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Precalculus for Team-Based Inquiry Learning 2025 Edition — Instructor Version

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Section 1.1 Linear Equations and Inequalities (EQ1)

Subsection 1.1.1 Activities

Remark 1.1.1.

Recall that when solving a linear equation, you use addition, subtraction, multiplication and division to isolate the variable.
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Activity 1.1.3.

Solve the linear equation.
\begin{equation*} \dfrac{2}{3}x-8=\dfrac{5x+1}{6} \end{equation*}
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(a)
Which equation is equivalent to \(\dfrac{2}{3}x-8=\dfrac{5x+1}{6}\) but does not contain any fractions?
  1. \(\displaystyle 12x-48=15x+3\)
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  2. \(\displaystyle 3x-24=10x+2\)
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  3. \(\displaystyle 4x-8=5x+1\)
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  4. \(\displaystyle 4x-48=5x+1\)
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Answer.
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(b)
Use the simplified equation from part (a) to solve \(\dfrac{2}{3}x-8=\dfrac{5x+1}{6}\text{.}\)
  1. \(\displaystyle x=-17\)
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  2. \(\displaystyle x=-\dfrac{26}{7}\)
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  3. \(\displaystyle x=-9\)
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  4. \(\displaystyle x=-49\)
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Answer.
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Activity 1.1.4.

It is not always the case that a linear equation has exactly one solution. Consider the equation
\begin{equation*} 3+5x=5(x+2)-7\text{.} \end{equation*}
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(a)
Solve the linear equation. Which of the following is your last step?
  1. \(\displaystyle 3+5x=5x+3\)
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  2. \(\displaystyle 0=0\)
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  3. \(\displaystyle 3=3\)
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  4. \(\displaystyle 3=10\)
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Answer.
A, B, or C. Although choice A may not be the last step, some students may not need to continue solving because they can see the left side is the same as the right side.
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(b)
Which of the following best describes the statement you got in part (a)?
  1. always true
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  2. sometimes true
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  3. never true
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Answer.
A. For students who chose choice A in part (a), ask how they can show the statement to always be true by considering different \(x\)-values.
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(c)
What do you think this means about the number of solutions?
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Answer.
This equation has infinitely many solutions. Students might have trouble with this idea, so one way to help them to think about the solutions is to have them try different values of \(x\text{.}\)
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Remark 1.1.5.

In Activity 1.1.4, we saw that a linear equation can have infinitely many solutions. Linear equations can also have a unique solution or no solution!
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Activity 1.1.6.

For each part in this activity, consider the conditions for when a linear equation has a unique solution, no solution, or infinitely many solutions.
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(d)
What happens to the \(x\) variable when a linear equation has no solution or many solutions?
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Answer.
The \(x\) variable cancels leaving only constants.
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Definition 1.1.7.

A linear equation with one unique solution is a conditional equation. A linear equation that is true for all values of the variable is an identity equation. A linear equation with no solutions is an inconsistent equation.
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Definition 1.1.8.

The remainder of this section will focus on linear inequalities. A linear inequality is an inequality that can be written in one of the following forms:
  • \begin{equation*} ax+b \gt 0 \end{equation*}
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  • \begin{equation*} ax+b \lt 0 \end{equation*}
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  • \begin{equation*} ax+b \ge 0 \end{equation*}
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  • \begin{equation*} ax+b \le 0 \end{equation*}
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where \(a\) and \(b\) are real numbers and \(a \neq 0\text{.}\)
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Activity 1.1.9.

In this activity, we explore the relationship between a linear equation and a linear inequality, as well as how to express solutions on a number line and in interval notation.
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(a)
What is the solution to the linear equation \(3x-1=5\text{?}\)
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Answer.
\(x=2\)
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(b)
Which of these values are solutions of the inequality \(3x-1 \ge 5\text{?}\)
  1. \(\displaystyle x=0\)
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  2. \(\displaystyle x=2\)
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  3. \(\displaystyle x=4\)
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  4. \(\displaystyle x=10\)
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Answer.
B, C, and D
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(c)
There are more solutions to the inequality than the ones found in part (b). How would you characterize all of them?
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Answer.
All numbers greater than \(2\) and \(2\) itself.
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(e)
Express the solution of the inequality \(3x-1 \ge 5\) in interval notation.
  1. \(\displaystyle (-\infty, 2]\)
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  2. \(\displaystyle (-\infty, 2)\)
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  3. \(\displaystyle (2,\infty) \)
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  4. \(\displaystyle [2,\infty)\)
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Hint.
Use the graph from part (d) to help you write the solution in interval notation.
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Answer.
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Activity 1.1.10.

