Mastery Guide: Linear & Quadratic Inequalities, Absolute Values & Feasible Regions
Mastery Guide: Linear & Quadratic Inequalities
Foundational Axioms, One-Variable Linear Models, Quadratic Interval Partitioning, 2D Half-Planes, Absolute Value Geometry, Systems of Inequalities & Applied Optimization Modeling
📖 Unit Overview & Target Learning Outcomes
Linear and quadratic inequalities, absolute value relationships, and multi-variable systems form the foundational language of mathematical optimization, engineering tolerances, and economic constraint modeling. Under the Federal Board (FBISE) and Single National Curriculum (SNC) standards, students master:
- Axiomatic Properties of Inequalities: Applying the Law of Trichotomy, Transitive Property, Additive Invariance, and the crucial Multiplicative Sign-Reversal Rule under negative scalars.
- Solving One-Variable Linear Inequalities: Transforming algebraic statements into isolated inequality intervals and graphing them on the real number line using precise open ($\circ$) and solid ($ullet$) endpoint conventions.
- Solving Quadratic Inequalities: Factorizing quadratic polynomials, identifying critical roots, constructing test interval matrices, and evaluating sign behaviors across interior valleys and exterior wings.
- Graphing Linear Inequalities in Two Variables: Constructing boundary lines (solid for inclusive $\le, \ge$; dashed for strict $<, >$), applying the origin test point rule, and shading the valid solution half-plane in $\mathbb{R}^2$.
- Absolute Value Equations & Inequalities: Interpreting absolute value as distance on the real number line, solving compound equations $|ax+b| = c$, and resolving bounded intervals ($|u| \le c$) versus disjoint infinite rays ($|u| \ge c$).
- Systems of Linear Inequalities & Feasible Regions: Graphing multiple simultaneous constraints on a single Cartesian plane to pinpoint the overlapping feasible solution polygon or identifying inconsistent (empty) systems.
- Real-World Mathematical Modeling: Formulating and solving applied inequality models in school zone vehicle braking distances, manufacturing quality control tolerances, sports equipment standards, and production optimization.
💡 Kid-Friendly Tips for Success & Memory Hooks
- The "Negative Flip" Warning (🚨 The Golden Rule of Inequalities): Whenever you multiply or divide an inequality by a negative number, the inequality sign MUST FLIP!
Example: If $-3x \le 12$, dividing by $-3$ gives $x \ge -4$. (Think of the negative sign as a mirror that reflects the entire number line!) - Open vs. Solid Points & Lines:
• Strict ($<, >$): Boundary is NOT included $\implies$ Draw an Open Circle ($\circ$) on the number line or a Dashed Line on the graph.
• Inclusive ($\le, \ge$): Boundary IS included $\implies$ Draw a Solid Circle ($ullet$) on the number line or a Solid Line on the graph. - The Origin $(0, 0)$ Test Point Trick: When graphing a line $ax + by \le c$, plug in $(0, 0)$! If $0 \le c$ is True, shade the side of the line that hugs the origin. If False, shade the opposite side! (Exception: If the line runs right through $(0,0)$, test $(1,0)$ or $(0,1)$ instead!)
- The Absolute Value "Prisoner vs. Explorer" Hook:
• $|x| \le a$ means you are trapped within distance $a$ $\implies$ Between: $-a \le x \le a$.
• $|x| \ge a$ means you are an explorer venturing beyond distance $a$ $\implies$ Disjoint: $x \le -a \text{ or } x \ge a$.
🌍 Real-World Connections & Practical Applications
- Civil Engineering & Road Safety: The stopping distance formula $d(v) = 0.05v^2 + 0.2v$ defines legal speed limits in school zones to ensure vehicles can brake within designated safety distances ($35\text{ m}$).
- Industrial Quality Assurance: In mechanical manufacturing, ball bearings and engine pistons must satisfy strict absolute tolerance bands like $|d - 25.00\text{ mm}| \le 0.02\text{ mm}$ to prevent mechanical seizure.
- Business Economics & Linear Programming: Factory production of multiple goods under limited labour hours and raw materials is modeled via systems of linear inequalities to identify the maximum profit feasible region.
- Sports Equipment Regulations: Professional FIFA footballs and NBA basketballs must meet exact air pressure tolerances (e.g., $|p - 8.0\text{ psi}| \le 0.5\text{ psi}$) before tournament play.
🌟 Section-by-Section Theoretical Foundations
Section 4.1: Foundations of Inequalities & One-Variable Linear Models
An inequality is a mathematical statement comparing two algebraic expressions using inequality symbols ($<, \le, >, \ge$). Unlike linear equations which yield discrete single roots, inequalities yield continuous intervals of solutions.
| Property Name | Algebraic Statement | Sign Rule | Concrete Numerical Example |
|---|---|---|---|
| Law of Trichotomy | For any $a, b \in \mathbb{R}$, exactly one holds: $a < b$, $a = b$, or $a > b$ | Mutually Exclusive | Between $5$ and $8$, strictly $5 < 8$ |
| Transitive Property | If $a < b$ and $b < c$, then $a < c$ | Direction Preserved | $2 < 7$ and $7 < 12 \implies 2 < 12$ |
| Additive Property | If $a < b$, then $a + c < b + c$ and $a - c < b - c$ | Direction Preserved | $3 < 9 \implies 3 + 5 < 9 + 5 \implies 8 < 14$ |
| Positive Multiplicative | If $a < b$ and $c > 0$, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$ | Direction Preserved | $4 < 6 \implies 4(3) < 6(3) \implies 12 < 18$ |
| Negative Multiplicative | If $a < b$ and $c < 0$, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$ | SIGN REVERSAL (Flips) | $-2 < 5 \implies -2(-3) > 5(-3) \implies 6 > -15$ |
Figure 4.1: Geometric Representation of Strict Intervals (Open) versus Inclusive Intervals (Closed) on the Real Number Line.
Section 4.2: Quadratic Inequalities in One Variable & Interval Sign Analysis
A quadratic inequality has the general standard form $ax^2 + bx + c > 0$ or $ax^2 + bx + c \le 0$ ($a \neq 0$). Solving quadratic inequalities involves finding the real critical roots $x_1, x_2$, partitioning the real number line into test intervals, and evaluating the algebraic sign ($+$ or $-$) in each region.
| Inequality Form | Factorized Form ($a < b$) | Critical Roots | Solution Set (Interval Form) | Geometric Number Line Pattern |
|---|---|---|---|---|
| $x^2 - (a+b)x + ab < 0$ | $(x - a)(x - b) < 0$ | $x = a, x = b$ | $(a, b) = \{x \mid a < x < b\}$ | Single open interior segment |
| $x^2 - (a+b)x + ab \le 0$ | $(x - a)(x - b) \le 0$ | $x = a, x = b$ | $[a, b] = \{x \mid a \le x \le b\}$ | Single closed interior segment |
| $x^2 - (a+b)x + ab > 0$ | $(x - a)(x - b) > 0$ | $x = a, x = b$ | $(-\infty, a) \cup (b, \infty)$ | Two disjoint open rays pointing outwards |
| $x^2 - (a+b)x + ab \ge 0$ | $(x - a)(x - b) \ge 0$ | $x = a, x = b$ | $(-\infty, a] \cup [b, \infty)$ | Two disjoint closed rays pointing outwards |
| Perfect Square $\ge 0$ | $(x - k)^2 \ge 0$ | $x = k$ | $\mathbb{R} = (-\infty, \infty)$ | Entire continuous real number line |
| Perfect Square $\le 0$ | $(x - k)^2 \le 0$ | $x = k$ | $\{k\}$ (Single isolated point) | Single isolated solid point at $x = k$ |
| Perfect Square $< 0$ | $(x - k)^2 < 0$ | None | $\emptyset$ (Empty Set / No Solution) | No points shaded on number line |
Figure 4.2: Quadratic Parabola Interval Partitioning showing Interior Negative Valley and Exterior Positive Wings.
