4.2 Linear Equations and Inequalities in One Variable

Linear Equations and Inequalities in One Variable

Once students understand how to manipulate algebraic expressions, the next logical step is to set expressions equal to each other or compare them using inequalities. This section covers solving multi-step linear equations and inequalities, exploring different types of equation outcomes, and handling absolute value cases.

Solving Multi-Step Linear Equations

A linear equation in one variable is an equation that can be written in the standard form $ax + b = c$. Solving a multi-step equation requires isolating the variable on one side of the equals sign using inverse operations. When an equation features variables on both sides, parentheses, and fractional coefficients, a systematic approach is necessary.

Here is a general sequence of steps to solve complex linear equations:

1. Clear fractions or decimals: If the equation has fractions, multiplying every single term by the Least Common Denominator (LCD) eliminates the fractions entirely, making the problem easier to manage.
2. Use the distributive property: Remove any parentheses by distributing coefficients into groups.
3. Combine like terms: Simplify each side of the equation independently.
4. Move variable terms to one side: Use addition or subtraction to get all terms containing the variable on the same side of the equals sign.
5. Isolate the variable: Use addition, subtraction, multiplication, and division to solve for the variable.

Example: Solve $\frac{1}{2}(4x - 6) = \frac{2}{3}x + 5$

1. Distribute the $\frac{1}{2}$: $2x - 3 = \frac{2}{3}x + 5$
2. Clear the fraction by multiplying the entire equation by $3$: $3(2x) - 3(3) = 3(\frac{2}{3}x) + 3(5) \rightarrow 6x - 9 = 2x + 15$
3. Subtract $2x$ from both sides: $4x - 9 = 15$
4. Add $9$ to both sides: $4x = 24$
5. Divide by $4$: $x = 6$

Identity vs. Contradictory Equations

Not all linear equations have exactly one unique solution. Sometimes, the variables cancel each other out entirely during the solving process. The remaining statement dictates the number of solutions.

• Identity Equations (Infinite Solutions): If simplifying the equation leads to a true mathematical statement (e.g., $5 = 5$ or $2x = 2x$), the equation is an identity. This means any real number substituted for the variable will make the equation true, resulting in infinitely many solutions.
• Contradictory Equations (No Solution): If simplifying leads to a definitively false statement (e.g., $3 = -7$), the equation is a contradiction. There is no real number that can satisfy the equation, meaning it has no solution.

Solving Linear Inequalities

Solving a linear inequality (like $2x - 3 < 7$) follows almost the exact same process as solving a linear equation, with one critical, vital exception.

Golden Rule of Inequalities: Whenever you multiply or divide both sides of an inequality by a negative number, you MUST flip the direction of the inequality symbol. If you fail to flip the sign, your solution set will be completely incorrect.

Example: Solve and describe the graph for $-3(x - 2) \ge 15$

1. Distribute: $-3x + 6 \ge 15$
2. Subtract $6$: $-3x \ge 9$
3. Divide by $-3$ AND FLIP THE SIGN: $x \le -3$

Graphing Solutions on a Number Line:

• Use an open circle for $<$ (less than) or $>$ (greater than) to indicate the boundary number is not included.
• Use a closed (solid) circle for $\le$ (less than or equal to) or $\ge$ (greater than or equal to) to indicate the boundary number is included.
• Shade the number line in the direction of the solution set.

Absolute Value Equations and Inequalities

The absolute value of a number represents its distance from zero on a number line, meaning it is always non-negative. To solve equations involving absolute value, you must account for both the positive and negative scenarios.

Absolute Value Equations: $|x| = a$ (where $a > 0$) implies $x = a$ or $x = -a$. Example: Solve $|2x - 5| = 7$ Set up two separate equations: $2x - 5 = 7$ and $2x - 5 = -7$. Solve the first: $2x = 12 \rightarrow x = 6$. Solve the second: $2x = -2 \rightarrow x = -1$. The solutions are $x = 6$ and $x = -1$.

Absolute Value Inequalities:

• "Less than" (AND inequalities): $|x| < a$ translates to $-a < x < a$. This creates a bounded segment on the number line. (Mnemonic: "Less th-AND")
• "Greater than" (OR inequalities): $|x| > a$ translates to $x > a$ OR $x < -a$. This creates two diverging rays on the number line. (Mnemonic: "Great-OR")

Detailed Worked Example: Solving a Multi-Step Equation

Let's solve a complex equation involving fractions, distribution, and variables on both sides: $\frac{3}{4}(8x - 12) = 2(3x - 1) - x$

Step 1: Distribute on both sides. $\frac{3}{4}(8x) - \frac{3}{4}(12) = 6x - 2 - x$ $6x - 9 = 6x - 2 - x$

Step 2: Combine like terms on the right side. $6x - 9 = 5x - 2$

Step 3: Move variables to one side by subtracting $5x$ from both sides. $x - 9 = -2$

Step 4: Isolate $x$ by adding $9$ to both sides. $x = 7$

Teaching Scenario: Analyzing Student Errors with Negative Inequalities

The most common error in solving inequalities is forgetting to flip the inequality sign when dividing or multiplying by a negative number. Consider the inequality $-2x > 10$. Students will often divide by $-2$ and write $x > -5$. To correct this conceptually, you can have students test a number in their incorrect solution set, such as $x = 0$. Plugging $0$ into the original inequality gives $-2(0) > 10$, which is $0 > 10$ (False). This proves the solution set is wrong. Then, demonstrate solving it by moving the variable to avoid negative division: $-2x > 10$ $0 > 2x + 10$ $-10 > 2x$ $-5 > x$ (which is $x < -5$). This shows exactly why the rule exists and solidifies understanding.

Step-by-Step Derivation of Absolute Value Inequalities

Why does $|x| < a$ become $-a < x < a$? Recall that absolute value is distance from zero. If $|x| < a$, the distance of $x$ from $0$ is less than $a$. On a number line, starting at $0$, if you move a distance less than $a$ to the right, you land on numbers less than $a$ ($x < a$). If you move a distance less than $a$ to the left, you land on numbers greater than $-a$ ($x > -a$). For $x$ to satisfy the condition, both must be true simultaneously. Thus, $x$ must be between $-a$ and $a$, which is written precisely as $-a < x < a$. Conversely, $|x| > a$ means the distance is greater than $a$. You must move further right than $a$ ($x > a$) or further left than $-a$ ($x < -a$). These cannot happen simultaneously, hence the use of "OR".