Linear equations and inequalities: Class 9 Mathematics Notes, Short Questions and Numericals

Linear equations in one variable, equations with fractions and absolute value, word problems, linear inequalities and their solution on the number line, and inequalities involving absolute value.

Key points

An equation states that two expressions are equal. A linear equation in one variable has the form ax + b = 0, where a is not zero, and the variable has power 1. The value of the variable that makes both sides equal is the solution or root. A linear equation in one variable has exactly one solution.

Rules for solving an equation (do the same thing to both sides):

Add or subtract the same number on both sides.
Multiply or divide both sides by the same non-zero number.
Simplify each side first: remove brackets and collect like terms.
Move all terms containing the variable to one side and the constants to the other, then divide by the coefficient.

Example: 3x − 7 = 2x + 5. Subtract 2x from both sides: x − 7 = 5. Add 7: x = 12. Always check by putting the answer in the original equation: left side 36 − 7 = 29, right side 24 + 5 = 29. Both sides match.

When an equation has fractions, multiply every term by the LCM of the denominators to clear them. For (x + 1)/2 + (x − 2)/3 = 4, the LCM is 6, so 3(x + 1) + 2(x − 2) = 24, which gives 5x − 1 = 24 and x = 5.

Some equations are not linear at first but become linear after simplifying. For example (x + 2)/(x − 1) = 3 can be written as x + 2 = 3(x − 1) provided x is not 1. A value that makes a denominator zero must be rejected, because the division is not allowed. Such a value is called an extraneous root.

The absolute value (or modulus) of a number is its distance from zero on the number line, so |a| = a if a ≥ 0 and |a| = −a if a < 0. Thus |5| = 5 and |−5| = 5. Distance is never negative, so |x| is never negative.

To solve |ax + b| = c with c ≥ 0, split into two linear equations: ax + b = c or ax + b = −c. If c is negative there is no solution, since an absolute value cannot be negative. For |2x − 3| = 7 we get 2x − 3 = 7, giving x = 5, or 2x − 3 = −7, giving x = −2.

Word problems are solved in four steps. Let a letter stand for the unknown, translate the statement into an equation, solve the equation, and check the answer against the story. Typical topics are consecutive numbers, ages, perimeters of figures, money and work. Write the unit in the final answer.

An inequality compares two expressions using <, >, ≤ or ≥. A linear inequality in one variable has a whole range of solutions, not a single value. The solution set can be shown on a number line: an open circle for < or > (the end value is not included) and a filled circle for ≤ or ≥ (the end value is included).

Rules for solving inequalities:

Adding or subtracting the same number on both sides keeps the sign unchanged.
Multiplying or dividing both sides by a positive number keeps the sign unchanged.
Multiplying or dividing both sides by a negative number reverses the sign. For example −3x < 9 gives x > −3.
Three-part inequalities such as −2 < 2x + 1 ≤ 7 are solved by working on all three parts at once.

Inequalities with absolute value (for a > 0):

|x| < a means −a < x < a (the distance from zero is less than a).
|x| > a means x < −a or x > a.
For |x − 2| < 3 we get −3 < x − 2 < 3, so −1 < x < 5.

Study tip: Do the same operation on both sides and keep each step on a separate line. Check every equation by substitution, which takes seconds and shows errors at once. For word problems, write the line 'Let x be ...' before anything else, because the examiner gives a mark for it. For inequalities, test one value from your solution set.

Definitions and flashcards

Say the meaning first, then open the card to check.

Linear equation

An equation in which the variable has power 1, written ax + b = 0.

Solution (root)

A value of the variable that makes the equation true.

Extraneous root

A value found in solving that does not satisfy the original equation.

Absolute value

The distance of a number from zero, written |x|.

Inequality

A statement using <, >, ≤ or ≥ to compare two expressions.

Solution set

The set of all values that satisfy an equation or inequality.

Consecutive integers

Integers that follow one another, such as n, n + 1, n + 2.

Equivalent equations

Equations that have the same solution.

