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Linear Inequalities - Inequalities

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Linear Inequalities are mathematical expressions involving variables where the relationship is expressed using inequality symbols like <<, >>, \le, or \ge. A linear inequality in one variable xx typically takes the form ax+b<0ax + b < 0, while in two variables xx and yy it looks like ax+by+c>0ax + by + c > 0.

Algebraic Rules: (1) Equal numbers can be added to or subtracted from both sides without changing the inequality sign. (2) Multiplying or dividing both sides by a positive number preserves the inequality sign. (3) Multiplying or dividing both sides by a negative number reverses the direction of the inequality sign (<< becomes >>, and vice versa).

Interval Notation: Solutions are often written as intervals. A closed interval [a,b][a, b] includes the endpoints (axba \le x \le b), whereas an open interval (a,b)(a, b) excludes them (a<x<ba < x < b). Semi-open/closed intervals like [a,b)[a, b) include one endpoint and exclude the other.

Representation on a Number Line: For inequalities in one variable, the solution set is graphed as a ray or segment. A 'hollow/open circle' \circ is used to denote that the boundary value is excluded (strict inequality like << or >>). A 'darkened/solid circle' \bullet is used to denote that the boundary value is included (slack inequality like \le or \ge).

Graphical Solution in Two Variables: A linear inequality ax+by<cax + by < c divides the Cartesian plane into two half-planes. To find the solution region, first graph the boundary line ax+by=cax + by = c. If the inequality is strict (<< or >>), draw a dashed or broken line; if it is slack (\le or \ge), draw a solid line. Visually, the solution set is represented by shading the appropriate half-plane.

Trial Point Method: To determine which half-plane to shade, pick a test point not on the boundary line (the origin (0,0)(0,0) is the simplest choice). If the point satisfies the inequality, shade the region containing that point. If not, shade the opposite side. Visually, this creates a shaded 'zone' of all possible solutions on the coordinate grid.

📐Formulae

Linear Inequality in one variable: ax+b<0,ax+b>0,ax+b0,ax+b0ax + b < 0, ax + b > 0, ax + b \le 0, ax + b \ge 0

Linear Inequality in two variables: ax+by<c,ax+by>c,ax+byc,ax+bycax + by < c, ax + by > c, ax + by \le c, ax + by \ge c

Property 1: a<b    a+c<b+ca < b \implies a + c < b + c for any cRc \in \mathbb{R}

Property 2: a<ba < b and k>0    ak<bkk > 0 \implies ak < bk

Property 3: a<ba < b and k<0    ak>bkk < 0 \implies ak > bk

Closed Interval: [a,b]={x:axb}[a, b] = \{x : a \le x \le b\}

Open Interval: (a,b)={x:a<x<b}(a, b) = \{x : a < x < b\}

💡Examples

Problem 1:

Solve the linear inequality 4x+3<6x+74x + 3 < 6x + 7 for xRx \in \mathbb{R}.

Solution:

4x+3<6x+74x + 3 < 6x + 7 Subtract 6x6x from both sides: 4x6x+3<74x - 6x + 3 < 7 2x+3<7-2x + 3 < 7 Subtract 33 from both sides: 2x<4-2x < 4 Divide by 2-2 and reverse the inequality sign: x>42x > \frac{4}{-2} x>2x > -2 The solution set is x(2,)x \in (-2, \infty).

Explanation:

The solution involves isolating the variable xx by applying algebraic operations. Crucially, when we divide by the negative number 2-2, the 'less than' sign flips to a 'greater than' sign.

Problem 2:

Solve 3x25x+83x - 2 \le 5x + 8 and represent the solution on a number line.

Solution:

3x25x+83x - 2 \le 5x + 8 Subtract 5x5x from both sides: 2x28-2x - 2 \le 8 Add 22 to both sides: 2x10-2x \le 10 Divide by 2-2 and reverse the inequality sign: x5x \ge -5 On the number line, draw a solid circle at 5-5 and shade the line to the right towards infinity.

Explanation:

The 'greater than or equal to' symbol means we use a solid circle at 5-5 to show it is included in the solution set. The shaded region extends to the right because the values are larger than 5-5.