Chapter 2.7 notes - College Algebra

Compound Inequalities and Interval Notation

• Plotting Compound "Or" Inequalities on a Number Line:

  • Treat compound "or" inequalities as two completely separate problems on the same number line, solving and graphing them individually before combining the visualization.

  • Example Problem: Graph $x \text{ is less than or equal to } 3$ or $x \text{ is greater than } 5$

    • Inequality 1 ($x \text{ less than or equal to } 3$):

      • Place a filled-in dot at $3$ because the inequality includes "or equal to" ($\text{or equal to}$ indicates inclusion).

      • Draw an arrow extending to the left (the negative direction) without a left bound, covering all numbers less than or equal to $3$

    • Inequality 2 ($x \text{ greater than } 5$):

      • Place an open dot at $5$ because $x$ cannot equal $5$

      • Draw an arrow extending to the right (the positive direction), covering all numbers strictly greater than $5$

  • Region Testing Strategy: To verify whether the correct side of a dot is shaded/highlighted, choose a test value located within the highlighted region and substitute it into the inequality.

    • Test: Choose $6$ from the right-hand highlighted section of the second inequality.

    • Check: Is $6 > 5$? Yes, this is a true statement, confirming the correct direction was shaded.

• Interval Notation for Disconnected Regions:

  • When an inequality representation consists of separate, disconnected regions on a number line, it cannot be written as a single unified interval; it requires multiple intervals.

  • Left Region ($x \text{ less than or equal to } 3$):

    • Right endpoint is $3$, which is included, so it receives a square bracket ].

    • Extends indefinitely to the left, indicated by negative infinity $-\text{infinity}$ with a parenthesis (.

    • Interval: $(-\text{infinity}, 3]$

  • Right Region ($x \text{ greater than } 5$):

    • Left endpoint is $5$, which is excluded, so it receives a parenthesis (.

    • Extends indefinitely to the right, indicated by positive infinity $\text{infinity}$ with a parenthesis ).

    • Interval: $(5, \text{infinity})$

  • Formatting Rules for Infinity:

    • Always place parentheses around positive or negative infinity ($-\text{infinity}$ or $\text{infinity}$); brackets are never used for infinity.

  • Combining Disconnected Intervals:

    • In standard online systems (such as ALEKS), place a comma between the individual intervals to indicate that $x$ can exist in either interval.

    • In standard mathematical literature and tutorials, the union symbol $\text{union}$ is placed between intervals (e.g., $(-\text{infinity}, 3] \text{ union } (5, \text{infinity})$).

Solving Algebraic Inequalities

• Inequalities vs. Equations:

  • Solving an inequality means isolating the variable $x$ on one side of the inequality symbol, utilizing inverse algebraic operations similar to solving standard linear equations.

  • Always read directions carefully: plotting an inequality versus solving an inequality require fundamentally different steps.

• The Sign-Flipping Rule:

  • Core Rule: Multiplying or dividing both sides of an inequality by a negative number requires flipping the direction of the inequality sign.

  • Adding or subtracting values (whether positive or negative) to both sides does not flip the sign.

  • Multiplying or dividing by positive values does not flip the sign.

• Demonstration Example 1: Solving with Sign Flip:

  • Given equation step: $-17x + 13 \text{ greater than or equal to } -4$

  • Subtracting $13$ from both sides yields: $-17x \text{ greater than or equal to } -17$ (sign remains unchanged during subtraction).

  • Dividing both sides by $-17$ triggers the flipping rule:

    • $\frac{-17x}{-17} = x$

    • $\frac{-17}{-17} = 1$

    • The inequality sign flips from $\text{greater than or equal to}$ to $\text{less than or equal to}$.

  • Final Solution: $x \text{ less than or equal to } 1$

• Demonstration Example 2: Step-by-Step Fraction Inequality Solution:

  • Solve: $-x + 4 \text{ less than } \frac{1}{2}x + 1$

  • Step 1: Collect variable terms on one side:

    • Add $x$ to both sides to move variable terms to the right.

    • Left side: $-x + 4 + x = 4$

    • Right side: $\frac{1}{2}x + 1x = \frac{3}{2}x$

    • Resulting equation: $4 \text{ less than } \frac{3}{2}x + 1$ (Inequality sign does not flip).

    • Note: College-level mathematics prefers working with improper fractions (e.g., $\frac{3}{2}$) over mixed numbers ($1 \frac{1}{2}$) or decimals ($1.5$).

  • Step 2: Collect constant terms on the opposite side:

    • Subtract $1$ from both sides.

    • Left side: $4 - 1 = 3$

    • Right side: $\frac{3}{2}x + 1 - 1 = \frac{3}{2}x$

    • Resulting equation: $3 \text{ less than } \frac{3}{2}x$ (Inequality sign does not flip).

  • Step 3: Isolate the variable $x$:

    • Divide both sides by positive $\frac{3}{2}$ (Sign does not flip because $\frac{3}{2}$ is positive).

    • Apply KFC Method (Keep, Flip, Change) for fraction division:

      • Keep the first number: $3$ (written as a fraction: $\frac{3}{1}$)

      • Flip the second fraction (reciprocal): $\frac{3}{2}$ becomes $\frac{2}{3}$

      • Change division to multiplication: $\frac{3}{1} \times \frac{2}{3}$

    • Multiply numerators and denominators: $\frac{3 \times 2}{1 \times 3} = \frac{6}{3} = 2$

  • Alternative Step 3 Method:

    • Multiply both sides by $2$ first: $3 \times 2 \text{ less than } 3x \text{ implies } 6 \text{ less than } 3x$

    • Divide both sides by $3$: $\frac{6}{3} \text{ less than } x \text{ implies } 2 \text{ less than } x$

  • Final Solution: $2 \text{ less than } x$ (or $x \text{ greater than } 2$)