Formula sheet

Name	Formula	What it means
Linear equation	ax + b = 0 ⇒ x = −b/a	Solution when a is not zero.
Absolute value	\|a\| = a if a ≥ 0, and \|a\| = −a if a < 0	Definition of modulus.
Absolute equation	\|u\| = c ⇒ u = c or u = −c	For c ≥ 0.
Absolute inequality (less)	\|u\| < a ⇒ −a < u < a	For a > 0.
Absolute inequality (greater)	\|u\| > a ⇒ u < −a or u > a	For a > 0.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Define a linear equation in one variable. How many solutions does it have? (2 marks)

A linear equation in one variable is an equation that can be written as ax + b = 0 with a not zero, where the variable has the highest power 1. It has exactly one solution, x = −b/a.

What is the absolute value of a number? Find |−8| and |3 − 7|. (2 marks)

The absolute value of a number is its distance from zero on the number line, which is never negative. |−8| = 8. Also 3 − 7 = −4, so |3 − 7| = |−4| = 4.

State when the sign of an inequality is reversed. (2 marks)

The sign of an inequality is reversed when both sides are multiplied or divided by a negative number. For example, from 2 < 5, multiplying by −1 gives −2 > −5.

Solve 5(x − 2) = 3x + 4. (3 marks)

Expand the left side: 5x − 10 = 3x + 4. Subtract 3x: 2x − 10 = 4. Add 10: 2x = 14. Divide by 2: x = 7. Check: left 5 × 5 = 25, right 21 + 4 = 25.

Solve |x − 2| < 3 and show the solution on a number line in words. (3 marks)

|x − 2| < 3 means −3 < x − 2 < 3. Add 2 to every part: −1 < x < 5. On the number line, draw open circles at −1 and 5 and shade the line between them. The end points are not included.

Solve the inequality 3 − 2x ≥ 11 and list the steps with the reason for each. (5 marks)

Start with 3 − 2x ≥ 11. Subtract 3 from both sides: −2x ≥ 8. Now divide both sides by −2. Because we divide by a negative number, the sign reverses: x ≤ −4. The solution set is all real numbers less than or equal to −4. On the number line this is a filled circle at −4 with shading to the left. Check with x = −5: 3 − 2(−5) = 13, and 13 ≥ 11 is true.

Numericals with worked steps

Solve 3x − 7 = 2x + 5.

Subtract 2x from both sides: x − 7 = 5.
Add 7 to both sides: x = 12.
Check: left 3 × 12 − 7 = 29, right 2 × 12 + 5 = 29.

Answer: x = 12

Solve (x + 1)/2 + (x − 2)/3 = 4.

LCM of 2 and 3 is 6. Multiply every term by 6.
3(x + 1) + 2(x − 2) = 24.
3x + 3 + 2x − 4 = 24, so 5x − 1 = 24.
5x = 25, so x = 5.
Check: 6/2 + 3/3 = 3 + 1 = 4.

Answer: x = 5

Solve |2x − 3| = 7.

Case 1: 2x − 3 = 7, so 2x = 10 and x = 5.
Case 2: 2x − 3 = −7, so 2x = −4 and x = −2.
Check: |10 − 3| = 7 and |−4 − 3| = 7.

Answer: x = 5 or x = −2

Solve −3x + 5 < 14.

Subtract 5: −3x < 9.
Divide by −3 and reverse the sign: x > −3.
Check x = 0: 5 < 14 is true.

Answer: x > −3

The sum of three consecutive integers is 72. Find them.

Let the integers be n, n + 1, n + 2.
n + (n + 1) + (n + 2) = 72, so 3n + 3 = 72.
3n = 69, so n = 23.
The numbers are 23, 24, 25. Check: 23 + 24 + 25 = 72.

Answer: 23, 24 and 25

Common mistakes

Not reversing the inequality sign when dividing or multiplying by a negative number.
Multiplying only one term by the LCM when clearing fractions; every term, including whole numbers, must be multiplied.
Solving |x| = c with only one case and missing the second solution, or giving a solution when c is negative.
Forgetting to check answers against the original equation, so an extraneous root is accepted.
Using a filled circle for < or > on the number line, or an open circle for ≤ or ≥.

Practise this chapter

Practice questions and timed test Printable worksheet