Section 4.3: Linear Inequalities in Two Variables & Half-Plane Geometry
A linear inequality in two variables has the standard form $ax + by \le c$, $ax + by < c$, $ax + by \ge c$, or $ax + by > c$. The corresponding linear equation $ax + by = c$ forms the boundary line which partitions the two-dimensional Cartesian coordinate plane $\mathbb{R}^2$ into two open half-planes.
| Inequality Type | Symbol | Boundary Line Style | Boundary Points Status | Half-Plane Shading Decision |
|---|---|---|---|---|
| Strict Greater | $>$ | Dashed / Dotted Line | Excluded ($\notin$ Solution Set) | Region where test point yields TRUE |
| Strict Less | $<$ | Dashed / Dotted Line | Excluded ($\notin$ Solution Set) | Region where test point yields TRUE |
| Inclusive Greater | $\ge$ | Solid Continuous Line | Included ($\in$ Solution Set) | Region containing true test point + boundary |
| Inclusive Less | $\le$ | Solid Continuous Line | Included ($\in$ Solution Set) | Region containing true test point + boundary |
| Horizontal Line | $y \ge c$ / $y < c$ | Solid / Dashed at $y = c$ | Points with ordinate $y = c$ | Above line ($>$) or Below line ($<$) |
| Vertical Line | $x \ge k$ / $x < k$ | Solid / Dashed at $x = k$ | Points with abscissa $x = k$ | Right of line ($>$) or Left of line ($<$) |
| Boundary Equation Form | Origin $(0,0)$ on Line? | Recommended Test Point | Substitution Evaluation | Shading Action |
|---|---|---|---|---|
| $ax + by \le c$ ($c \neq 0$) | No ($a(0)+b(0) = 0 \neq c$) | $(0, 0)$ (Origin) | Compute $0 + 0 = 0 \le c$ | If True, shade side containing $(0,0)$ |
| $ax + by > 0$ ($c = 0$) | Yes ($a(0)+b(0) = 0$) | $(1, 0)$ or $(0, 1)$ | Evaluate $a(1) + b(0) = a > 0$ | If True, shade side containing $(1,0)$ |
| $y > mx$ | Yes (Passes through origin) | $(0, 1)$ on $y$-axis | Evaluate $1 > m(0) \implies 1 > 0$ (True) | Shade half-plane above line |
| $x \ge k$ ($k \neq 0$) | No (Vertical line) | $(0, 0)$ | Check $0 \ge k$ | Shade right if True, left if False |
| $y < k$ ($k \neq 0$) | No (Horizontal line) | $(0, 0)$ | Check $0 < k$ | Shade region containing origin if True |
Figure 4.3: 2D Cartesian Coordinate Plane partitioned into two half-planes by a solid boundary line.
Section 4.4: Absolute Value Equations & Inequality Tolerances
The absolute value of a real number $x$, denoted by $|x|$, represents the non-negative distance between $x$ and the origin $0$ on the real number line: $$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$
| Absolute Value Form | Condition on $c$ | Equivalent Algebraic Structure | Geometric Interpretation | Solution Set |
|---|---|---|---|---|
| $|ax + b| = c$ | $c > 0$ | $ax + b = c$ or $ax + b = -c$ | Points at exact distance $c$ from origin | Two discrete roots |
| $|ax + b| = c$ | $c < 0$ | Impossible ($|u| \ge 0$) | Distance cannot be negative | $\emptyset$ (No Solution) |
| $|ax + b| < c$ | $c > 0$ | $-c < ax + b < c$ | Bounded within distance $c$ (Between) | Single bounded open interval |
| $|ax + b| \le c$ | $c > 0$ | $-c \le ax + b \le c$ | Bounded within distance $c$ (Inclusive) | Single bounded closed interval |
| $|ax + b| > c$ | $c > 0$ | $ax + b > c$ or $ax + b < -c$ | Farther than distance $c$ (Disjoint) | Two infinite open rays |
| $|ax + b| \ge c$ | $c > 0$ | $ax + b \ge c$ or $ax + b \le -c$ | Farther than distance $c$ (Inclusive) | Two infinite closed rays |
| Applied Problem Domain | Target Variable | Mathematical Constraint Inequality | Algebraic Bound | Practical Decision Meaning |
|---|---|---|---|---|
| School Zone Stopping Distance | Vehicle Speed $x\text{ (km/h)}$ | $0.05x^2 + 0.2x \le 35$ | $0 \le x \le 24.54\text{ km/h}$ | Speed must not exceed $24.54\text{ km/h}$ for safety |
| Sports Ball Pressure Tolerance | Air Pressure $p\text{ (psi)}$ | $|p - 8| \le 0.5$ | $7.5 \le p \le 8.5\text{ psi}$ | Inflation between $7.5$ and $8.5\text{ psi}$ passes inspection |
| Essay Contest Word Count | Word Count $w$ | $|w - 500| \le 30$ | $470 \le w \le 530\text{ words}$ | Entries between $470$ and $530$ words are accepted |
| Field Garden Perimeter & Area | Garden Width $w\text{ (ft)}$ | $w(34 - w) \ge 240$ | $10 \le w \le 24\text{ ft}$ | Width between $10\text{ ft}$ and $24\text{ ft}$ guarantees area $\ge 240\text{ sq ft}$ |
| Fencing Dimension Constraint | Width $x\text{ (m)}$ ($x = 2L$) | $2(L + x) \le 60 \implies 3x \le 60$ | $x \le 20\text{ m}, L \le 10\text{ m}$ | Maximum width $20\text{ m}$ and length $10\text{ m}$ within $60\text{ m}$ wire |
Section 4.5: Systems of Linear Inequalities & Feasible Solution Polygons
A system of linear inequalities consists of two or more simultaneous inequalities in the same variables. The solution of the system is the set of all ordered pairs $(x, y)$ that simultaneously satisfy every single inequality in the system, forming the Feasible Region.
| System Configuration | Boundary Lines Relation | Half-Planes Intersection | Feasible Region Geometry | Solution Nature |
|---|---|---|---|---|
| Intersecting Lines | Non-parallel ($m_1 \neq m_2$) | Intersecting angular sector | Infinite convex polygonal sector | Infinitely many solutions |
| Parallel (Disjoint) | Parallel ($m_1 = m_2$), opposite shades | Empty intersection | No overlapping area | $\emptyset$ (No Solution) |
| Parallel (Strip) | Parallel ($m_1 = m_2$), inward shades | Overlapping corridor | Infinite parallel strip between lines | Infinitely many solutions |
| Multi-Constraint System | Three or more linear boundaries | Simultaneous polygon overlap | Bounded convex polygon (e.g. Triangle, Quad) | Closed Feasible Polygon |
| Non-Negativity System | $ax+by \le c, x \ge 0, y \ge 0$ | First Quadrant restrictions | Region confined strictly to Quadrant I | Applied Economic Feasible Region |
| Interval Classification | Inequality Notation | Set-Builder Notation | Interval Notation | Number Line Endpoints |
|---|---|---|---|---|
| Closed Bounded Interval | $a \le x \le b$ | $\{x \in \mathbb{R} \mid a \le x \le b\}$ | $[a, b]$ | Solid circles at $a$ and $b$ |
| Open Bounded Interval | $a < x < b$ | $\{x \in \mathbb{R} \mid a < x < b\}$ | $(a, b)$ | Open circles at $a$ and $b$ |
| Half-Open / Half-Closed | $a \le x < b$ | $\{x \in \mathbb{R} \mid a \le x < b\}$ | $[a, b)$ | Solid at $a$, Open at $b$ |
| Half-Open / Half-Closed | $a < x \le b$ | $\{x \in \mathbb{R} \mid a < x \le b\}$ | $(a, b]$ | Open at $a$, Solid at $b$ |
| Unbounded Ray (Right) | $x \ge a$ | $\{x \in \mathbb{R} \mid x \ge a\}$ | $[a, \infty)$ | Solid circle at $a$, arrow to $+\infty$ |
| Unbounded Ray (Left) | $x < b$ | $\{x \in \mathbb{R} \mid x < b\}$ | $(-\infty, b)$ | Open circle at $b$, arrow to $-\infty$ |
| Disjoint Union Rays | $x \le a \text{ or } x \ge b$ | $\{x \mid x \le a \text{ or } x \ge b\}$ | $(-\infty, a] \cup [b, \infty)$ | Two solid circles with opposing outward arrows |
Figure 4.4: Simultaneous Convex Feasible Region formed by the intersection of multiple linear half-planes.
Part 2: Solved Textbook Exercises (Exhaustive Step-by-Step Manual)
All exercises solved in complete arithmetic detail adhering strictly to FBISE marking schemes.
Exercise 4.1: Linear & Quadratic Inequalities in One Variable
Solve the inequality and graph the solution on a number line:
$$5x - 12 \le 3x - 4$$
- Step 1 (Isolate variable terms): Subtract $3x$ from both sides:
$$5x - 3x - 12 \le -4 \implies 2x - 12 \le -4$$
- Step 2 (Isolate constant terms): Add $12$ to both sides:
$$2x \le -4 + 12 \implies 2x \le 8$$
- Step 3 (Solve for $x$): Divide both sides by $2$ (inequality sign remains unchanged):
$$x \le \frac{8}{2} \implies x \le 4$$
- Step 4 (Number Line Representation):
- Place a solid (closed) circle ($\bullet$) at $x = 4$ because of the non-strict inequality $\le$.
- Shade the number line to the left of $4$ extending towards $-\infty$.
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid x \le 4\}$ or $(-\infty, 4]$.
