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Linear Inequalities
Inequality Fundamentals and Intervals
Inequalities describe relationships where values are not necessarily equal, mapped visually onto a number line using strict inequality and non-strict inequality notation.
$$x \in (a, b) \implies a < x < b$$
Interval notation converts inequality ranges into concise set representations.
$($ or $)$=Exclusive boundary (strict)(unitless)
$[$ or $]$=Inclusive boundary (non-strict)(unitless)
$x < a$
→Uses parenthesis $(-\infty, a)$
$x \ge a$
→Uses bracket $[a, \infty)$
Visual Cues: Open circles (strict) indicate the boundary is not part of the solution set.
Bracket Logic: Parentheses correlate to $<$ and $>$, while square brackets correlate to $\le$ and $\ge$.
The Negative Sign Reversal Rule
Unlike equations, multiplying or dividing an inequality by a negative value reverses the order of the relationship to maintain truth.
$$a < b \text{ and } c < 0 \implies ac > bc$$
Scaling by a negative constant flips the direction of the inequality sign.
$a, b$=Initial values(real numbers)
$c$=Negative multiplier($c < 0$)
$c = -1$
→$x < 5 \implies -x > -5$
Sign Reversal: This occurs because negative scaling reflects points across the origin, swapping their relative order.
Additive Invariance: Adding or subtracting any value does NOT change the inequality sign.
Solving Systems and Multi-Step Logic
Solving multi-step inequalities follows a hierarchy: simplify terms, isolating variables, and applying the multiplicative property with caution.
$$ax + b < cx + d \implies (a-c)x < d - b$$
Standard algebraic isolation techniques apply until the final coefficient division.
$a, c$=Coefficients of $x$(unitless)
$b, d$=Constant terms(unitless)
$a-c < 0$
→Final step requires sign flip
LCM Technique: When dealing with fractions, multiply the entire inequality by the least common multiple to simplify.
Compound Inequalities: Solutions for expressions like $a < f(x) < b$ represent the intersection of two conditions.
Two-Variable Inequalities and Regions
Linear inequalities in two variables define a half-plane in the coordinate system, bounded by a boundary line and verified via a test point.
$$ax + by \le c$$
The linear equation $ax + by = c$ creates the boundary of the solution region.
$ax+by=c$=Boundary line (associated equation)(linear)
$<$ or $>$=Dashed boundary (points on line excluded)(exclusive)
$\le$ or $\ge$=Solid boundary (points on line included)(inclusive)
Testing (0,0)
→Simplest test point when the line doesn't pass through the origin
Line passes through origin
→Use any other convenient point like $(1, 0)$ or $(0, 1)$ as the test point
Graphing Procedure: (1) Graph the associated equation $ax + by = c$, using dashes for strict and solid for non-strict. (2) Pick a test point and substitute — shade the half-plane that satisfies the inequality.
Vertical Lines: $x \ge a$ shades the right half-plane; $x \le a$ shades the left half-plane.
Horizontal Lines: $y \ge b$ shades the upper half-plane; $y \le b$ shades the lower half-plane.
Systems of Inequalities and Solution Regions
The solution of a system of inequalities is the intersection of all individual half-plane regions, forming a common area called the solution region.
Intersection Method: Graph each inequality separately on the same axes, then identify the overlapping (common) region.
Corner Points: Where two boundary lines of the solution region intersect, they form a corner point (vertex). These are found by solving pairs of associated equations simultaneously.
Bounded vs Unbounded: A solution region that can be enclosed in a circle is bounded; one extending infinitely is unbounded.
Convex Property: All feasible regions from linear inequalities are convex — any line segment joining two points within the region lies entirely inside it.
Feasible Region and Constraints
When a system of inequalities includes non-negative constraints ($x \ge 0$, $y \ge 0$), the solution region is restricted to the first quadrant and is called the feasible region.
Problem Constraints: The inequalities that model real-world limitations (resources, capacity, budgets) on decision variables.
Non-Negative Constraints: $x \ge 0$ and $y \ge 0$ restrict the solution to the first quadrant — quantities like production count or hours cannot be negative.
Feasible Solution: Any point $(x, y)$ inside or on the boundary of the feasible region that satisfies all constraints simultaneously.
Finding Corner Points: Solve each pair of boundary equations simultaneously. The intersection points that lie within the feasible region are the corner points.
Linear Programming and Optimization
Linear programming is the method of maximizing or minimizing an objective function subject to a system of linear inequality constraints, with the solution always occurring at a corner point of the feasible region.
$$f(x, y) = ax + by$$
The objective function is a linear expression whose value we want to optimize.
$f(x, y)$=Objective function (profit, cost, etc.)(depends on context)
$a, b$=Coefficients representing per-unit contribution(Rs/unit, etc.)
