Linear Inequalities

Linear Inequalities in One Variable

Linear Inequations: Mathematical expressions help us convert problem statements into entities and thus, help solve them. If the expression equates two expressions or values, then it is called an equation. For e.g. 3x + 5y = 8. On the other hand, if an expression relates two expressions or values with a ‘<’ (less than) sign, ‘>’ (greater than) sign, ‘≤’ (less than or equal) sign or ‘≥’ (greater than or equal) sign, then it is called as an Inequality.

An inequality which involves a linear function is a linear inequality. It looks like a linear equation, except that the ‘=’ sign is replaced by an inequality sign, called linear inequations. In this article, we will look at linear inequalities with one or two variables.

Suggested Videos

Play
Play
Play
previous arrow
next arrow
previous arrownext arrow
Slider

 

Understanding Linear Inequations

Example 1

Let’s say that your mother sends you to a shop to buy rice. She gives you Rs 200 and instructs you to buy the maximum quantity possible. In the shop, rice is available at Rs 30 per kg and in packets of 1 kg each. Let’s convert this statement into an expression:

If ‘x’ is the number of packets purchased by you, then the total amount spent will be 30x (since one packet of rice weighing 1 kg costs Rs 30). Now, you have to buy the maximum amount of rice within Rs 200. Since 200 is not divisible by 30, you will not be able to spend the entire amount. Hence, the expression will be:

30x < 200 … (1)

Note, that this expression does not have an equal sign and hence is not an equation. It is an inequality.

Example 2

You go to a shop to buy some chocolates and biscuits. You want to spend up to Rs 120. One chocolate costs Rs 40 and one packet of biscuits costs Rs 20. Let’s convert this to an expression too:

If ‘x’ is the number of chocolates and ‘y’ is the number of biscuits purchased by you, then the total amount spent is (40x + 20y). Notice that 120 is divisible by 60 (40+20). Also, the statement says that you want to spend ‘up to Rs 120’. Hence,

40x +20y ≤ 120 … (2)

Observe that this expression has two parts,

  • 40x + 20y < 120 … (3) … not an equation. It is an inequality.
  • 40x + 20y = 120 … (4) … an equation.

Definition 1

Two real numbers or two algebraic expressions related by the symbol ‘<’, ‘>’, ‘≤’ or ‘≥’ form an inequality. Here are some examples of inequalities:

  • Numerical inequalities: 3 < 5; 7 > 5.
  • Literal inequalities or inequalities which involves variables: x>=2; y <=5.
  • Double inequalities: 3 < 5 < 7; 5 < x < 9.
  • Strict inequalities or inequalities that have either ‘<’ or ‘>’ in the equation: ax + b < 0; ax + b > 0; ax + by < c; ax + by > c.
  • Slack inequalities or inequalities that have a ‘≤’ or ‘≥’ sign: ax + b ≤ 0; ax + b ≥ 0; ax + by ≤ c; ax + by ≥ c.
  • Linear inequalities in one variable – ‘x’ where a ≠0: ax + b >0; ax + b ≤ 0.
  • Linear inequalities in two variables – ‘x’ and ‘y’ where a ≠ 0 and b ≠ 0: ax + by < c; ax + by ≥ c.

Remark: ax2 + bx + c < 0 is an example of a quadratic inequality in ‘x’ where a ≠ 0.

Linear Inequations in One Variable – Algebraic Solutions and Graphical Representation

Let’s look at example 1 from above. From equation (1), we have

30x < 200

Now, since x is the number of packets purchased, it cannot be a fraction or a negative integer. In this inequality, LHS = 30x and RHS = 200. Therefore, let’s calculate the value of LHS for different values of ‘x’ and try to find the solution to the inequality. We have,

  • For x = 0, L.H.S. = 30 (0) = 0. ∴ LHS < RHS (0 < 200) is true.
  • For x = 1, L.H.S. = 30 (1) = 30 ∴ LHS < RHS (30 < 200) is true.
  • For x = 2, L.H.S. = 30 (2) = 60 ∴ LHS < RHS (60 < 200) is true.
  • For x = 3, L.H.S. = 30 (3) = 90 ∴ LHS < RHS (90 < 200) is true.
  • For x = 4, L.H.S. = 30 (4) = 120 ∴ LHS < RHS (120 < 200) is true.
  • For x = 5, L.H.S. = 30 (5) = 150 ∴ LHS < RHS (150 < 200) is true.
  • For x = 6, L.H.S. = 30 (6) = 180 ∴ LHS < RHS (180 < 200) is true.
  • For x = 7, L.H.S. = 30 (7) = 210 ∴ LHS < RHS (210 < 200) is false.