Solve the inequality and determine the nature of the solution set:
$$1 - 8x \le -4(2x - 1)$$
- Step 1 (Expand the right-hand side): Apply the distributive property to $-4(2x - 1)$:
$$1 - 8x \le -8x + 4$$
- Step 2 (Combine variable terms): Add $8x$ to both sides:
$$1 - 8x + 8x \le -8x + 8x + 4 \implies 1 \le 4$$
- Step 3 (Mathematical Interpretation):
- The variable $x$ cancels out completely, leaving the constant statement $1 \le 4$.
- Since $1 \le 4$ is an unconditionally TRUE identity for all $x \in \mathbb{R}$, every real number satisfies the inequality.
- Final Answer: Solution Set = $\mathbb{R} = (-\infty, \infty)$ (All Real Numbers).
Solve the inequality with rational coefficients and graph the solution on a number line:
$$-\frac{2}{3}x - 2 < -\frac{1}{3}x + 8$$
- Step 1 (Clear denominators): Multiply the entire inequality by $3$ ($\text{LCM} = 3$):
$$3\left(-\frac{2}{3}x - 2\right) < 3\left(-\frac{1}{3}x + 8\right) \implies -2x - 6 < -x + 24$$
- Step 2 (Group variable terms): Add $x$ to both sides:
$$-2x + x - 6 < 24 \implies -x - 6 < 24$$
- Step 3 (Isolate variable): Add $6$ to both sides:
$$-x < 24 + 6 \implies -x < 30$$
- Step 4 (Divide by $-1$ and REVERSE sign): Multiplying or dividing by a negative number flips the inequality symbol:
$$x > \frac{30}{-1} \implies x > -30$$
- Step 5 (Number Line Representation):
- Place an open circle ($\circ$) at $x = -30$ (boundary point excluded).
- Shade the ray to the right extending towards $+\infty$.
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid x > -30\}$ or $(-30, \infty)$.
Solve the inequality and graph the solution:
$$8 - \frac{2}{5}x \ge -14 + 2x$$
- Step 1 (Clear fractions): Multiply through by $5$:
$$5\left(8 - \frac{2}{5}x\right) \ge 5(-14 + 2x) \implies 40 - 2x \ge -70 + 10x$$
- Step 2 (Isolate variable terms):
$$40 + 70 \ge 10x + 2x \implies 110 \ge 12x \implies 12x \le 110$$
- Step 3 (Solve for $x$):
$$x \le \frac{110}{12} = \frac{55}{6} \approx 9.17$$
- Step 4 (Number Line Representation): Solid circle ($\bullet$) at $x = \frac{55}{6}$, shaded leftwards to $-\infty$.
- Final Answer: Solution Set = $\left\{x \in \mathbb{R} \mid x \le \frac{55}{6}\right\}$ or $\left(-\infty, \frac{55}{6}\right]$.
Solve the decimal inequality and graph the solution on a number line:
$$-0.6(x - 5) \le 15$$
- Step 1 (Divide by negative decimal): Divide both sides by $-0.6$ and reverse the inequality symbol:
$$x - 5 \ge \frac{15}{-0.6} = -25$$
- Step 2 (Isolate $x$): Add $5$ to both sides:
$$x \ge -25 + 5 \implies x \ge -20$$
- Step 3 (Number Line Representation):
- Place a solid circle at $x = -20$.
- Shade the ray to the right towards $+\infty$.
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid x \ge -20\}$ or $[-20, \infty)$.
Solve the multi-step fractional inequality:
$$\frac{3x - 1}{4} - \frac{x + 2}{3} \ge 1$$
- Step 1 (Clear denominators): Multiply entire inequality by $\text{LCM}(4, 3) = 12$:
$$12\left(\frac{3x - 1}{4}\right) - 12\left(\frac{x + 2}{3}\right) \ge 12(1)$$
$$3(3x - 1) - 4(x + 2) \ge 12$$
- Step 2 (Expand and combine like terms):
$$9x - 3 - 4x - 8 \ge 12 \implies 5x - 11 \ge 12$$
- Step 3 (Solve for $x$):
$$5x \ge 12 + 11 \implies 5x \ge 23 \implies x \ge \frac{23}{5} = 4.6$$
- Final Answer: Solution Set = $\left\{x \in \mathbb{R} \mid x \ge \frac{23}{5}\right\}$ or $\left[\frac{23}{5}, \infty\right)$.
Translate the verbal phrase into an algebraic inequality, then solve and graph the solution on a number line:
*"Four more than the product of 3 and $x$ is less than 40."*
- Step 1 (Formulate algebraic statement):
- "The product of $3$ and $x$": $3x$
- "Four more than...": $3x + 4$
- "is less than 40": $< 40$
$$\text{Inequality: } 3x + 4 < 40$$
- Step 2 (Solve the inequality):
$$3x < 40 - 4 \implies 3x < 36 \implies x < \frac{36}{3} \implies x < 12$$
- Step 3 (Graph):
- Place an open circle at $x = 12$.
- Shade the line extending to the left towards $-\infty$.
- Final Answer: Inequality: $3x + 4 < 40$; Solution: $x < 12$ or $(-\infty, 12)$.
Translate into an algebraic inequality, solve, and graph the solution:
*"Twice the sum of $x$ and 8 appears less than or equal to $-2$."*
- Step 1 (Formulate algebraic statement):
- "Sum of $x$ and 8": $(x + 8)$
- "Twice the sum": $2(x + 8)$
- "less than or equal to $-2$": $\le -2$
$$\text{Inequality: } 2(x + 8) \le -2$$
- Step 2 (Solve):
$$2x + 16 \le -2 \implies 2x \le -2 - 16 \implies 2x \le -18 \implies x \le -9$$
- Step 3 (Graph): Solid circle at $x = -9$, shaded leftwards to $-\infty$.
- Final Answer: Inequality: $2(x + 8) \le -2$; Solution: $x \le -9$ or $(-\infty, -9]$.
Write and solve an inequality to find the possible values of $x$ for a rectangular region with width $9\text{ ft}$ and length $(x + 2)\text{ ft}$ such that:
$$\text{Area} > 81\text{ sq ft}$$
- Step 1 (Area Formula): For a rectangle, $\text{Area} = \text{width} \times \text{length}$.
$$\text{Area} = 9(x + 2) > 81$$
- Step 2 (Solve for $x$):
$$9x + 18 > 81 \implies 9x > 81 - 18 \implies 9x > 63 \implies x > \frac{63}{9} \implies x > 7$$
- Final Answer: $x > 7\text{ ft}$ (or $(7, \infty)$).
Write and solve an inequality to find the possible values of $x$ for a rectangle with width $8\text{ cm}$ and length $(x + 1)\text{ cm}$ such that:
$$\text{Area} \le 44\text{ sq cm}$$
- Step 1 (Area Formula):
$$\text{Area} = 8(x + 1) \le 44$$
- Step 2 (Solve upper bound):
$$8x + 8 \le 44 \implies 8x \le 36 \implies x \le \frac{36}{8} \implies x \le 4.5\text{ cm}$$
- Step 3 (Physical dimension constraint): Length must be strictly positive:
$$x + 1 > 0 \implies x > -1\text{ cm}$$
- Final Answer: $-1 < x \le 4.5\text{ cm}$ or $(-1, 4.5]$.
Solve the quadratic inequality by factoring:
$$x^2 + 4x - 21 < 0$$
- Step 1 (Find Critical Roots): Set the corresponding quadratic equation to zero:
$$x^2 + 4x - 21 = 0 \implies (x + 7)(x - 3) = 0 \implies x = -7 \text{ and } x = 3$$
- Step 2 (Partition Number Line into Test Intervals): The critical points divide $\mathbb{R}$ into:
1. Interval 1: $(-\infty, -7)$
2. Interval 2: $(-7, 3)$
3. Interval 3: $(3, \infty)$
- Step 3 (Sign Testing):
- For $x = -8 \in (-\infty, -7)$: $(-8+7)(-8-3) = (-1)(-11) = +11 > 0$ (False)
- For $x = 0 \in (-7, 3)$: $(0+7)(0-3) = (7)(-3) = -21 < 0$ (TRUE)
- For $x = 4 \in (3, \infty)$: $(4+7)(4-3) = (11)(1) = +11 > 0$ (False)
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid -7 < x < 3\}$ or $(-7, 3)$.
Solve the quadratic inequality:
$$x^2 - 2x - 8 \le 0$$
- Step 1 (Factorize):
$$(x - 4)(x + 2) \le 0$$
Critical roots are $x = -2$ and $x = 4$.
- Step 2 (Interval Sign Analysis):
A monic quadratic polynomial $(x - a)(x - b)$ is non-positive ($\le 0$) between and including its real roots $[a, b]$ where $a \le b$.
