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Linear Inequalities - Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

Grade 11CBSE

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

🔑Concepts

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A linear inequality in one variable is an expression involving algebraic terms with a degree of 1, separated by inequality signs such as <,>,≤,<, >, \le, or ≥\ge. Solving it means finding all real values of xx that make the statement true.

Number line representation of x > 2 using an open circle at 2 and a bold line to the right.
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Equal numbers can be added to or subtracted from both sides of an inequality without changing the sign of the inequality. For example, if x−3<5x - 3 < 5, then x<8x < 8.

Diagram
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When both sides of an inequality are multiplied or divided by a positive number, the inequality sign remains the same. However, if multiplied or divided by a negative number, the inequality sign MUST be reversed.

Instructional box showing the reversal of the inequality sign when dividing by a negative number.
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Graphical representation on a number line uses a dark circle (dot) for inclusive inequalities (≤,≥\le, \ge) and an open circle for strict inequalities (<,><, >). The solution set is denoted by a thick line or arrow in the direction of the valid values.

Number line representation of x <= -1 with a closed dot and a shaded line to the left.

📐Formulae

General forms: ax+b<0ax + b < 0, ax+b>0ax + b > 0, ax+b≤0ax + b \le 0, ax+b≥0ax + b \ge 0

If a<ba < b and cc is any real number, then a+c<b+ca + c < b + c and a−c<b−ca - c < b - c

If a<ba < b and c>0c > 0, then ac<bcac < bc and ac<bc\frac{a}{c} < \frac{b}{c}

If a<ba < b and c<0c < 0, then ac>bcac > bc and ac>bc\frac{a}{c} > \frac{b}{c}

Solution set for x>ax > a: (a,∞)(a, \infty)

Solution set for x≤ax \le a: (−∞,a](-\infty, a]

💡Examples

Problem 1:

Solve the inequality 4x+3<6x+74x + 3 < 6x + 7 and represent the solution on a number line.

Solution:

Step 1: Subtract 4x4x from both sides: 3<6x−4x+73 < 6x - 4x + 7 3<2x+73 < 2x + 7 Step 2: Subtract 7 from both sides: 3−7<2x3 - 7 < 2x −4<2x-4 < 2x Step 3: Divide by 2: −2<x-2 < x or x>−2x > -2 The solution set in interval notation is (−2,∞)(-2, \infty).

Explanation:

To solve, we isolate the variable xx by performing inverse operations. Since we did not multiply or divide by a negative number, the inequality sign direction remains the same throughout. On a number line, this would be shown by an open circle at −2-2 and a shaded arrow pointing to the right.

Problem 2:

Solve 3(x−2)5≤5(2−x)3\frac{3(x-2)}{5} \le \frac{5(2-x)}{3} for real xx.

Solution:

Step 1: Multiply both sides by the LCM of 5 and 3, which is 15: 3[3(x−2)]≤5[5(2−x)]3[3(x-2)] \le 5[5(2-x)] 9(x−2)≤25(2−x)9(x-2) \le 25(2-x) Step 2: Expand the brackets: 9x−18≤50−25x9x - 18 \le 50 - 25x Step 3: Add 25x25x to both sides: 34x−18≤5034x - 18 \le 50 Step 4: Add 18 to both sides: 34x≤6834x \le 68 Step 5: Divide by 34: x≤2x \le 2 The solution set is (−∞,2](-\infty, 2].

Explanation:

We first eliminate the fractions by multiplying by the positive LCM. After simplifying the terms, we isolate xx. The 'less than or equal to' sign means 2 is included in the solution, represented by a solid circle on the number line at 2 with shading to the left.

Problem 3:

Solve the inequality 3x−5<x+73x - 5 < x + 7 and show the solution on a number line for real xx.

Number line from 0 to 10 with an open circle at 6 and a bold line extending to the left.

Solution:

3x−5<x+73x - 5 < x + 7 3x−x<7+53x - x < 7 + 5 2x<122x < 12 x<6x < 6 The solution set is (−∞,6)(-\infty, 6).

Explanation:

We group the variable terms on the left and the constant terms on the right. Dividing by a positive coefficient (2) does not change the inequality sign. Since xx is strictly less than 6, we use an open circle at 6.

Problem 4:

Solve the inequality 5−2x3≤x6−5\frac{5 - 2x}{3} \le \frac{x}{6} - 5 for real xx.

Number line with a solid dot at 8 and a bold line extending to the right representing x >= 8.

Solution:

5−2x3≤x−306\frac{5 - 2x}{3} \le \frac{x - 30}{6} Multiply both sides by 6 (LCM of 3 and 6): 2(5−2x)≤x−302(5 - 2x) \le x - 30 10−4x≤x−3010 - 4x \le x - 30 −4x−x≤−30−10-4x - x \le -30 - 10 −5x≤−40-5x \le -40 Divide by -5 and reverse the inequality sign: x≥8x \ge 8 The solution set is [8,∞)[8, \infty).

Explanation:

After clearing fractions by multiplying by 6, we isolate xx. The final step involves dividing by a negative number (-5), which requires flipping the inequality sign from ≤\le to ≥\ge. A solid dot is used at 8 to show it is included.