Not all equations in maths denote equality on both sides. In cases where there is a shared relation between the two sides, namely the right-hand side and the left-hand side, the relations are said to be in linear inequation. Moreover, the equal sign in a linear equation is replaced with  ‘<’, ‘>’, ‘<=’, ‘>=’ in such linear inequations. However, simplification of these expressions is also done in a similar manner. 

Solving linear inequalities in one variable is simple. The process is similar to that of linear equations. The only additional rule is that when you multiply or divide both sides with a negative integer or number, the sign of inequality reverses. Moreover, you may also be asked to plot such a solution on a number line.

 

For instance, consider the equation 2x + 2 < 12. For this equation, we need to subtract 2 from both sides first to arrive at 2x < 10. Further, if we divide both sides with +2, we get x < 5. This means that x has assumed a value less than that of 5. If you are asked to express the solution in terms of real numbers, your answer will be -infinity to less than 5. On the graph, a slight bulge drawn next to the 5 and marking the range leftwards is the correct graphical representation.

To graphically represent this solution would mean to plot and mark the intersection of the solution of the two equations on a 2D cartesian plane.  This is a little more complicated than linear equations in one variable. One linear inequation in 2 variables can give you one straight line equation. You can use test values of the two variables and check where the values are being satisfied after plotting the straight lines on a graph sheet. Multiple inequalities may exist simultaneously. And, you will also have to assume equality for the inequation before arriving at the final simplified solution set. 

If you have two equations such that upon solving them, the variables have a value x > 6 and y < 11. And you are asked for all real numbers within that range; you can draw a straight line just after 6 on the x-axis and another straight line from just before 11 on the y-axis. The resultant area formed by the intersection of the lines will give you the required solution.

1. Solve for ‘a’ when 3a + 8 >2 for:

(i) a is an integer.

(ii) a is a real number

 

Solution 1:

 

Given 3a + 8 > 2

First, we subtract 8 from both sides we get,

3a + 8 – 8 > 2 – 8

Further simplifying, the above inequality becomes

3a > – 6

Next we divide both the sides by 3 we get,

3a/3 > -6/3

We arrive at the solution:

a > -2

i) Thus if a is an integer then the solution set is {-1, 0, 1, 2,..,.∞}

ii) Thus if a is a real number then the solution set is a ∈ (-2, ∞)

 

2. Given equation has a real number ‘p’, Solve for p : 4p + 3 < 5p + 7

 

Solution 2:

Given that 4p + 3 < 5p + 7

First, we subtract 7 from both the sides, and we get:

4p + 3 – 7 < 5p + 7 – 7

Further, simplifying, the inequality it becomes:

4p – 4 < 5p

Next, we subtract 4p from both the sides and we get:

4p – 4 – 4p < 5p – 4p

We arrive at the solution

p > – 4

Thus, the required solutions of the given inequality are all the real numbers lying in the range greater than -4 or (-4, ∞).

 

3. Solve for z when z is a real number: 3(z – 1) ≤ 2 (z – 3)

 

Solution 3:

Given that 3(z – 1) ≤ 2 (z – 3)

Using BODMAS, we first open the parentheses to arrive at:

3z – 3 ≤ 2z – 6

Now we add 3 to both the sides, and we get:

3z – 3+ 3 ≤ 2z – 6+ 3

Which upon further simplifying, gives:

3z ≤ 2z – 3

Next, we subtract 2z from both the sides to get:

3z – 2z ≤ 2z – 3 – 2z

Therefore,

z ≤ -3

Thus, the required solutions of the given inequality are all the real numbers lying in the range less than or equal to -3 or  (-∞, -3]

 

4. Represent the following linear inequation graphically: x+2y >9

 

Solution 4:

Given that x + 2y > 9

We first need to replace the inequality sign with an equal sign and plot the graph according to the formula for a straight line. 

 

Next, we take any point above the straight line we have just drawn. For instance, if we take (10, 10) or x = 10 and y =10 and check if it satisfies the given equation. And then, we take a point below the straight line and repeat the same steps. 

 

The region where the equation tests true is our required graphical solution.

In case of Slack inequality (≥ or ≤), use a solid line, as the points on the straight line are also included in the final set and in case of Strict inequality (> or <), use a dotted line, as the points on the straight line are not included in the solution set.

In case of multiple linear inequalities, the common region shared by all the inequalities is the solution region.

What are the important topics covered in the NCERT Solutions for Class 11 Maths Chapter 6?

 

The NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities covers all the subtopics under linear inequalities. These subtopics include basic inequalities operations, linear inequalities in two variables and graphical representation of linear inequations. 

 

How to plot a linear inequality on a graph?

 

You can graphically represent a linear inequality once you have found the value of the variables. Then you just need to draw a number line and plot the range as per the given instructions. For integer values, drawing dots on the integers within the range is good enough.

 

Is Class 11 Maths Chapter 6 hard?

 

No, Class 11 Maths Chapter 6 Linear Inequalities that contains the basic concepts and problems on linear inequalities, is not tough. It is one of the easiest chapters to attempt during your NCERT exams. The chapter is not lengthy and has only a few basic concepts.