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid -2 \le x \le 4\}$ or $[-2, 4]$.
Solve the quadratic inequality:
$$x^2 - x - 90 > 0$$
- Step 1 (Factorize):
$$(x - 10)(x + 9) > 0$$
Critical roots are $x = -9$ and $x = 10$.
- Step 2 (Interval Sign Analysis):
For a strict greater-than inequality ($> 0$), the solution set consists of the two exterior intervals (wings) outside the roots:
$$x < -9 \quad \text{or} \quad x > 10$$
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid x < -9 \text{ or } x > 10\}$ or $(-\infty, -9) \cup (10, \infty)$.
Solve the quadratic inequality:
$$x^2 - 4x - 12 > 0$$
- Step 1 (Factorize):
$$(x - 6)(x + 2) > 0$$
Critical roots are $x = -2$ and $x = 6$.
- Step 2 (Interval Sign Analysis):
The expression is strictly positive for $x < -2$ and for $x > 6$.
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid x < -2 \text{ or } x > 6\}$ or $(-\infty, -2) \cup (6, \infty)$.
Solve the perfect square quadratic inequality:
$$x^2 + 8x + 16 \ge 0$$
- Step 1 (Factorize as a binomial square):
$$(x + 4)^2 \ge 0$$
- Step 2 (Algebraic Analysis):
The square of any real number $(x + 4) \in \mathbb{R}$ is always greater than or equal to zero for all real numbers $x \in \mathbb{R}$.
- Final Answer: Solution Set = $\mathbb{R} = (-\infty, \infty)$ (All Real Numbers).
Solve the quadratic inequality:
$$9x^2 - 6x + 1 \le 0$$
- Step 1 (Factorize):
$$(3x - 1)^2 \le 0$$
- Step 2 (Analyze conditions):
1. For any real number $x$, $(3x - 1)^2 \ge 0$.
2. Therefore, $(3x - 1)^2 < 0$ has no real solutions.
3. The only possible solution occurs where equality holds: $(3x - 1)^2 = 0 \implies 3x - 1 = 0 \implies x = \frac{1}{3}$.
- Final Answer: Solution Set = $\left\{\frac{1}{3}\right\}$ (Single Isolated Real Solution).
Mr. Khalid has a field and wants to make a rectangular garden with a perimeter of $68\text{ ft}$. He would like the area of the garden to be at least $240\text{ sq ft}$. What are the possible values for the width of the garden?
- Step 1 (Define Variables and Perimeter Constraint):
Let the length of the garden be $L$ and the width be $W$.
$$\text{Perimeter } P = 2(L + W) = 68 \implies L + W = 34 \implies L = 34 - W$$
- Step 2 (Formulate Area Inequality):
$$\text{Area } A = L \times W = (34 - W)W \ge 240$$
$$34W - W^2 \ge 240$$
- Step 3 (Convert to Standard Quadratic Inequality):
$$-W^2 + 34W - 240 \ge 0 \implies W^2 - 34W + 240 \le 0$$
- Step 4 (Factorize to Find Critical Roots):
$$(W - 10)(W - 24) \le 0$$
Critical values are $W = 10\text{ ft}$ and $W = 24\text{ ft}$.
- Step 5 (Establish Feasible Interval):
The product is $\le 0$ within the closed bounded interval $[10, 24]$.
- Final Answer: The width must be between $10\text{ ft}$ and $24\text{ ft}$ inclusive ($10 \le W \le 24\text{ ft}$).
Determine whether the numbers $\{-2, -1, 0, 1, 2\}$ are solutions of the quadratic inequality:
$$2x^2 + 3x + 1 \le 0$$
- Step 1 (Find the algebraic solution range):
Factor $2x^2 + 3x + 1 = (2x + 1)(x + 1) \le 0 \implies -1 \le x \le -0.5$.
- Step 2 (Test each given candidate number):
- $x = -2$: $2(-2)^2 + 3(-2) + 1 = 8 - 6 + 1 = 3 \le 0$ $\implies$ FALSE
- $x = -1$: $2(-1)^2 + 3(-1) + 1 = 2 - 3 + 1 = 0 \le 0$ $\implies$ TRUE
- $x = 0$: $2(0)^2 + 3(0) + 1 = 1 \le 0$ $\implies$ FALSE
- $x = 1$: $2(1)^2 + 3(1) + 1 = 6 \le 0$ $\implies$ FALSE
- $x = 2$: $2(2)^2 + 3(2) + 1 = 8 + 6 + 1 = 15 \le 0$ $\implies$ FALSE
- Final Answer: Only $x = -1$ is a solution.
The stopping distance $d(x)$ (in meters) of a car traveling at $x\text{ km/h}$ is modeled by $d(x) = 0.05x^2 + 0.2x$. If the maximum stopping distance allowed in a school zone is $35\text{ meters}$, find the speed range that ensures the stopping distance does not exceed this limit.
- Step 1 (Formulate Inequality):
$$d(x) \le 35 \implies 0.05x^2 + 0.2x \le 35 \implies 0.05x^2 + 0.2x - 35 \le 0$$
- Step 2 (Multiply by $20$ to clear decimals):
$$x^2 + 4x - 700 \le 0$$
- Step 3 (Solve using Quadratic Formula):
$$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-700)}}{2} = \frac{-4 \pm \sqrt{16 + 2800}}{2} = \frac{-4 \pm \sqrt{2816}}{2} \approx \frac{-4 \pm 53.07}{2}$$
$$x_1 \approx \frac{49.07}{2} \approx 24.54\text{ km/h}, \quad x_2 \approx \frac{-57.07}{2} \approx -28.54\text{ km/h}$$
- Step 4 (Physical Speed Constraints):
Speed cannot be negative ($x \ge 0$). Thus: $0 \le x \le 24.54\text{ km/h}$.
- Final Answer: Speed Range = $0 \le x \le 24.54\text{ km/h}$ (or $[0, 24.54]$).
A rectangular garden is to be enclosed with a fence. The total length of the fence is at most $60\text{ meters}$ and the width $x$ of the garden is twice its length ($x = 2L$). Find the possible dimensions of the garden.
- Step 1 (Relate Dimensions): Given width $x = 2L \implies L = \frac{x}{2}$.
- Step 2 (Perimeter Inequality):
$$P = 2(L + x) \le 60 \implies 2\left(\frac{x}{2} + x\right) \le 60 \implies 2\left(\frac{3x}{2}\right) \le 60 \implies 3x \le 60 \implies x \le 20\text{ m}$$
- Step 3 (Compute Length):
$$L = \frac{x}{2} \le \frac{20}{2} = 10\text{ m}$$
- Final Answer: Length $0 < L \le 10\text{ m}$ and Width $0 < x \le 20\text{ m}$.
Exercise 4.2: Linear Inequalities in Two Variables & Half-Planes
Tell whether the ordered pair $(0, 0)$ is a solution of the linear inequality:
$$x + y < -4$$
- Step 1 (Substitute the coordinates):
Substitute the point $(0, 0)$ into $x + y < -4$:
$$0 + 0 = 0 < -4 \text{ (False)}$$
- Step 2 (Conclusion):
Since the resulting numerical statement is FALSE, the point $(0, 0)$ IS NOT a solution of the inequality.
- Final Answer: No, it is not a solution.
Tell whether the ordered pair $(-2, -3)$ is a solution of the linear inequality:
$$y - x > -2$$
- Step 1 (Substitute the coordinates):
Substitute the point $(-2, -3)$ into $y - x > -2$:
$$-3 - (-2) = -1 > -2 \text{ (True)}$$
- Step 2 (Conclusion):
Since the resulting numerical statement is TRUE, the point $(-2, -3)$ IS a solution of the inequality.
- Final Answer: Yes, it is a solution.
Tell whether the ordered pair $(5, 2)$ is a solution of the linear inequality:
$$2x + 3y \ge 14$$
- Step 1 (Substitute the coordinates):
Substitute the point $(5, 2)$ into $2x + 3y \ge 14$:
$$2(5) + 3(2) = 10 + 6 = 16 \ge 14 \text{ (True)}$$
- Step 2 (Conclusion):
Since the resulting numerical statement is TRUE, the point $(5, 2)$ IS a solution of the inequality.
- Final Answer: Yes, it is a solution.
Tell whether the ordered pair $(-1, 5)$ is a solution of the linear inequality:
$$4x - 7y > 28$$
- Step 1 (Substitute the coordinates):
Substitute the point $(-1, 5)$ into $4x - 7y > 28$:
$$4(-1) - 7(5) = -4 - 35 = -39 > 28 \text{ (False)}$$
- Step 2 (Conclusion):
Since the resulting numerical statement is FALSE, the point $(-1, 5)$ IS NOT a solution of the inequality.
