to take a very close look at them. But for now, we’re going to focus on just some basic algebra involving fractions.
Many students are at least vaguely familiar with something called "cross multiplication." This is a technique for working with the specific situation when you have two fractions equal to each other and you’re trying to solve for the variable. Here is how it’s often portrayed:
Many teachers teach this method, but there are some downsides to it. It creates an additional rule for students to remember, it’s often applied incorrectly, and it doesn’t teach or reinforce any specific algebraic concepts.
Fractions in equations introduce their own set of difficulties because fraction manipulations involve their own concepts (for example, common denominators and reducing) which make things more complicated. Unfortunately, this leads to the tendency to create even more manipulations for students to learn so that you have one set of rules for fraction equations and a different set of rules for non-fraction equations. (If you’re not comfortable with fractions, there’s a very brief fraction review in Subsection 15.1.1, and a more thorough discussion that runs through Section 17.1, Section 18.1, Section 19.1, and Section 20.1.)
If you look back at Definition 1.1, you might notice that there’s one algebraic step that we did not use in this section. We never multiplied both sides of the equation by a value in our attempts to solve for the variable.
A very broad category of algebraic manipulations is known as "clearing the denominators." And this technique is built on multiplying both sides of the equation by the same value. The basic premise of this method is that it’s often much easier to solve equations when there aren’t fractions, and so it can make sense to eliminate them from the equation as the first step.
\begin{equation*} \begin{aligned} \frac{3x}{4} \amp = \frac{5}{3} \\ 12 \cdot \frac{3x}{4} \amp = 12 \cdot \frac{5}{3} \amp \eqnspacer \amp \text{Multiply both sides by $12$} \\ 3 \cdot \cancel{4} \cdot \frac{3x}{\cancel{4}} \amp = \cancel{3} \cdot 4 \cdot \frac{5}{\cancel{3}} \amp \amp \text{Reduce} \\ 9x \amp = 20 \amp \amp \text{Simplify} \end{aligned} \end{equation*}
You can argue that cross-multiplication is "faster" in this case, and you would probably be correct. But that doesn’t mean that cross-multiplication isn’t without its drawbacks. The blind application of cross-multiplication can lead to numbers that are larger than necessary. Consider the following:
\begin{equation*} \begin{aligned} \frac{5x}{12} \amp = \frac{3}{8} \\ 24 \cdot \frac{5x}{12} \amp = 24 \cdot \frac{3}{8} \amp \eqnspacer \amp \text{Multiply both sides by $24$} \\ 2 \cdot \cancel{12} \cdot \frac{5x}{\cancel{12}} \amp = 3 \cdot \cancel{8} \cdot \frac{3}{\cancel{8}} \amp \amp \text{Reduce} \\ 10x \amp = 9 \amp \amp \text{Simplify} \end{aligned} \end{equation*}
If we were to use cross-multiplication, we would have ended up with \(40x = 36\text{.}\) Then after dividing, we would have had to reduce the fraction, which creates more opportunities for error.
And then there’s the challenge of more complex equations. Consider the following:
\begin{equation*} \frac{3x}{4} + \frac{1}{3} = \frac{5}{6} \end{equation*}
Students that have learned cross-multiplication will often try to apply it to this situation, even though it doesn’t apply. When students learn rule-based mathematics, they will go through all sorts of interesting machinations to try to apply a rule where it doesn’t belong because they simply don’t know what else to do.
We could solve this using fraction manipulations, but most people (including instructors) prefer not to do that if they don’t have to. And clearing the denominator is the way around that. The trick is to multiply both sides of the equation by a number that causes all the denominators to cancel out.
\begin{equation*} \begin{aligned} \frac{3x}{4} + \frac{1}{3} \amp = \frac{5}{6} \\ 12 \cdot \left( \frac{3x}{4} + \frac{1}{3} \right) \amp = 12 \cdot \frac{5}{6} \amp \eqnspacer \amp \text{Multiply both sides by $12$} \\ 12 \cdot \frac{3x}{4} + 12 \cdot \frac{1}{3} \amp = 12 \cdot \frac{5}{6} \amp \amp \text{Distribute} \\ 3 \cdot \cancel{4} \cdot \frac{3x}{\cancel{4}} + \cancel{3} \cdot 4 \cdot \frac{1}{\cancel{3}} \amp = 2 \cdot \cancel{6} \cdot \frac{5}{\cancel{6}} \amp \amp \text{Reduce} \\ 9x + 4 \amp = 10 \amp \amp \text{Simplify} \end{aligned} \end{equation*}
This technique is used much further along in mathematics. For example, there’s a technique called "partial fraction decomposition" that is used in calculus to break apart a fraction into simpler components. It comes down to working with an equation that might look like the following:
\begin{equation*} \frac{A}{x-2} + \frac{B}{x+1} = \frac{x+4}{(x-2)(x+1)} \end{equation*}
The goal is to solve for \(A\) and \(B\text{.}\) A common first step is to clear the denominators so that you don’t have to deal with fractions, and that manipulation looks very similar to the one above.
\begin{equation*} \begin{aligned} \frac{A}{x-2} + \frac{B}{x+1} \amp = \frac{x+4}{(x-2)(x+1)} \\ (x-2)(x+1) \cdot \left( \frac{A}{x-2} + \frac{B}{x+1} \right) \amp = (x-2)(x+1) \cdot \frac{x+4}{(x-2)(x+1)} \\ (x-2)(x+1) \cdot \frac{A}{x-2} + (x-2)(x+1) \cdot \frac{B}{x+1} \amp = (x-2)(x+1) \cdot \frac{x+4}{(x-2)(x+1)} \\ \cancel{(x-2)}(x+1) \cdot \frac{A}{\cancel{x-2}} + (x-2)\cancel{(x+1)} \cdot \frac{B}{\cancel{x+1}} \amp = \cancel{(x-2)}\cancel{(x+1)} \cdot \frac{x+4}{\cancel{(x-2)}\cancel{(x+1)}} \\ (x+1)A + (x-2)B \amp = x+4 \end{aligned} \end{equation*}
This is only the first step of this part of the problem. In practice, the next step would be to determine the values of \(A\) and \(B\text{,}\) then use this equation to substitute for the integrand of an integral, which then needs to be integrated.
Fortunately, you don’t need to worry about this right now. The point is that the algebraic techniques you’re learning right now are the same algebraic techniques you’re going to see down the line, especially if you are on a track that’s taking you towards calculus. It is important to do your best to build a solid foundation now so that you will be ready when you see more complicated manipulations in the future. It is an incredibly difficult task to do both at the same time.