We have already studied the basics of inequality for class 11 CBSE/NCERT. In this section we will learn how to solve the question when multiple inequality equations are given.

One basic rule to solve multiple inequality equation is to find solution that are common among the given equations. Let us understand the concept with the help of example

**(01) Solve the inequality graphically**

x + y ≥ 5

x – y ≤ 3

**Solution**

Let us find answer set of equation x + y ≥ 5

The graph of x + y= 5 is drawn as follows

Let’s take point (6,0) from Area 1 to check for inequality

6 + 0 ≥ 5

6 ≥ 5

Condition is satisfied, which means that point under area 1 are the solution of inequality **x+y ≥ 5****Solution for Inequality ****x+y ≥ 5**

Let us now look at the another given equation

x – y ≤ 3

Graph of x – y ≤ 3

Taking point (4,0) from Area 1 and check whether it is the solution or not**x – y ≤ 3****4 – 0** **≤ 3****4 ≤ 3**

This statement is false, which means that area 01 is not the part of solution of equation.

Hence Area 2 is the solution and its graph is shown below**Solution for Inequality x – y ≤ 3**

Now we know the individual solution of both the given inequality.

In order to find the combined answer, we have to look for areas in graph which are common for both the inequality

On comparing the above two graph we found that the red shaded portion shown below is the common solution for both the equation, hence is the answer for our question **Combined Solution of Inequality****x – y ≤ 3****x + y ≥ 5**

**(02) Solve the following system of inequality graphically**

5x + 4y ≤ 40

x ≥ 2

y ≥ 3

**Solution**

Let’s start with first inequality 5x + 4y ≤ 40

Let’s find the solution area of inequality 5x + 4y ≤ 40

We take point (10, 0) from Area 1

5 (10) + 4(0) ≤ 40

50 ≤ 40

The condition is false which means that Area 1 is not the solution area.

So Area 2 is the solution area for inequality 5x + 4y ≤ 40**Solution for Inequality 5x + 4y ≤ 40**

Now let’s find the solution area for other inequality

x ≥ 2

y ≥ 3

Common Area for x ≥ 2, y ≥ 3 is shown in below diagram

**Solution for Inequalityx ≥ 2y ≥ 3**

Now combining all the solution area of the given inequality we get

Hence the triangle red portion is the solution of our answer

**(03)Solve the inequality using graphical representation**

2x + y ≥ 6,

3x + 4y < 12

**Solution**

Plotting the first inequality

Checking the solution area

Taking point (6,0) from area 1

2x + y ≥ 6,

2(6)+0≥ 6,

12 ≥ 6

Correct, Hence Area 1 is the solution area of inequality 2x + y ≥ 6**Solution of inequality 2x + y ≥ 6**

Now plotting other inequality

3x + 4y < 12

Now let us check for the solution area

Taking point (5,0) from the area 1

3x + 4y < 12

3(5) + 4(0) < 12

15 < 12

False, Hence Area 1 is not the solution plane

The other plane Area 2 is the solution area

**Solution of inequality 3x + 4y < 12**

Taking common solution area from both the inequality, we get the following solution

The red shaded region is the solution of given inequality