- Final Answer: No, it is not a solution.
Tell whether the ordered pair $(0, 0)$ is a solution of the linear inequality:
$$-3y - 2x < 12$$
- Step 1 (Substitute the coordinates):
Substitute the point $(0, 0)$ into $-3y - 2x < 12$:
$$-3(0) - 2(0) = 0 < 12 \text{ (True)}$$
- Step 2 (Conclusion):
Since the resulting numerical statement is TRUE, the point $(0, 0)$ IS a solution of the inequality.
- Final Answer: Yes, it is a solution.
Tell whether the ordered pair $(-4, 0)$ is a solution of the linear inequality:
$$x \ge -3$$
- Step 1 (Substitute the coordinates):
Substitute the point $(-4, 0)$ into $x \ge -3$:
$$-4 \ge -3 \text{ (False)}$$
- Step 2 (Conclusion):
Since the resulting numerical statement is FALSE, the point $(-4, 0)$ IS NOT a solution of the inequality.
- Final Answer: No, it is not a solution.
Graph the linear inequality in the Cartesian coordinate plane:
$$y > 2x$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$y = 2x$$
- Line style: DASHED line because the inequality is strict ($<, >$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 1)$:
$$1 > 2(0) \implies 1 > 0 \text{ (True)}$$
- Step 3 (Shade the Feasible Half-Plane):
Region above the line $y = 2x$ (containing $(0,1)$).
- Final Answer: Boundary line is drawn dashed; feasible region is Region above the line $y = 2x$ (containing $(0,1)$).
Graph the linear inequality in the Cartesian coordinate plane:
$$x - y \ge 2$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$x - y = 2$$
- Line style: SOLID line because the inequality is inclusive ($\le, \ge$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$0 - 0 = 0 \ge 2 \text{ (False)}$$
- Step 3 (Shade the Feasible Half-Plane):
Region below the line $x - y = 2$ (not containing $(0,0)$).
- Final Answer: Boundary line is drawn solid; feasible region is Region below the line $x - y = 2$ (not containing $(0,0)$).
Graph the linear inequality in the Cartesian coordinate plane:
$$-3y - 2x < 12$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$-3y - 2x = 12$$
- Line style: DASHED line because the inequality is strict ($<, >$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$-3(0) - 2(0) = 0 < 12 \text{ (True)}$$
- Step 3 (Shade the Feasible Half-Plane):
Region above the line $-2x - 3y = 12$ (containing $(0,0)$).
- Final Answer: Boundary line is drawn dashed; feasible region is Region above the line $-2x - 3y = 12$ (containing $(0,0)$).
Graph the linear inequality in the Cartesian coordinate plane:
$$2x + y \ge 8$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$2x + y = 8$$
- Line style: SOLID line because the inequality is inclusive ($\le, \ge$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$2(0) + 0 = 0 \ge 8 \text{ (False)}$$
- Step 3 (Shade the Feasible Half-Plane):
Region above and to the right of line $2x + y = 8$.
- Final Answer: Boundary line is drawn solid; feasible region is Region above and to the right of line $2x + y = 8$.
Graph the linear inequality in the Cartesian coordinate plane:
$$x - y \le -11$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$x - y = -11$$
- Line style: SOLID line because the inequality is inclusive ($\le, \ge$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$0 - 0 = 0 \le -11 \text{ (False)}$$
- Step 3 (Shade the Feasible Half-Plane):
Region above the line $x - y = -11$.
- Final Answer: Boundary line is drawn solid; feasible region is Region above the line $x - y = -11$.
Graph the linear inequality in the Cartesian coordinate plane:
$$y < -5$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$y = -5$$
- Line style: DASHED line because the inequality is strict ($<, >$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$0 < -5 \text{ (False)}$$
- Step 3 (Shade the Feasible Half-Plane):
Half-plane strictly below the horizontal line $y = -5$.
- Final Answer: Boundary line is drawn dashed; feasible region is Half-plane strictly below the horizontal line $y = -5$.
Graph the linear inequality in the Cartesian coordinate plane:
$$x \ge 4$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$x = 4$$
- Line style: SOLID line because the inequality is inclusive ($\le, \ge$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$0 \ge 4 \text{ (False)}$$
- Step 3 (Shade the Feasible Half-Plane):
Half-plane to the right of the vertical line $x = 4$.
- Final Answer: Boundary line is drawn solid; feasible region is Half-plane to the right of the vertical line $x = 4$.
Graph the linear inequality in the Cartesian coordinate plane:
$$-(x + 2) + 3y < 8$$
- Step 1 (Plot the Boundary Line):
Graph the corresponding linear equation:
$$-(x + 2) + 3y = 8$$
- Line style: DASHED line because the inequality is strict ($<, >$).
- Step 2 (Select a Test Point):
Choose a convenient test point not lying on the line, such as $(0, 0)$:
$$-(0 + 2) + 3(0) = -2 < 8 \text{ (True)}$$
- Step 3 (Shade the Feasible Half-Plane):
Region below the boundary line $-x + 3y = 10$.
- Final Answer: Boundary line is drawn dashed; feasible region is Region below the boundary line $-x + 3y = 10$.
Can we use $(0, 0)$ as a test point when graphing $x + y > 0$? Explain with reason.
- Analysis:
1. The boundary equation for the inequality $x + y > 0$ is $x + y = 0$.
2. Substituting $(x, y) = (0, 0)$ into the boundary equation gives $0 + 0 = 0$, which means the origin $(0, 0)$ lies directly on the boundary line.
3. A valid test point must lie strictly inside one of the two separate open half-planes so that it can unambiguously decide which side satisfies the inequality.
4. Therefore, $(0, 0)$ cannot be used as a test point for this line.
- Recommendation:
Use an off-line test point such as $(1, 0)$ or $(0, 1)$ instead. Substituting $(1, 0)$ gives $1 + 0 = 1 > 0$ (True), so shade the half-plane containing $(1, 0)$.
- Final Answer: No, because $(0, 0)$ lies on the boundary line. A point not on the line (like $(1, 0)$) must be chosen.
Write the verbal sentence as an inequality, then graph the inequality:
*"Three less than $x$ is greater than or equal to $y$."*
- Step 1 (Algebraic Translation):
$$x - 3 \ge y \iff y \le x - 3$$
- Step 2 (Graphing the Half-Plane):
Solid line $y = x - 3$, shaded below the line (contains $(0,0)$).
- Final Answer: Inequality: $x - 3 \ge y \iff y \le x - 3$; Graph: Solid line $y = x - 3$, shaded below the line (contains $(0,0)$).
Write the verbal sentence as an inequality, then graph the inequality:
*"The product of $-2$ and $y$ is less than or equal to the sum of $x$ and 6."*
- Step 1 (Algebraic Translation):
$$-2y \le x + 6 \iff y \ge -\frac{1}{2}x - 3$$
- Step 2 (Graphing the Half-Plane):
Solid line $y = -rac{1}{2}x - 3$, shaded above the line (contains $(0,0)$).
- Final Answer: Inequality: $-2y \le x + 6 \iff y \ge -\frac{1}{2}x - 3$; Graph: Solid line $y = -rac{1}{2}x - 3$, shaded above the line (contains $(0,0)$).
Write the verbal sentence as an inequality, then graph the inequality:
*"The sum of $x$ and the product of 4 and $y$ is less than $-2$."*
- Step 1 (Algebraic Translation):
$$x + 4y < -2 \iff y < -\frac{1}{4}x - \frac{1}{2}$$
- Step 2 (Graphing the Half-Plane):
Dashed line $x + 4y = -2$, shaded below the line (not containing $(0,0)$).
- Final Answer: Inequality: $x + 4y < -2 \iff y < -\frac{1}{4}x - \frac{1}{2}$; Graph: Dashed line $x + 4y = -2$, shaded below the line (not containing $(0,0)$).
Exercise 4.3: Absolute Value Equations & Inequalities
Solve the absolute value equation:
$$8|x - 3| = 88$$
- Step 1 (Isolate the absolute value expression): Divide both sides by $8$:
$$|x - 3| = \frac{88}{8} = 11$$
- Step 2 (Split into two linear equations):
$$x - 3 = 11 \quad \text{or} \quad x - 3 = -11$$
- Step 3 (Solve both cases):
- Case 1: $x = 11 + 3 = 14$
- Case 2: $x = -11 + 3 = -8$
- Final Answer: Solution Set = $\{-8, 14\}$
Solve the absolute value equation:
$$|2x + 9| = 30$$
- Step 1 (Split into positive and negative cases):
$$2x + 9 = 30 \quad \text{or} \quad 2x + 9 = -30$$
- Step 2 (Solve each equation):
- Case 1: $2x = 30 - 9 = 21 \implies x = \frac{21}{2} = 10.5$
- Case 2: $2x = -30 - 9 = -39 \implies x = -\frac{39}{2} = -19.5$
- Final Answer: Solution Set = $\left\{-\frac{39}{2}, \frac{21}{2}\right\}$
Solve the absolute value equation:
$$|x - 3| = -26$$
- Step 1 (Apply definition of absolute value):
For any real number $u$, the absolute value $|u|$ represents geometric distance from the origin on the number line, which is inherently non-negative:
$$|u| \ge 0 \quad \forall u \in \mathbb{R}$$
- Step 2 (Analyze the equation):
Since $|x - 3| \ge 0$, it can never be equal to the negative constant $-26$.
