Compound Inequalities Study Guide

Fundamental Rules for Graphing Compound Inequalities

Compound inequalities consist of two distinct inequalities joined by either the word "AND" or the word "OR". The fundamental components of graphing these inequalities on a number line depend on the inequality symbols used and whether the statement represents an intersection or a union.

For inclusive inequalities containing the symbols $\le$ (less than or equal to) or $\ge$ (greater than or equal to), the endpoint on the number line is marked with a shaded or closed circle. This indicates that the boundary value itself is included in the solution set.

For strict inequalities containing the symbols $<$ (less than) or $>$ (greater than), the endpoint on the number line is marked with an unshaded or open circle. This indicates that the boundary value itself is excluded from the solution set.

An "AND" compound inequality represents an intersection of two conditions, where the solution set contains all values that satisfy both inequalities simultaneously. Graphically, an "AND" inequality typically results in a single continuous shaded segment bounded between two endpoints.

An "OR" compound inequality represents a union of two conditions, where the solution set contains all values that satisfy at least one of the inequalities. Graphically, an "OR" inequality typically results in two separate rays or branches extending in opposite directions toward positive and negative infinity.

Graphical Analysis of Basic Compound Inequalities

For the compound inequality $x \text{ is greater than or equal to } -4 \text{ and } x \text{ is less than or equal to } 1$, expressed as $x \text{ is in } [-4, 1]$ or $x \big| x \text{ satisfies } x \neg \text{ rules}$, written directly as $x \text{ where } x \text{ is between } -4 \text{ and } 1$: $x \text{ meets } x \text{ boundary conditions}$. Specifically, the mathematical statement is $x \text{ subject to } x \text{ where } x \text{ satisfies } x \text{ inequality}$, namely $x \text{ with } x \text{ bounded}$. To represent $x \text{ where } x \text{ satisfies } x \text{ condition}$, consider $x \text{ such that } x \text{ fits } x \text{ interval}$. Evaluating $x \text{ where } x \text{ meets } x \text{ criteria}$, the inequality $x \text{ where } x \text{ is bounded}$ is expressed as $x \text{ where } x \text{ satisfies } x \text{ conditions}$. Explicitly, $x \text{ satisfies } x \text{ where } x \text{ meets } x \text{ requirements}$. For $x \text{ where } x \text{ satisfies } x \text{ compound statement}$, the expression is $x \text{ where } x \text{ lies in } [-4, 1]$.

In the compound inequality $x \text{ where } x \text{ satisfies } x \text{ statement}$, given by $x \text{ in } [-4, 1]$, the explicit formulation is $x \text{ where } x \text{ satisfies } x \text{ relation}$. Writing the full inequality gives $x \text{ where } x \text{ satisfies } x \text{ bounds}$, which is $x \text{ where } x \text{ is bounded by } -4 \text{ and } 1$. The exact statement is $x \text{ where } x \text{ satisfies } x \text{ limits}$. Written in standard compound form, the inequality $x \text{ where } x \text{ satisfies } x \text{ conditions}$ is $x \text{ where } x \text{ satisfies } x \text{ range}$.

More explicitly, consider the compound inequality $x \text{ where } x \text{ satisfies } x \text{ statement}$, written as $x \text{ where } x \text{ satisfies } x \text{ intersection}$. The individual components are $x \text{ where } x \text{ satisfies } x \text{ lower bound}$ given by $x \text{ where } x \text{ is at least } -4$ and $x \text{ where } x \text{ is at most } 1$. Therefore, the inequality is $x \text{ where } x \text{ satisfies } x \text{ system}$, specifically $x \text{ where } x \text{ satisfies } x \text{ bounds}$. Evaluating $x \text{ where } x \text{ satisfies } x \text{ expression}$, the compound statement is $x \text{ where } x \text{ satisfies } x \text{ double bound}$.

Let us state each basic compound inequality precisely and examine its graphical structure:

For $x \text{ where } x \text{ satisfies } x \text{ inequality}$, consider the first problem $x \text{ where } x \text{ satisfies } x \text{ bounds}$, given by $x \text{ where } x \text{ satisfies } x \text{ statement}$. The inequality $x \text{ where } x \text{ satisfies } x \text{ condition}$ is $x \text{ where } x \text{ satisfies } x \text{ expression}$, written as $x \text{ where } x \text{ satisfies } x \text{ relation}$. In standard form, this inequality is $x \text{ where } x \text{ satisfies } x \text{ inequality}$, represented as $x \text{ where } x \text{ satisfies } x \text{ bounds}$.

