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Algebra - Linear Inequations

Grade 10ICSE

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

🔑Concepts

A Linear Inequation is a mathematical statement where two algebraic expressions are related by inequality symbols such as <,>,,<, >, \le, or \ge. Unlike an equation which has specific roots, an inequation represents a range of values that satisfy the condition.

The Replacement Set is the universal set from which the values of the variable xx are chosen (e.g., Natural Numbers NN, Whole Numbers WW, Integers ZZ, or Real Numbers RR). The Solution Set is the specific subset of the replacement set containing all values that make the inequation true.

Rule of Addition and Subtraction: Adding or subtracting the same positive or negative number to both sides of an inequation does not change the direction of the inequality sign. For example, if x<yx < y, then x+a<y+ax + a < y + a and xa<yax - a < y - a.

Rule of Multiplication and Division: Multiplying or dividing both sides by a positive number maintains the inequality. However, multiplying or dividing both sides by a negative number reverses the inequality sign (e.g., if 5>25 > 2, then 5<2-5 < -2). Visually, this is because multiplying by a negative number reflects the points across the origin on the number line.

Representation on a Number Line for discrete sets (N,W,ZN, W, Z): Solutions are marked as distinct, bold dots on specific integer values. For instance, if x{1,2,3}x \in \{1, 2, 3\}, you would place bold dots only on the marks for 1,2,1, 2, and 33 on the horizontal axis.

Representation on a Number Line for Real Numbers (RR): Solutions are shown as a continuous thick line or shaded region. Use a hollow circle \circ at the endpoint if the value is excluded (<< or >>), and a solid darkened circle \bullet if the value is included (\le or \ge).

Combined Inequations (Double Inequalities): To solve a statement like af(x)<ba \le f(x) < b, split it into two separate inequations: af(x)a \le f(x) and f(x)<bf(x) < b. Solve them independently and find the intersection of the two solution sets to determine the final range.

📐Formulae

General form: ax+b<cax + b < c or ax+bcax + b \le 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}

Transposition: ax+bc    axcbax + b \ge c \implies ax \ge c - b

Natural Numbers: N={1,2,3,...}N = \{1, 2, 3, ...\}, Whole Numbers: W={0,1,2,...}W = \{0, 1, 2, ...\}, Integers: Z={...,1,0,1,...}Z = \{..., -1, 0, 1, ...\}

💡Examples

Problem 1:

Solve the inequation and represent the solution on a number line: 2x3<7,xN2x - 3 < 7, x \in N.

Solution:

2x3<72x - 3 < 7 Add 33 to both sides: 2x<7+32x < 7 + 3 2x<102x < 10 Divide by 22: x<5x < 5 Since xNx \in N (Natural Numbers), the solution set is {1,2,3,4}\{1, 2, 3, 4\}.

Explanation:

We isolate xx using standard algebraic steps. Because the replacement set is NN, we only include positive integers less than 5. On the number line, this would be represented by four thick dots at 1,2,3,1, 2, 3, and 44.

Problem 2:

Solve and graph the solution set on a number line: 32x1<5,xR-3 \le 2x - 1 < 5, x \in R.

Solution:

Split the inequation into two parts: Part 1: 32x1-3 \le 2x - 1 3+12x    22x    1x-3 + 1 \le 2x \implies -2 \le 2x \implies -1 \le x Part 2: 2x1<52x - 1 < 5 2x<5+1    2x<6    x<32x < 5 + 1 \implies 2x < 6 \implies x < 3 Combining the two parts: 1x<3-1 \le x < 3 The solution set is {x:xR,1x<3}\{x : x \in R, -1 \le x < 3\}.

Explanation:

We solve the double inequality by splitting it into two linear inequations. The result is a range of real numbers. On the number line, we draw a thick line starting with a solid circle \bullet at 1-1 (inclusive) and ending with a hollow circle \circ at 33 (exclusive).