- Final Answer: Solution Set = $\emptyset$ (Empty Set / No Solution)
Solve and verify extraneous roots for the equation:
$$3|x + 6| = 9x - 6$$
- Step 1 (Isolate absolute value): Divide both sides by $3$:
$$|x + 6| = 3x - 2$$
- Step 2 (Solve Case 1):
$$x + 6 = 3x - 2 \implies 6 + 2 = 3x - x \implies 8 = 2x \implies x = 4$$
Check: $|4 + 6| = 10$; $3(4) - 2 = 10$ ($10 = 10 \implies$ VALID).
- Step 3 (Solve Case 2):
$$x + 6 = -(3x - 2) \implies x + 6 = -3x + 2 \implies 4x = -4 \implies x = -1$$
Check: $|-1 + 6| = 5$; $3(-1) - 2 = -5$ ($5 \neq -5 \implies$ EXTRANEOUS).
- Final Answer: Solution Set = $\{4\}$
Solve and check for extraneous roots:
$$|x - 3| = 2x$$
- Step 1 (Case 1: $x - 3 = 2x$):
$$x - 2x = 3 \implies -x = 3 \implies x = -3$$
Check: $|-3 - 3| = |-6| = 6$; $2(-3) = -6 \implies 6 \neq -6$ (Extraneous).
- Step 2 (Case 2: $x - 3 = -2x$):
$$x + 2x = 3 \implies 3x = 3 \implies x = 1$$
Check: $|1 - 3| = |-2| = 2$; $2(1) = 2 \implies 2 = 2$ (Valid).
- Final Answer: Solution Set = $\{1\}$
Solve the absolute value equation:
$$\frac{|4 - 5x|}{6} = 3.5$$
- Step 1 (Clear denominator):
$$|4 - 5x| = 6 \times 3.5 = 21$$
- Step 2 (Solve Case 1):
$$4 - 5x = 21 \implies -5x = 17 \implies x = -\frac{17}{5} = -3.4$$
- Step 3 (Solve Case 2):
$$4 - 5x = -21 \implies -5x = -25 \implies x = 5$$
- Final Answer: Solution Set = $\left\{-\frac{17}{5}, 5\right\}$
Explain why the equation $|3x - 6| + 7 = 4$ has no real solution.
- Step 1 (Isolate the absolute value term): Subtract $7$ from both sides:
$$|3x - 6| = 4 - 7 \implies |3x - 6| = -3$$
- Step 2 (Analyze by definition):
The absolute value of any real algebraic quantity is always $\ge 0$. It can never equal a negative value like $-3$.
- Final Answer: No solution because $|3x - 6| = -3$ is impossible in the real number system.
Before the start of a professional basketball game, a basketball must be inflated to an air pressure of $8\text{ pounds per square inch (psi)}$ with an allowable absolute error (deviation) of at most $0.5\text{ psi}$. Find the minimum and maximum air pressure acceptable for the basketball.
- Step 1 (Formulate Inequality):
Let the actual air pressure be $p$. The absolute error constraint is:
$$|p - 8| \le 0.5$$
- Step 2 (Decompose into compound inequality):
$$-0.5 \le p - 8 \le 0.5$$
- Step 3 (Add $8$ to all parts):
$$8 - 0.5 \le p \le 8 + 0.5 \implies 7.5 \le p \le 8.5\text{ psi}$$
- Final Answer: Minimum Acceptable Pressure = $7.5\text{ psi}$, Maximum = $8.5\text{ psi}$
Solve the absolute value inequality and graph the solution on a number line:
$$|9x - 1| \le 10$$
- Step 1 (Decompose $|u| \le c$):
$$-10 \le 9x - 1 \le 10$$
- Step 2 (Add $1$ to all parts):
$$-9 \le 9x \le 11$$
- Step 3 (Divide by $9$):
$$-1 \le x \le \frac{11}{9}$$
- Final Answer: Solution Set = $\left\{x \in \mathbb{R} \mid -1 \le x \le \frac{11}{9}\right\}$ or $\left[-1, \frac{11}{9}\right]$
Solve the absolute value inequality:
$$|2x - 7| \le 1$$
- Step 1 (Decompose):
$$-1 \le 2x - 7 \le 1$$
- Step 2 (Add $7$):
$$6 \le 2x \le 8$$
- Step 3 (Divide by $2$):
$$3 \le x \le 4$$
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid 3 \le x \le 4\}$ or $[3, 4]$
Solve the inequality and graph the solution:
$$|5 - x| + 3 > 6$$
- Step 1 (Isolate absolute value):
$$|5 - x| > 6 - 3 \implies |5 - x| > 3$$
- Step 2 (Split into disjoint cases):
1. $5 - x > 3 \implies -x > -2 \implies x < 2$
2. $5 - x < -3 \implies -x < -8 \implies x > 8$
- Final Answer: Solution Set = $\{x \in \mathbb{R} \mid x < 2 \text{ or } x > 8\}$ or $(-\infty, 2) \cup (8, \infty)$
Solve the absolute value inequality:
$$3|14 - x| > 6$$
- Step 1 (Divide by $3$):
$$|14 - x| > 2$$
- Step 2 (Solve cases):
1. $14 - x > 2 \implies -x > -12 \implies x < 12$
2. $14 - x < -2 \implies -x < -16 \implies x > 16$
- Final Answer: Solution Set = $(-\infty, 12) \cup (16, \infty)$
Solve the inequality:
$$\frac{|3x - 2|}{5} \le 4$$
- Step 1 (Clear denominator):
$$|3x - 2| \le 20 \implies -20 \le 3x - 2 \le 20$$
- Step 2 (Add $2$):
$$-18 \le 3x \le 22$$
- Step 3 (Divide by $3$):
$$-6 \le x \le \frac{22}{3}$$
- Final Answer: Solution Set = $\left[-6, \frac{22}{3}\right]$
Solve the inequality:
$$\frac{|2 - 5x|}{3} \ge 4$$
- Step 1 (Clear denominator):
$$|2 - 5x| \ge 12$$
- Step 2 (Solve cases):
1. $2 - 5x \ge 12 \implies -5x \ge 10 \implies x \le -2$
2. $2 - 5x \le -12 \implies -5x \le -14 \implies x \ge \frac{14}{5} = 2.8$
- Final Answer: Solution Set = $(-\infty, -2] \cup \left[\frac{14}{5}, \infty\right)$
An essay contest requires that essay entries consist of $500\text{ words}$ with an absolute deviation of at most $30\text{ words}$. What are the possible word counts that an acceptable essay can have?
- Step 1 (Formulate Absolute Inequality):
Let $w$ be the number of words.
$$|w - 500| \le 30$$
- Step 2 (Solve Compound Inequality):
$$-30 \le w - 500 \le 30 \implies 500 - 30 \le w \le 500 + 30 \implies 470 \le w \le 530$$
- Final Answer: Possible word count ranges from $470$ to $530$ words inclusive ($[470, 530]$).
Exercise 4.4: Systems of Linear Inequalities in Two Variables
Describe the steps you would take to graph the system of linear inequalities shown below:
$$\begin{cases} x - y < 6 \\ y \ge 3 \end{cases}$$
- Step 1 (Graph Inequality 1: $x - y < 6$):
- Draw the boundary line $x - y = 6$ as a dashed line because of strict inequality $<$.
- Test the origin $(0, 0)$: $0 - 0 = 0 < 6$ is True. Shade the region containing $(0, 0)$ (above and to the left of the line).
- Step 2 (Graph Inequality 2: $y \ge 3$):
- Draw the horizontal boundary line $y = 3$ as a solid line because of inclusive inequality $\ge$.
- Test $(0, 0)$: $0 \ge 3$ is False. Shade the region above the line $y = 3$.
- Step 3 (Determine the Solution of the System):
- The solution set is the common overlapping shaded region where both half-planes intersect simultaneously.
- Final Answer: Feasible region is the mutual intersection of the upper half-plane of $y=3$ and the left half-plane of $x-y=6$.