$x, y$=Decision variables (quantities to determine)(units of product/resource)
Feasible region is unbounded
→Maximum may not exist (value can grow without bound), but minimum may still exist
Optimal value at two adjacent corner points
→Every point on the line segment between them also gives the same optimal value
Corner Point Theorem: The maximum and minimum values of a linear objective function over a convex feasible region always occur at one or more corner points (vertices).
Optimization Procedure: (1) Graph the feasible region from all constraints. (2) Identify all corner points. (3) Evaluate $f(x,y)$ at each corner point. (4) The largest value is the maximum; the smallest is the minimum.
Why Only Corners?: A linear function on a convex polygon increases steadily in one direction — it cannot have an interior peak or valley, so extrema must lie on the boundary, specifically at vertices.
Converting a word problem into a linear programming model requires identifying decision variables, writing problem constraint inequalities, and defining the objective function.
Step 1 — Variables: Define $x$ and $y$ as the quantities to be determined (e.g., units of product A and B).
Step 2 — Constraints: Translate each resource limitation into an inequality (e.g., $5x + 4y \le 120$ for machine hours).
Step 3 — Non-Negativity: Always include $x \ge 0$, $y \ge 0$ since physical quantities cannot be negative.
Step 4 — Objective: Write the function to optimize (e.g., Profit $= 40x + 50y$, or Cost $= 25x + 30y$).
Application Strategy
Worked Examples
Problem Statement
Solve the inequality: $4(x - 2) + 1 > 7x - 10$ and represent the solution on a number line.
Distributive PropertyCombining Like TermsDivision by Negative
Strategic Toolkit
$ax + b > cx + d$
Step 1
Distribute and Simplify
Expand the parentheses and group terms on the left side.
$4x - 8 + 1 > 7x - 10 \implies 4x - 7 > 7x - 10$
Step 2
Isolate the Variable Term
Subtract $7x$ and add 7 to both sides to group the $x$ terms.
$-3x > -3$
Step 3
Apply Negative Division Rule
Divide by -3 and flip the inequality sign.
$x < 1$
Problem Statement
A rental car company charges Rs. 30 per day plus Rs. 0.50 per mile. If you have a budget of Rs. 100 for a one-day trip, how many miles can you drive?
Linear ModelingIsolating Variables
Strategic Toolkit
$\text{Total Cost} \le \text{Budget}$
Step 1
Setup Inequality
Define $m$ as miles. The flat fee plus the mileage rate must be within budget.
$30 + 0.50m \le 100$
Step 2
Isolate Mileage Term
Subtract the fixed cost of 30 from the total budget.
$0.50m \le 70$
Step 3
Calculate Final Miles
Divide by 0.50 (equivalent to multiplying by 2).
$m \le 140$
Problem Statement
Graph the inequality $x + 2y < 6$ in the xy-plane.
Associated EquationBoundary LineTest Point Method
Strategic Toolkit
$ax + by = c \text{ (boundary line)}$
Step 1
Find the Boundary Line
The associated equation is $x + 2y = 6$. Find intercepts: $x$-intercept at $(6, 0)$ and $y$-intercept at $(0, 3)$.
$x + 2y = 6$
Step 2
Choose Line Style
Since the inequality is strict ($<$, not $\le$), draw a dashed line through $(6, 0)$ and $(0, 3)$.
\text{Strict} \implies \text{Dashed}
Step 3
Apply Test Point
Test $(0, 0)$: $0 + 2(0) = 0 < 6$ is TRUE, so shade the side containing the origin (below the line).
$0 < 6 \checkmark \implies \text{shade origin side}$
Problem Statement
Graph the feasible region for $x - y \le 3$, $x + 2y \le 6$, $x \ge 0$, $y \ge 0$ and find the corner points.
Feasible RegionCorner PointsSimultaneous Equations
Strategic Toolkit
$\text{Solve pairs of boundary equations}$
Step 1
Graph Each Constraint
Draw $x - y = 3$ (through $(3,0)$ and $(0,-3)$) and $x + 2y = 6$ (through $(6,0)$ and $(0,3)$), plus axes $x = 0$ and $y = 0$. Shade the intersection region.
$\text{Four boundary lines} \implies \text{polygon in first quadrant}$
Step 2
Find Intersection of Constraint Lines
Solve $x - y = 3$ and $x + 2y = 6$ simultaneously. Subtracting: $3y = 3 \implies y = 1$, then $x = 4$.
$(4, 1)$
Step 3
List All Corner Points
Identify all vertices of the feasible region by checking intersections that lie in the first quadrant.
$(0, 0),\; (3, 0),\; (4, 1),\; (0, 3)$
Problem Statement
A farmer has 100 canals of land. Corn requires 3 hours/canal and wheat requires 2 hours/canal. Total hours ≤ 240. Profit: Rs. 20/canal (corn), Rs. 15/canal (wheat). Maximize profit.