Let’s consider what happens to the inequality when the variable has a negative coefficient.
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(a)
Which of these values is a solution of the inequality \(-x\lt 8\text{?}\)
  1. \(\displaystyle x=-10\)
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  2. \(\displaystyle x=-8\)
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  3. \(\displaystyle x=4\)
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  4. \(\displaystyle x=10\)
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Answer.
C and D
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(b)
Solve the linear inequality \(-x\lt 8\text{.}\) How does your solution compare to the values chosen in part (a)?
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Answer.
\(x>-8\)
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(c)
Expression the solution of the inequality \(-x\lt 8\) in interval notation.
  1. \(\displaystyle (-\infty,-8]\)
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  2. \(\displaystyle (-\infty,-8)\)
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  3. \(\displaystyle (-8, \infty)\)
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  4. \(\displaystyle [-8,\infty)\)
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Answer.
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Remark 1.1.11.

You can treat solving linear inequalities, just like solving an equation. The one exception is when you multiply or divide by a negative value, reverse the inequality symbol.
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Activity 1.1.12.

Solve the following inequalities. Express your solution in interval notation and graphically on a number line.
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(a)
\(-3x-1 \le 5\)
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Answer.
\(x \ge -2\text{,}\) \([-2,\infty]\)
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(b)
\(3(x+4) \gt 2x-1\)
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Answer.
\(x \gt -13\text{,}\) \((-13,\infty)\)
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(c)
\(-\dfrac{1}{2}x \ge -\dfrac{3}{4}+\dfrac{5}{4}x\)
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Answer.
\(x \le \dfrac{3}{7}\text{,}\) \(\left(-\infty, \dfrac{3}{7} \right]\)
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Definition 1.1.13.

A compound inequality includes multiple inequalities in one statement.
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Activity 1.1.14.

Consider the statement \(3 \le x \lt 8\text{.}\) This really means that \(3 \le x\) and \(x \lt 8\text{.}\)
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(a)
Which of the following inequalities are equivalent to the compound inequality \(3 \le 2x-3 \lt 8 \text{?}\)
  1. \(\displaystyle 3 \le 2x-3\)
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  2. \(\displaystyle 3 \ge 2x-3\)
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  3. \(\displaystyle 2x-3 \lt 8\)
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  4. \(\displaystyle 2x-3 \gt 8\)
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Answer.
A and C
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(c)
Solve the inequality \(2x-3 \lt 8\text{.}\)
  1. \(\displaystyle x \gt \dfrac{11}{2}\)
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  2. \(\displaystyle x \lt \dfrac{11}{2}\)
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  3. \(\displaystyle x \gt \dfrac{5}{2}\)
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  4. \(\displaystyle x \lt \dfrac{5}{2}\)
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Answer.
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(d)
Which compound inequality describes how the two solutions overlap?
  1. \(\displaystyle 0 \le x \lt \dfrac{11}{2}\)
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  2. \(\displaystyle 0 \le x \lt \dfrac{5}{2}\)
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  3. \(\displaystyle \dfrac{5}{2} \lt x \le 3\)
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  4. \(\displaystyle 3 \le x \lt \dfrac{11}{2}\)
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Answer.
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Remark 1.1.15.

Solving a compound linear inequality, uses the same methods as a single linear inequality ensuring that you perform the same operations on all three parts. Alternatively, you can break the compound inquality up into two and solve separately.
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Activity 1.1.16.

Solve the following inequalities. Express your solution in interval notation and graphically on a number line.
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(a)
\(8 \lt -3x-1 \le 11\)
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Answer.
\(-4 \le x \lt -3\text{,}\) \([-4,-3)\)
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(b)
\(-6 \le \dfrac{x-12}{4} \lt -2\)
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Answer.
\(-12 \le x \lt 4\text{,}\) \([-12,4)\)
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Subsection 1.1.2 Exercises

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