Chapter 5 - Linear Inequalities

Linear inequality is a statement where two expressions are not equal but one is less than or greater than the other using symbols < less than, > greater than, ≤ less than or equal, ≥ greater than or equal. Example x<5 means x is less than 5, x≥3 means x is 3 or more than 3.

Difference from equation: Equation x=5 has one solution 5, inequality x<5 has infinite solutions all numbers less than 5 ...2,3,4,4.9 etc.

Linear means power of variable is 1, no x². Example 2x+3<7 is linear inequality one variable, x+y≤6 is linear inequality two variables.

Solution of inequality shown on number line. Open circle ○ means endpoint not included < or >. Filled dot ● means endpoint included ≤ or ≥.

Intervals: Closed interval [a,b] means a≤x≤b both endpoints included a and b included filled dots. Example [2,5] = {x|2≤x≤5} 2 to 5 including 2 and 5.

Open interval (a,b) means a<x<b both endpoints not included open circles. Example (1,4) = {x|1<x<4} between 1 and 4 not including 1 and 4.

Half-open (a,b] = a<x≤b a excluded b included, [a,b) = a≤x<b a included b excluded.

Infinite intervals: (-∞,5] = x≤5 all numbers less than or equal to 5 arrow to -∞, (2,∞) = x>2 all numbers greater than 2 arrow to ∞. Note ∞ always uses parenthesis ( because infinity is not a real number.

Rules to solve linear inequality like equation but with care:

1. Add or subtract same number both sides inequality stays same direction. Example x+3<7 subtract 3 both sides x<4 direction same.

2. Multiply or divide by positive number both sides direction stays same. Example 2x<8 divide by 2 positive x<4 same direction.

3. Multiply or divide by negative number both sides direction flips < becomes > and > becomes <. Example -2x<8 divide by -2 negative flip x>-4 because -2 negative direction reverses. This is most important rule sign flip when multiply divide by negative.

Example solve 3x-5<7: add 5 both sides 3x<12 divide by 3 positive x<4 solution (-∞,4) open interval.

Example solve -3x+2≥8: subtract 2 -3x≥6 divide by -3 negative flip x≤-2 solution (-∞,-2] closed at -2.

Compound: 2x+3≤11 and 2x+3≥7 solve both 2x≤8 x≤4 and 2x≥4 x≥2 combined 2≤x≤4 interval [2,4].

Linear inequality in two variables ax+by≤c or ≥ or < >. Boundary line ax+by=c solid line if ≤ or ≥ included, dotted line if < or > not included.

Steps to graph: 1. Draw boundary line ax+by=c find two points. Example x+y≤6 points (0,6) and (6,0) join line solid because ≤ included.

2. Choose test point not on line usually origin (0,0) if origin not on line. Put in inequality. If true shade side containing test point, if false shade opposite side.

Example x+y≤6 test (0,0): 0+0≤6 true 0≤6 true so shade side containing origin (0,0) below line. So half-plane containing origin is solution.

Example y≥0.5x+1 boundary y=0.5x+1 solid line slope 0.5 intercept 1 test (0,0): 0≥0.5×0+1? 0≥1 false so shade opposite side not containing origin above line.

Strict inequalities < > use dotted boundary line not included, ≤ ≥ use solid boundary included.

System means more than one inequality together. Solution is common region where all inequalities satisfied simultaneously overlapping shaded regions feasible region.

Steps: Graph each inequality separately shade its half-plane, overlap of all shadings is feasible region common solution. Boundaries solid if ≤ ≥ included.

Example system: (1) x+y≤6, (2) y≥0.5x+1, (3) y≤2. Graph line 1 x+y=6 slope -1 intercept 6 solid, line 2 y=0.5x+1 slope 0.5 intercept 1 solid, line 3 y=2 horizontal solid at y=2. Shade: inequality 1 shade below containing origin, inequality 2 shade above not containing origin, inequality 3 shade below y≤2. Overlap is triangle feasible region with vertices A=(1,1), B=(4,1) approx etc. Any point inside feasible region satisfies all three simultaneously.

Feasible region may be bounded polygon triangle quadrilateral or unbounded infinite goes to infinity.

Linear inequalities used daily for limits and constraints.

Budgeting: You have Rs 1000 budget. Item A costs Rs 200, Item B costs Rs 150. Inequality 200x+150y≤1000 number of items x and y you can buy within budget.

Diet problem: Need at least 60g protein. Food A has 10g protein per unit, Food B has 15g. Inequality 10x+15y≥60 to meet protein requirement.

Factory production: Machine can produce at most 100 units per day x≤100, needs at least 2 hours per unit time constraint.

Profit: Profit = 20x+30y need to maximize profit subject to inequalities constraints feasible region optimization later in linear programming.

Real life: Taxi fare distance limit, exam marks passing criteria marks≥33, height requirement for ride height≥120cm, speed limit speed≤80 km/h all are inequalities.

Key tricks for exams:

Sign flip: Whenever multiply or divide by negative number inequality flips <↔> ≤↔≥. Remember negative flips.

Open vs Closed: < > open interval ( ) open circle ○ not included, ≤ ≥ closed interval [ ] filled dot ● included.

Infinite intervals: -∞ always parenthesis ( because -∞ not real number, ∞ always parenthesis ). Example (-∞,5] x≤5, (2,∞) x>2.

Graphical test: For ax+by≤c test origin (0,0) if inequality true at origin shade towards origin, if false shade away from origin. If line passes through origin choose another test point like (1,0).

System feasible region: Intersection of all shadings common region. If feasible region bounded polygon corner points give extreme values.

Compound: a≤x≤b means interval [a,b] closed both included, a<x<b means (a,b) open both excluded.

Trick: < and > without equal dotted boundary line not included, ≤ and ≥ with equal solid boundary included.

Complete summary for quick revision before exam.