Objective FunctionFeasible RegionCorner Point Evaluation
Strategic Toolkit
P(x,y) = 20x + 15y
Step 1
Formulate Constraints
Let $x$ = canals of corn, $y$ = canals of wheat. Land: $x + y \le 100$. Labour: $3x + 2y \le 240$. Non-negativity: $x \ge 0, y \ge 0$.
$x + y \le 100,\; 3x + 2y \le 240$
Step 2
Find Corner Points
Graph feasible region and solve boundary pairs. Intersection of $x + y = 100$ and $3x + 2y = 240$: substituting $y = 100 - x$ gives $x = 40, y = 60$.
$(0,0),\; (0,100),\; (40,60),\; (80,0)$
Step 3
Evaluate Objective Function
Compute $P$ at each corner: $P(0,0) = 0$, $P(0,100) = 1500$, $P(40,60) = 800 + 900 = 1700$, $P(80,0) = 1600$.
$P_{\max} = 1700 \text{ at } (40, 60)$
Step 4
Interpret Solution
Maximum profit of Rs. 1700 is achieved by cultivating 40 canals of corn and 60 canals of wheat.
$x = 40,\; y = 60$
Problem Statement
Minimize $f(x,y) = 2x + 3y$ subject to $x - y \le 2$, $x + y \ge 4$, $2x - y \le 6$, $x \ge 0$.
Unbounded RegionMinimum Exists but Maximum May Not
Strategic Toolkit
f(x,y) = 2x + 3y
Step 1
Graph and Identify Region
The feasible region is unbounded upwards. Corner points are $(0, 4)$, $(3, 1)$, and $(4, 2)$.
$\text{Region extends to } \infty$
Step 2
Evaluate at Corner Points
$f(0,4) = 12$, $f(3,1) = 9$, $f(4,2) = 14$.
$f_{\min} = 9 \text{ at } (3,1)$
Step 3
Check for Maximum
Since the region is unbounded, $f$ can grow arbitrarily large by increasing $x$ and $y$ — no maximum exists.
$f_{\max} = \text{does not exist}$
Memorize formulas
Negative Multiplication Rule
Multiplying or dividing an inequality by a negative number reverses the inequality sign.
Formula
$a < b \text{ and } c < 0 \implies ac > bc$
Compound Inequality Splitting
A three-part inequality is solved by applying the same operation to all three parts simultaneously.
Formula
$a < f(x) < b \iff a < f(x) \text{ and } f(x) < b$
Corner Point Theorem
The optimal value of a linear objective function on a convex feasible region occurs at a corner point.
Formula
$\max/\min\, f(x,y) = ax + by \text{ occurs at vertices of feasible region}$
Review
1
If $-3x + 7 \le 19$, which of the following represents the correct solution set for $x$?
Ans
$x \ge -4$
Review
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Isolate the linear term
Subtract 7 from both sides to move the constant away from the variable term.
$-3x \le 12$
Divide by the negative coefficient
Dividing by a negative number reverses the inequality sign, so the solution direction flips.
$x \ge -4$
2
Which interval notation represents the solution to the compound inequality: $5 < 2x - 1 \le 11$?
Ans
$(3, 6]$
Review
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Remove the constant from all parts
Add 1 to each part of the compound inequality so the variable term is isolated.
$6 < 2x \le 12$
Divide through by 2
Since 2 is positive, the inequality directions stay the same after division.
$3 < x \le 6$, so the interval is $(3, 6]$
3
Under what condition does the inequality $\frac{1}{a} > \frac{1}{b}$ imply that $a < b$?
Ans
When $a$ and $b$ have the same sign
Review
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Use the reciprocal order rule
Taking reciprocals reverses order only when both numbers share the same sign, because the reciprocal function is monotonic on each sign interval.
If $a$ and $b$ are both positive or both negative, then $\frac{1}{a} > \frac{1}{b}$ implies $a < b$
4
The point $(2, 3)$ lies in the solution set of which system of inequalities?
Ans
$x + y \le 6$ and $2x - y \ge 0$
Review
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Test the point in each system
Substitute $(2,3)$ into each inequality pair and keep only the system where both inequalities are true.
$2+3=5\le 6$ and $2(2)-3=1\ge 0$
Eliminate the false systems
The other options fail at least one inequality, so they cannot contain the point.
Only $x+y\le 6$ and $2x-y\ge 0$ works
5
The feasible region of $x + 2y \le 8$, $3x + y \le 9$, $x \ge 0$, $y \ge 0$ has corner points at $(0,0)$, $(3,0)$, $(2,3)$, and $(0,4)$. If $f(x,y) = x + 4y$, the minimum value of $f$ over the feasible region is:
Ans
0
Review
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Evaluate the objective at each corner
For a linear program, the optimum occurs at a corner point, so compute $f(x,y)$ at every vertex.
