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Linear Inequalities - Graphical Solution of Linear Inequalities in Two Variables

Grade 11CBSE

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

🔑Concepts

A linear inequality in two variables is an algebraic expression involving variables xx and yy related by inequality signs like <,>,,<, >, \le, or \ge. Visually, while a linear equation ax+by=cax + by = c represents a straight line on a Cartesian plane, a linear inequality represents an entire region or 'half-plane' bounded by that line.

The line ax+by=cax + by = c acts as the boundary of the solution set. If the inequality is strict (<< or >>), the boundary line is represented as a dashed or dotted line to show that points on the line are not included in the solution. If the inequality is non-strict (\le or \ge), the boundary is drawn as a solid line, indicating points on the line are part of the solution set.

A vertical line x=ax = a or a non-vertical line ax+by=cax + by = c divides the XYXY-plane into two distinct regions called half-planes. A non-vertical line divides the plane into a 'lower half-plane' and an 'upper half-plane', while a vertical line divides it into a 'left half-plane' and a 'right half-plane'.

To determine which half-plane to shade, the 'Test Point Method' is used. Select a point not lying on the boundary line (the origin (0,0)(0,0) is the simplest choice if the line doesn't pass through it) and substitute its coordinates into the inequality. If the resulting statement is true, the region containing that point is the solution; if false, the opposite half-plane is the solution.

The graphical solution of a system of linear inequalities is the region of the plane that is common to all individual inequalities in the system. Visually, this is represented by the 'overlapping' or 'intersection' of the shaded regions of each inequality.

The solution region of a system of inequalities can be 'bounded' (enclosed by boundary lines on all sides like a polygon) or 'unbounded' (extending infinitely in at least one direction).

Linear inequalities often include 'non-negative constraints' like x0x \ge 0 and y0y \ge 0. Visually, these constraints restrict the solution region to the first quadrant of the coordinate system.

📐Formulae

General form of linear inequalities: ax+by<cax + by < c, ax+by>cax + by > c, ax+bycax + by \le c, ax+bycax + by \ge c

Boundary line equation: ax+by=cax + by = c

Slope-intercept form for plotting: y=mx+cy = mx + c, where m=abm = -\frac{a}{b}

Intercept form for plotting: xxintercept+yyintercept=1\frac{x}{x_{intercept}} + \frac{y}{y_{intercept}} = 1

Horizontal line inequality: yky \le k (region below the line y=ky=k) or yky \ge k (region above the line y=ky=k)

Vertical line inequality: xhx \le h (region to the left of line x=hx=h) or xhx \ge h (region to the right of line x=hx=h)

💡Examples

Problem 1:

Solve the linear inequality 3x+2y>63x + 2y > 6 graphically.

Solution:

Step 1: Convert the inequality into an equation to find the boundary line: 3x+2y=63x + 2y = 6. \nStep 2: Find the intercepts. When x=0,y=3x = 0, y = 3 (point (0,3)(0, 3)). When y=0,x=2y = 0, x = 2 (point (2,0)(2, 0)). \nStep 3: Draw a dashed line passing through (0,3)(0, 3) and (2,0)(2, 0) because the inequality is strict (>>). \nStep 4: Use (0,0)(0,0) as a test point. Substitute into 3x+2y>63x + 2y > 6: 3(0)+2(0)>60>63(0) + 2(0) > 6 \Rightarrow 0 > 6. This is False. \nStep 5: Since the origin does not satisfy the inequality, shade the half-plane that does not contain the origin.

Explanation:

The solution is the region above the line 3x+2y=63x + 2y = 6, excluding the points on the line itself.

Problem 2:

Find the graphical solution for the system of inequalities: x+y5x + y \le 5, 4x+y44x + y \ge 4, x0x \ge 0, y0y \ge 0.

Solution:

Step 1: For x+y=5x + y = 5, intercepts are (5,0)(5,0) and (0,5)(0,5). Draw a solid line. Testing (0,0)(0,0) gives 050 \le 5 (True), so shade towards the origin. \nStep 2: For 4x+y=44x + y = 4, intercepts are (1,0)(1,0) and (0,4)(0,4). Draw a solid line. Testing (0,0)(0,0) gives 040 \ge 4 (False), so shade away from the origin. \nStep 3: x0x \ge 0 and y0y \ge 0 restrict the solution to the first quadrant. \nStep 4: Identify the common region where all conditions overlap.

Explanation:

The solution is a bounded quadrilateral region in the first quadrant with vertices determined by the intersection of the boundary lines.