Tell whether the ordered pairs $(0, 0)$, $(-3, 1)$, and $(3, 2)$ are solutions of the system of linear inequalities:
$$\begin{cases} 2x - y \le 5 \\ x + 2y > 2 \end{cases}$$
- Testing $(0, 0)$:
- $2(0) - 0 = 0 \le 5$ (True)
- $0 + 2(0) = 0 > 2$ (False) $\implies$ NOT a solution of the system.
- Testing $(-3, 1)$:
- $2(-3) - 1 = -7 \le 5$ (True)
- $-3 + 2(1) = -1 > 2$ (False) $\implies$ NOT a solution of the system.
- Testing $(3, 2)$:
- $2(3) - 2 = 4 \le 5$ (True)
- $3 + 2(2) = 7 > 2$ (True) $\implies$ VALID SOLUTION of the system.
- Final Answer: Only $(3, 2)$ satisfies both inequalities simultaneously.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} x + y > 3 \\ y - 2x \ge 3 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Dashed line $x+y=3$ shaded above; solid line $y-2x=3$ shaded above-left. Solution is the intersection cone containing points like $(0,4)$.
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} x + y < 1 \\ y \le -2 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Dashed line $x+y=1$ shaded below; solid horizontal line $y=-2$ shaded below. Feasible region is the overlap below $y=-2$.
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} y - 2x \le 3 \\ x \ge 2 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Solid line $y - 2x = 3$ shaded below-right; solid vertical line $x = 2$ shaded to the right ($x \ge 2$). Overlapping region forms an unbounded polygon.
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} x + y \le 3 \\ x - y \le 4 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Solid line $x + y = 3$ shaded below-left; solid line $x - y = 4$ shaded above-left. Intersection contains origin $(0,0)$.
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} 2x + y < 3 \\ y \ge 2x - 1 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Dashed line $2x + y = 3$ shaded below-left; solid line $y = 2x - 1$ shaded above-left. Feasible region is the open sector between both lines.
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} x - 4y \le -8 \\ x \ge 0 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Solid line $x - 4y = -8$ shaded above; solid vertical line $x = 0$ ($y$-axis) shaded to right ($x \ge 0$). First quadrant feasible strip.
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} x + y \ge 4 \\ x + y \le 1 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Parallel lines with $m = -1$. Line $x+y=4$ shaded above-right; line $x+y=1$ shaded below-left. The half-planes do not intersect $\implies$ Solution Set = $\emptyset$ (No Solution).
- Final Answer: Feasible region is the common intersection described above.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} 3x - y \ge 6 \\ x - y < 1 \end{cases}$$
- Step 1 (Graph Inequality 1):
- Plot boundary line 1 with appropriate line style (solid for inclusive, dashed for strict).
- Test origin $(0,0)$ or convenient point to establish shading direction.
- Step 2 (Graph Inequality 2):
- Plot boundary line 2 on the same Cartesian coordinate plane.
- Determine shading direction via test point.
- Step 3 (Find Common Intersection / Feasible Region):
- Solid line $3x - y = 6$ shaded below-right; dashed line $x - y = 1$ shaded above-left. Overlapping region forms an acute angular wedge.
- Final Answer: Feasible region is the common intersection described above.
Does the system of linear inequalities have any solution? Justify geometrically:
$$\begin{cases} x - y > 5 \\ x - y < 1 \end{cases}$$
- Step 1 (Analyze Slopes of Boundary Lines):
- Line 1: $x - y = 5 \implies y = x - 5$ (Slope $m_1 = 1$, $y$-intercept $= -5$).
- Line 2: $x - y = 1 \implies y = x - 1$ (Slope $m_2 = 1$, $y$-intercept $= -1$).
- Since $m_1 = m_2 = 1$ and $y$-intercepts are distinct, the boundary lines are strictly parallel.
- Step 2 (Evaluate Shading Regions):
- The inequality $x - y > 5 \implies y < x - 5$ shades all points strictly below the lower line $y = x - 5$.
- The inequality $x - y < 1 \implies y > x - 1$ shades all points strictly above the upper line $y = x - 1$.
- Step 3 (Geometric Conclusion):
- The two shaded half-planes point in opposite outward directions away from each other and have zero overlapping points.
- Final Answer: No solution ($\text{Solution Set} = \emptyset$) because the two parallel half-planes do not intersect.
Describe a real-world situation that can be modeled by a system of linear inequalities. Then write and graph the system of inequalities.
- Real-World Scenario:
A bakery produces two types of cakes: Chocolate Cakes ($x$) and Vanilla Cakes ($y$).
1. Oven Capacity Constraint: Total cakes per day cannot exceed $50 \implies x + y \le 50$.
2. Labor Preparation Time: Making a chocolate cake takes $2\text{ hours}$ and a vanilla cake takes $3\text{ hours}$, with a daily maximum of $120\text{ labor hours} \implies 2x + 3y \le 120$.
3. Non-Negativity: Production cannot be negative $\implies x \ge 0, y \ge 0$.
- System of Inequalities:
$$\begin{cases} x + y \le 50 \\ 2x + 3y \le 120 \\ x \ge 0 \\ y \ge 0 \end{cases}$$
- Graph & Feasible Region:
Bounded convex polygon in the First Quadrant with vertices $(0, 0)$, $(50, 0)$, $(30, 20)$, and $(0, 40)$.
- Final Answer: Feasible polygon in First Quadrant bounded by vertices $(0,0), (50,0), (30,20), (0,40)$.
Miscellaneous Exercise 4: Comprehensive Review & Mixed Problems
The solution of the inequality $6x - 7 \ge 2x + 17$ is:
- Algebraic Derivation:
Subtract $2x$ and add $7$: $4x \ge 24 \implies x \ge 6$.
- Final Answer: $x \ge 6$
The solution of the inequality $x < 3$ for $x \in \mathbb{N}$ is:
- Algebraic Derivation:
Natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$. The natural numbers strictly less than $3$ are $\{1, 2\}$.
- Final Answer: $\{1, 2\}$
In general, we use which test point when graphing a linear inequality whose boundary line does not pass through the origin?
- Algebraic Derivation:
The origin $(0, 0)$ is the standard, simplest test point because substituting zeros eliminates all variable terms immediately.
- Final Answer: $(0, 0)$
The solution of the inequality $\frac{x}{2} < 3 - x$ is:
- Algebraic Derivation:
Multiply by $2$: $x < 6 - 2x \implies 3x < 6 \implies x < 2$.
- Final Answer: $x < 2$
Which ordered pair is a solution of the inequality $4x - y \ge 3$?
- Algebraic Derivation:
Testing $(1, 0)$: $4(1) - 0 = 4 \ge 3$ is TRUE.
- Final Answer: $(1, 0)$
Which ordered pair is a solution of the system $\begin{cases} 2x - y \le 5 \\ x + 2y > 2 \end{cases}$?
- Algebraic Derivation:
Testing $(3, 2)$: $2(3)-2 = 4 \le 5$ (True) and $3+2(2) = 7 > 2$ (True).
- Final Answer: $(3, 2)$
The solution for $-5x \ge -80$ is:
- Algebraic Derivation:
Divide by $-5$ and reverse inequality symbol: $x \le \frac{-80}{-5} \implies x \le 16$.
- Final Answer: $\{x \mid x \le 16\}$
$(7, -2)$ is a solution of which inequality?
- Algebraic Derivation:
Testing $(7, -2)$ in $x + 10y < 1$: $7 + 10(-2) = 7 - 20 = -13 < 1$ is TRUE.
- Final Answer: $x + 10y < 1$
Solve the inequality and graph the solution on a number line:
$$\frac{2}{3}x - 4 \ge 1$$
- Step 1:
$$\frac{2}{3}x \ge 5 \implies x \ge 5 \times \frac{3}{2} = \frac{15}{2}$$
- Final Answer: $x \ge \frac{15}{2} = 7.5$
Solve the inequality and graph the solution on a number line:
$$1 - 3x \le -14 + 2x$$
- Step 1:
$$1 + 14 \le 2x + 3x \implies 15 \le 5x \implies x \ge 3$$
- Final Answer: $x \ge 3$
Solve the inequality and graph the solution on a number line:
$$-3(2x - 1) < 27$$
- Step 1:
$$-6x + 3 < 27 \implies -6x < 24 \implies x > \frac{24}{-6} = -4$$
- Final Answer: $x > -4$
Solve the inequality and graph the solution on a number line:
$$\frac{x - 1}{2} - \frac{x - 2}{3} \ge 1$$
- Step 1:
$$3(x - 1) - 2(x - 2) \ge 6 \implies 3x - 3 - 2x + 4 \ge 6 \implies x + 1 \ge 6 \implies x \ge 5$$
- Final Answer: $x \ge 5$
Graph the linear inequality in the Cartesian coordinate plane:
$$y > 2x$$
- Step 1: Plot dashed line $y = 2x$.