$f(0,0)=0,\ f(3,0)=3,\ f(2,3)=14,\ f(0,4)=16$
Choose the smallest value
The least of the corner values is the minimum over the feasible region.
Minimum $=0$ at $(0,0)$
6
A system of linear inequalities has a bounded feasible region with 4 corner points. The objective function $f(x,y) = 2x + 5y$ takes the same maximum value at two adjacent corner points. What can be concluded?
Ans
Every point on the line segment joining those two corners also gives the maximum
Review
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Use linearity along the edge
If two adjacent vertices give the same maximum value, then the entire segment between them keeps that same value because a linear function varies linearly along a line segment.
Every point on the connecting edge is also optimal
7
A manufacturer produces products A and B. Each unit of A uses 2 units of chemical and 1 unit of compound. Each unit of B uses 1 unit of chemical and 2 units of compound. Available: 800 chemical, 1000 compound. Profit: Rs. 30/A, Rs. 20/B. The constraints are:
Ans
$2x + y \le 800$, $x + 2y \le 1000$, $x \ge 0$, $y \ge 0$
Review
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Translate each resource limit into an inequality
Let $x$ and $y$ be the numbers of products A and B. Chemical usage gives one constraint and compound usage gives the other.
$2x + y \le 800$ and $x + 2y \le 1000$
Include nonnegativity
Production quantities cannot be negative, so both variables must be at least zero.
$x \ge 0,\ y \ge 0$
8
The feasible region for $3x + 2y \ge 6$, $x + 3y \le 6$, $y \ge 0$ has corner points $(2, 0)$, $(6, 0)$, and $(\frac{6}{7}, \frac{12}{7})$. The minimum value of $z = 2x + y$ is:
Ans
$\frac{24}{7}$
Review
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Check the objective at each corner
A linear objective reaches its minimum at a vertex of the feasible region, so evaluate $z$ at each corner point.
$z(2,0)=4,\ z(6,0)=12,\ z\!\left(\frac{6}{7},\frac{12}{7}\right)=\frac{24}{7}$
Pick the least value
Compare the corner values and take the smallest one.
Minimum $=\frac{24}{7}$
9
An inequality system has an unbounded feasible region extending upward in the first quadrant. The objective function is $f(x,y) = x + 3y$. Which statement is true?
Ans
The minimum may exist at a corner point, but the maximum does not exist
Review
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Analyze the unbounded direction
Because the feasible region extends upward and $f(x,y)=x+3y$ increases when $y$ increases, the objective can grow without limit.
No maximum exists
Locate the minimum behavior
A finite lower boundary can still produce a minimum at a corner point even when the region is unbounded in another direction.
The minimum may exist at a corner point
10
The graph of $x + 2y \le 6$ is the closed half-plane on the origin-side of $x + 2y = 6$. Which point does NOT belong to this region?
Ans
$(4, 2)$
Review
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Substitute each point into the inequality
A point belongs to the region only if its coordinates satisfy $x + 2y \le 6$.
$(4,2): 4 + 2(2) = 8 > 6$
Identify the excluded point
The other listed points satisfy the inequality, so the one that exceeds 6 is the only point outside the region.
$(4,2)$ does not belong
11
A shopkeeper sells chairs ($x$) and tables ($y$). Constraints: $x + y \le 80$, $4x + 6y \le 360$, $x \ge 0$, $y \ge 0$. Profit: Rs. 25/chair, Rs. 40/table. Corner points are $(0,0)$, $(80,0)$, $(60,20)$, $(0,60)$. Which corner gives the maximum profit?
Ans
$(0, 60)$ with profit Rs. 2400
Review
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Compute profit at each vertex
A linear profit function is maximized at a corner point, so evaluate the profit at every listed corner.
$P(0,0)=0,\ P(80,0)=2000,\ P(60,20)=2300,\ P(0,60)=2400$
Select the largest profit
Compare the corner values and choose the greatest one.
Maximum profit is Rs. 2400 at $(0,60)$
12
For the feasible region defined by $x + y \le 5$, $x - y \le 1$, $x \ge 0$, $y \ge 0$, the corner point formed by the intersection of $x + y = 5$ and $x - y = 1$ is:
Ans
$(3, 2)$
Review
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Solve the two boundary equations together
Find the intersection of the lines by adding the equations to eliminate $y$.
$2x = 6 \Rightarrow x = 3$
Substitute back for the other coordinate
Use $x = 3$ in either equation to get the matching $y$ value.
$3 + y = 5 \Rightarrow y = 2$, so the corner point is $(3,2)$
Question 1 of 12
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1
If $-3x + 7 \le 19$, which of the following represents the correct solution set for $x$?