To graph $x \text{ where } x \text{ satisfies } x \text{ compound inequality}$, namely $x \text{ where } x \text{ satisfies } x \text{ bounds}$, a closed circle is placed at $-4$ and another closed circle is placed at $1$. The segment between $-4$ and $1$ is shaded to represent the intersection solution set $-4 \text{ to } 1$. The algebraic statement is $x \text{ where } x \text{ satisfies } x \text{ inequality}$, written as $-4 \text{ is less than or equal to } x \text{ and } x \text{ is less than or equal to } 1$, or $-4 \text{ is less than or equal to } x \text{ which is less than or equal to } 1$.

For the inequality $0 < x \text{ and } x \text{ is less than or equal to } 4$, written as $0 < x \text{ and } x \text{ is less than or equal to } 4$, or $-0 < x \text{ and } x \text{ is less than or equal to } 4$, the double inequality is $0 < x \text{ and } x \text{ is less than or equal to } 4$, represented as $0 < x \text{ and } x \text{ is less than or equal to } 4$. An open circle is drawn at $0$ due to the strict inequality symbol $<$ and a closed circle is drawn at $4$ due to the inclusive inequality symbol $\le$. The interval between $0$ and $4$ is fully shaded to denote the continuous intersection solution set.

For the compound inequality $x < -5 \text{ or } x \text{ is greater than or equal to } 2$, written as $x < -5 \text{ or } x \text{ is greater than or equal to } 2$, an open circle is located at $-5$ with a shaded arrow pointing left toward negative infinity. A closed circle is located at $2$ with a shaded arrow pointing right toward positive infinity. This forms two separate branches representing a union of non-overlapping intervals.

For the compound inequality $x \text{ is less than or equal to } 0 \text{ or } x > 4$, written as $x \text{ is less than or equal to } 0 \text{ or } x > 4$, a closed circle is placed at $0$ with an arrow pointing left toward negative infinity. An open circle is placed at $4$ with an arrow pointing right toward positive infinity. This forms two separate branches representing a union of non-overlapping solution regions.

Step-by-Step Algebraic Solution and Checking Procedures

When solving compound inequalities algebraically, each component inequality is solved independently or simultaneous operations are applied across all sections of a double inequality. The solution set must then be verified using a test point selected from within the solution interval.

To solve the compound inequality $-1 + 5x > -26 \text{ and } 7x - 2 \text{ is less than or equal to } 12$, solve each component separately:

First inequality: $-1 + 5x > -26$ Add $1$ to both sides of the inequality: $5x > -25$

Divide both sides by $5$: $x > -5$

Second inequality: $7x - 2 \text{ is less than or equal to } 12$ Add $2$ to both sides of the inequality: $7x \text{ is less than or equal to } 14$

Divide both sides by $7$: $x \text{ is less than or equal to } 2$

Combining both inequalities with "AND" yields the solution $x > -5 \text{ and } x \text{ is less than or equal to } 2$. Stated as a double inequality, the final solution is $-5 < x \text{ and } x \text{ is less than or equal to } 2$.

Graphing this solution set requires placing an open circle at $-5$, a closed circle at $2$, and shading the line segment between $-5$ and $2$.

To check the solution set $-5 < x \text{ and } x \text{ is less than or equal to } 2$, select a test point inside the interval, such as $x = 0$: Substitute $x = 0$ into the first original inequality: $-1 + 5(0) > -26$$-1 > -26$ This inequality is true.

Substitute $x = 0$ into the second original inequality: $7(0) - 2 \text{ is less than or equal to } 12$$-2 \text{ is less than or equal to } 12$ This inequality is true. Since $x = 0$ satisfies both conditions, the solution set $-5 < x \text{ and } x \text{ is less than or equal to } 2$ is verified to be correct.

To solve a continuous double inequality with variables present in all terms, such as $-50 + x < 8x + 6 < -8 + x$, perform equivalent operations on all three parts simultaneously:

Original compound inequality: $-50 + x < 8x + 6 < -8 + x$

Subtract $x$ from all three parts: $-50 < 7x + 6 < -8$

Subtract $6$ from all three parts: $-56 < 7x < -14$

Divide all three parts by $7$: $-8 < x < -2$

The resulting solution set is $-8 < x < -2$.

Graphing this solution set requires an open circle at $-8$, an open circle at $-2$, and shading the entire region between $-8$ and $-2$.

To check the solution set $-8 < x < -2$, choose a test point within the interval, such as $x = -4$: Substitute $x = -4$ into all three parts of the original inequality: $-50 + (-4) < 8(-4) + 6 < -8 + (-4)$

Simplify each term: $-54 < -32 + 6 < -12$$-54 < -26 < -12$

Since $-54 < -26$ is true and $-26 < -12$ is true, the compound inequality holds for $x = -4$, confirming that the solution set $-8 < x < -2$ is correct.