- Step 2: Test point evaluation determines that the region is shaded above line (contains $(0, 1)$).
- Final Answer: Boundary is dashed line $y = 2x$, shaded shaded above line (contains $(0, 1)$).
Graph the linear inequality in the Cartesian coordinate plane:
$$y < 3x$$
- Step 1: Plot dashed line $y = 3x$.
- Step 2: Test point evaluation determines that the region is shaded below line (contains $(1, 0)$).
- Final Answer: Boundary is dashed line $y = 3x$, shaded shaded below line (contains $(1, 0)$).
Graph the linear inequality in the Cartesian coordinate plane:
$$y \ge -5$$
- Step 1: Plot solid horizontal line $y = -5$.
- Step 2: Test point evaluation determines that the region is shaded above line (contains $(0, 0)$).
- Final Answer: Boundary is solid horizontal line $y = -5$, shaded shaded above line (contains $(0, 0)$).
Graph the linear inequality in the Cartesian coordinate plane:
$$x \le 6$$
- Step 1: Plot solid vertical line $x = 6$.
- Step 2: Test point evaluation determines that the region is shaded to the left of line (contains $(0, 0)$).
- Final Answer: Boundary is solid vertical line $x = 6$, shaded shaded to the left of line (contains $(0, 0)$).
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} 3x + 4y \ge 12 \\ 5x + 6y \le 30 \end{cases}$$
- Step 1: Graph all boundary lines in the system on the same plane.
- Step 2: Apply test points to shade individual half-planes.
- Step 3: The feasible region is the mutual overlapping area: Feasible region is bounded by solid lines $3x+4y=12$ and $5x+6y=30$ in the first quadrant strip..
- Final Answer: Feasible Region: Feasible region is bounded by solid lines $3x+4y=12$ and $5x+6y=30$ in the first quadrant strip.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} y - x \ge 1 \\ y - 4 \le 4 \end{cases}$$
- Step 1: Graph all boundary lines in the system on the same plane.
- Step 2: Apply test points to shade individual half-planes.
- Step 3: The feasible region is the mutual overlapping area: Solid lines $y - x = 1$ and $y \le 8$; feasible region lies below $y = 8$ and above $y = x + 1$..
- Final Answer: Feasible Region: Solid lines $y - x = 1$ and $y \le 8$; feasible region lies below $y = 8$ and above $y = x + 1$.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} 8x + 5y \le 40 \\ x \ge 0 \\ y \ge 0 \end{cases}$$
- Step 1: Graph all boundary lines in the system on the same plane.
- Step 2: Apply test points to shade individual half-planes.
- Step 3: The feasible region is the mutual overlapping area: Bounded triangular region in First Quadrant with vertices $(0, 0)$, $(5, 0)$, and $(0, 8)$..
- Final Answer: Feasible Region: Bounded triangular region in First Quadrant with vertices $(0, 0)$, $(5, 0)$, and $(0, 8)$.
Graph the system of linear inequalities in the Cartesian coordinate plane:
$$\begin{cases} 2x + y \le 12 \\ x \ge 0 \\ y \ge 0 \end{cases}$$
- Step 1: Graph all boundary lines in the system on the same plane.
- Step 2: Apply test points to shade individual half-planes.
- Step 3: The feasible region is the mutual overlapping area: Bounded triangular region in First Quadrant with vertices $(0, 0)$, $(6, 0)$, and $(0, 12)$..
- Final Answer: Feasible Region: Bounded triangular region in First Quadrant with vertices $(0, 0)$, $(6, 0)$, and $(0, 12)$.
Write a system of linear inequalities for the shaded region described:
The shaded region is a rectangle with vertices $(2, 1)$, $(2, 4)$, $(6, 4)$, and $(6, 1)$.
- Step 1 (Identify Horizontal Bounds):
- Left boundary: vertical line $x = 2$.
- Right boundary: vertical line $x = 6$.
$$\implies 2 \le x \le 6$$
- Step 2 (Identify Vertical Bounds):
- Bottom boundary: horizontal line $y = 1$.
- Top boundary: horizontal line $y = 4$.
$$\implies 1 \le y \le 4$$
- Final Answer: $\begin{cases} 2 \le x \le 6 \\ 1 \le y \le 4 \end{cases}$ (or $x \ge 2, x \le 6, y \ge 1, y \le 4$)
Write a system of linear inequalities for the shaded region described:
The shaded region is a triangle with vertices at $(3, 0)$, $(3, 2)$, and $(0, -2)$.
- Step 1 (Vertical Right Boundary):
Vertical line passing through $(3, 0)$ and $(3, 2) \implies x \le 3$.
- Step 2 (Lower Boundary Line through $(0, -2)$ and $(3, 0)$):
Slope $m = \frac{0 - (-2)}{3 - 0} = \frac{2}{3}$; $y$-intercept $c = -2 \implies y = \frac{2}{3}x - 2$.
Region is above this line $\implies y \ge \frac{2}{3}x - 2$.
- Step 3 (Upper Boundary Line through $(0, -2)$ and $(3, 2)$):
Slope $m = \frac{2 - (-2)}{3 - 0} = \frac{4}{3}$; $y$-intercept $c = -2 \implies y = \frac{4}{3}x - 2$.
Region is below this line $\implies y \le \frac{4}{3}x - 2$.
- Final Answer: $\begin{cases} x \le 3 \\ y \ge \frac{2}{3}x - 2 \\ y \le \frac{4}{3}x - 2 \end{cases}$
Part 3: Extra High-Yield Objective Booster (SNC Question Bank)
Which operation always requires reversing the direction of an inequality symbol?
Final Answer: $Multiplying or dividing by a negative number$
The inequality $|x - 5| < 0$ has how many real solutions?
Final Answer: $No solution ($\emptyset$)$
The quadratic inequality $(x - 3)(x - 7) < 0$ has solution set:
Final Answer: $3 < x < 7$
When graphing the inequality $2x - 5y > 10$, the boundary line must be drawn as:
Final Answer: $Dashed / dotted line$
The solution set of $|2x - 3| \ge 7$ is:
Final Answer: $x \le -2 \text{ or } x \ge 5$
The Law of Trichotomy states that for any two real numbers $a$ and $b$, exactly one of $a < b$, $a = b$, or ________ holds true.
Final Answer: a > b
On a number line, a strict inequality ($<$ or $>$) is represented at the boundary by an ________ circle.
Final Answer: open
The region containing all points that simultaneously satisfy every linear inequality in a system is called the ________ region.
Final Answer: feasible
The solution of the absolute value inequality $|x| \le 4$ in interval notation is ________.
Final Answer: [-4, 4]
If $(x - a)^2 \le 0$ for a real number $x$, then the only real solution is $x =$ ________.
Final Answer: a
State True or False: If $a < b$ and $c < 0$, then $ac < bc$.
Final Answer: False
State True or False: The equation $|x + 3| = -5$ has two real solutions.
Final Answer: False
State True or False: The origin $(0, 0)$ is a valid test point for $3x - 4y \le 12$.
Final Answer: True
State True or False: The quadratic inequality $x^2 + 6x + 9 \ge 0$ is true for all $x \in \mathbb{R}$.
Final Answer: True
Match each algebraic inequality in Column A with its corresponding solution set in Column B:
- (a) $|x| \le 5 \implies -5 \le x \le 5$ (ii)
- (b) $|x| > 5 \implies x < -5 \text{ or } x > 5$ (iv)
- (c) $x^2 - 25 < 0 \implies (x-5)(x+5) < 0 \implies -5 < x < 5$ (iii)
- (d) $x^2 - 25 \ge 0 \implies x \le -5 \text{ or } x \ge 5$ (i)
- (e) $(x - 5)^2 \le 0 \implies x = 5 \implies \{5\}$ (v)
- Final Answer: (a) -> (ii), (b) -> (iv), (c) -> (iii), (d) -> (i), (e) -> (v)
Match each 2D linear inequality in Column A with its geometric boundary line description in Column B:
- (a) $2x + 3y > 6$: Strict inequality $\implies$ Dashed line through intercepts $(3,0), (0,2)$ (ii)
- (b) $2x + 3y \le 6$: Inclusive inequality $\implies$ Solid line through intercepts $(3,0), (0,2)$ (iv)
- (c) $y \ge 4$: Single variable $y$ $\implies$ Solid horizontal line at $y = 4$ (i)
- (d) $x < -2$: Single variable $x$ $\implies$ Dashed vertical line at $x = -2$ (iii)
- (e) $x + y = 0$: Passes through origin $(0,0)$ (v)
- Final Answer: (a) -> (ii), (b) -> (iv), (c) -> (i), (d) -> (iii), (e) -> (v)
More Chapter Notes for Class 10 (FBISE)
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