Algebraic Solution of Linear Inequalities in One Variable

We know different types of linear inequalities. Let’s learn how to solve the inequalities.

For example, consider the linear inequality given below,

3x < 6, where x is a whole number

LHS of the inequality is 3x and RHS of the inequality is 6

When x = 0,

LHS = 3 × 0 = 0, RHS = 6

Since 0 is less than 6, x = 0 satisfies the inequality.

When x = 1,

LHS = 3 × 1 = 3, RHS = 6

Since 3 is less than 6, x = 1 satisfies the inequality.

When x = 2,

LHS = 3 × 2 = 6, RHS = 6

Since 6 is not less than 6, x = 2 will not satisfy the inequality.

It is understood that, x should be less than 2 to satisfy the inequality.

“Values of the variable which make the statement true are called solutions of the inequality”.

Therefore, solutions of the above inequality are 0 and 1 and {0, 1} is called the solution set.

In the above example we have got the solutions by trial and error method which is not always effective. It becomes time consuming when the numbers are big and there are many conditions.There are two rules to be followed while solving inequalities. That will be explained with the following example.

Rule 1: Equal numbers can be added to or subtracted from both the sides of the inequality without changing the sign of inequality.

Rule 2: Same positive number can be multiplied or divided on the both sides of the inequality. Sign of inequality is reversed if both sides are multiplied or divided by a negative number.

Example: 3 – 2x < 1 , solve for x if x is an integer less than 6.

3 – 2x < 1

Subtract 3 from both the sides,

3 – 3 – 2x < 1 – 3

-2x < -2

Dividing both sides of the inequality by -2,

(Note that -2 is a negative number, sign of the inequality has to be reversed)

\(~~~~~~~~~~~~~~~~~~~~\)

\(~~~~~~~~~~~~~~~~~~~~\)

Since x is an integer less than 6, solution set of x is {2, 3, 4, 5}

Let’s take another example and represent the solution of inequality on a number line.

Example: Solve 2x + 3 ≤ 9 , represent ‘x’ on a number line if x is positive.

\(~~~~~~~~~~~~~~~~~~~~~~\)

Subtract 3 from both sides,

2x + 3 – 3 ≤ 9 – 3

2x ≤ 6

Dividing both sides by 2 gives,

\( \frac {2x}{2} ≤ \frac {6}{2} \)

x ≤ 3

It is given that x is positive.

Therefore x is greater than zero and less than or equal to 3. i.e.,x ∈(0, 3]

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