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Precalculus for Team-Based Inquiry Learning: 2024 Development 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.

Activity 1.1.2.

Solve the linear equations.
(a)
\(3x-8=5x+2\)
  1. \(\displaystyle x=2\)
  2. \(\displaystyle x=5\)
  3. \(\displaystyle x=-5\)
  4. \(\displaystyle x=-2\)
(b)
\(5(3x-4)=2x-(x+3)\)
  1. \(\displaystyle x=\frac{17}{14}\)
  2. \(\displaystyle x=\frac{14}{17}\)
  3. \(\displaystyle x=\frac{23}{14}\)
  4. \(\displaystyle x=\frac{14}{23}\)

Activity 1.1.3.

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

Activity 1.1.4.

It is not always the case that a linear equation has exactly one solution. Consider the following linear equations which appear similar, but their solutions are very different.
(a)
Which of these equations has one unique solution?
  1. \(\displaystyle 4(x-2)=4x+6\)
  2. \(\displaystyle 4(x-1)=4x+4\)
  3. \(\displaystyle 4(x-1)=x+4\)
(b)
Which of these equations has no solutions?
  1. \(\displaystyle 4(x-2)=4x+6\)
  2. \(\displaystyle 4(x-1)=4x+4\)
  3. \(\displaystyle 4(x-1)=x+4\)
(c)
Which of these equations has many solutions?
  1. \(\displaystyle 4(x-2)=4x+6\)
  2. \(\displaystyle 4(x-1)=4x+4\)
  3. \(\displaystyle 4(x-1)=x+4\)
(d)
What happens to the \(x\) variable when a linear equation has no solution or many solutions?

Definition 1.1.5.

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.

Activity 1.1.6.

An inequality is a relationship between two values that are not equal.
(a)
What is the solution to the linear equation \(3x-1=5\text{?}\)
(b)
Which of these values is a solution of the inequality \(3x-1 \ge 5\text{?}\)
  1. \(\displaystyle x=0\)
  2. \(\displaystyle x=2\)
  3. \(\displaystyle x=4\)
  4. \(\displaystyle x=10\)
(c)
Express the solution of the inequality \(3x-1 \ge 5\) in interval notation.
  1. \(\displaystyle (-\infty, 2]\)
  2. \(\displaystyle (-\infty, 2)\)
  3. \(\displaystyle (2,\infty) \)
  4. \(\displaystyle [2,\infty)\)
(d)
Draw the solution to the inequality on a number line.

Activity 1.1.7.

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

Remark 1.1.8.

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.

Activity 1.1.9.

Solve the following inequalities. Express your solution in interval notation and graphically on a number line.
(a)
\(-3x-1 \le 5\)
(b)
\(3(x+4) \gt 2x-1\)
(c)
\(-\frac{1}{2}x \ge -\frac{2}{4}+\frac{5}{4}x\)

Definition 1.1.10.

A compound inequality includes multiple inequalities in one statement.

Activity 1.1.11.

Consider the statement \(3 \le x \lt 8\text{.}\) This really means that \(3 \le x\) and \(x \lt 8\text{.}\)
(a)
Which of the following inequalities are equivalent to the compound inequality \(3 \le 2x-3 \lt 8 \text{?}\)
  1. \(\displaystyle 3 \le 2x-3\)
  2. \(\displaystyle 3 \ge 2x-3\)
  3. \(\displaystyle 2x-3 \lt 8\)
  4. \(\displaystyle 2x-3 \gt 8\)
(b)
Solve the inequality \(3 \le 2x-3\text{.}\)
  1. \(\displaystyle x \le 0\)
  2. \(\displaystyle x \ge 0\)
  3. \(\displaystyle x \le 3\)
  4. \(\displaystyle x \ge 3\)
(c)
Solve the inequality \(2x-3 \lt 8\text{.}\)
  1. \(\displaystyle x \gt \frac{11}{2}\)
  2. \(\displaystyle x \lt \frac{11}{2}\)
  3. \(\displaystyle x \gt \frac{5}{2}\)
  4. \(\displaystyle x \lt \frac{5}{2}\)
(d)
Which compound inequality describes how the two solutions overlap?
  1. \(\displaystyle 0 \le x \lt \frac{11}{2}\)
  2. \(\displaystyle 0 \le x \lt \frac{5}{2}\)
  3. \(\displaystyle \frac{5}{2} \lt x \le 3\)
  4. \(\displaystyle 3 \le x \lt \frac{11}{2}\)
(e)
Draw the solution to the inequality on a number line.

Remark 1.1.12.

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.

Activity 1.1.13.

Solve the following inequalities. Express your solution in interval notation and graphically on a number line.
(a)
\(8 \lt -3x-1 \le 11\)
(b)
\(-6 \le \frac{x-12}{4} \lt -2\)

Subsection 1.1.2 Videos

It would be great to include videos down here, like in the Calculus book!
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