Therefore, we find that the inequality is true for x = 0, 1, 2, 3, 4, 5, and 6. These values are called solutions of the inequality and the set {0, 1, 2, 3, 4, 5, 6} is called the solution set. Hence, we have:

Any solution of an inequality in one variable is a value of the variable which makes it a true statement.

While solving linear equations, we followed the following rules:

  • Rule 1: Equal numbers may be added to (or subtracted from) both sides of an equation.
  • Rule 2: Both sides of an equation may be multiplied (or divided) by the same non-zero number.

For solving an inequality, we follow similar rules with slight modifications as given below

Rule 1

Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality.

Rule 2

Both sides of an inequality can be multiplied (or divided) by the same positive number. But when both sides are multiplied or divided by a negative number, then the sign of inequality is reversed. For example, 3 > 2 while – 3 < – 2. Also, – 8 < – 7, whereas (– 8)( – 2) > (– 7)( – 2) i.e. 16 > 14.

To understand the application of the rules, let’s solve 30x < 200 in two scenarios:

  1. ‘x’ is a natural number
  2. ‘x’ is an integer.

We are given, 30 x < 200. By Rule 2, we know that we can divide both sides of an inequality by the same positive number. Hence, 30x/30 < 200/30. Or, x < 20/3

  • If ‘x’ is a natural number, then the solution set of the inequality is {0, 1, 2, 3, 4, 5, 6}
  • If ‘x’ is an integer, then the solution set is {…, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

Learn linear inequalities in two variables here in detail. 

Graphical Representation of the Solution of Linear Inequations

Example: Solve 7x + 3 < 5x + 9. Show the graph of the solutions on the number line.

Solution: We have, 7x + 3 < 5x + 9

Using Rule 1, we have,
7x – 5x < 9 – 3
⇒ 2x < 6
⇒ x < 3

This can be represented graphically, as follows:

linear inequations

More Solved Examples for You

Question 1: Solve \( \frac {x}{3} \) > \( \frac {x}{2} \) + 1 where x is a real number.

Answer. We have,

\( \frac {x}{3} \) > \( \frac {x}{2} \) + 1
⇒ \( \frac {x}{3} \) – \( \frac {x}{2} \) > 1 … using Rule 1
⇒ \( \frac {2x – 3x}{6} \) > 1
⇒ – \( \frac {x}{6} \) > 1
⇒ – x > 6
Or, x < – 6 … using Rule 2.

Therefore, the solution set of the inequality consists of all real numbers less than (- 6) and is denoted as {….., -8, -7, -6}.

Question 2: What is a linear inequality?

Answer: An inequality involving a linear function refers to a linear inequality. It resembles a linear equation, except that the inequality sign replaces the ‘=’, which we call linear inequations.

Question 3: What is the difference between linear equation and linear inequality?

Answer: The graph of linear inequalities consists of a dashed line if they are greater than or less than but not equal to. On the other hand, linear equations consist of a solid line in every condition. Furthermore, linear inequalities contain shaded regions while linear equations do not.

Question 4: How do you solve a linear inequality with absolute value?

Answer: An absolute value equation will have no solution if the absolute value expression will equal a negative number because an absolute value may never be negative. We can write an absolute value inequality as a compound inequality. This is applicable for all absolute value inequalities. You just need to replace > above with ≥ and < with ≤.

Question 5: What makes an equation linear?

Answer: A linear equation resembles any other equation. It is composed of two expressions which are set equal to each other.  This equation is special because it consists of one or two variables. No variable in a linear equation is raised to a power greater than 1 or we use it as the denominator of a fraction.

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

Browse

Linear Inequalities

Leave a Reply

Your email address will not be published. Required fields are marked *

Browse

Linear Inequalities

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.