Question

Solve the following system of inequalities graphically: x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1

Answer 1

Given : x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1  

To find : Solve the following system of inequalities graphically

Solution :

x - 2y ≤ 3

3x + 4y ≥ 12

x ≥ 0  , y ≥1

First of all we will solve the equations   x - 2y = 3  , 3x + 4y =  12   and draw in x - y plane

x - 2y = 3

x       y

0     -3/2

0     3

0 - 0 < 3

origin lies with in  x - 2y  ≤ 3    

Hence Right sides of area  denotes  x - 2y ≤ 3      

3x + 4y = 12

x       y

0     3

4      0

0 + 0 < 12

origin does not lie in 3x + 4y ≥ 12  

Area right sides of   3x + 4y ≥ 12    निरूपित करता है  

x ≥ 0  , y ≥1  

Shaded area is solution

Check attached graph

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