Question

Question 3 Solve the given inequality graphically in two-dimensional plane: 3x + 4y ≤ 12

Class X1 - Maths -Linear Inequalities Page 127

Answer 1

steps for solving,
3x + 4y ≤ 12
step1:- consider the inequation as strict equation . e.g 3x + 4y = 12

step2:- find the points on the co-ordinate axes.
e.g when x = 0, y = 3
when y = 0, x = 4

step3:- plot the graph using above data's.

step4:- take a point (0,0) .and put it in inequation .we get,
3×0+4×0≤ 12 => 0 ≤ 12 , which is true,so the shaded region will be towards the origin .
see attachment, shaded region shows inequality 3x + 4y ≤ 12

Answer 2

$\mathtt{\huge{ \fbox{SOLUTION :}}}$

Liner equation of given inequation is

3x + 4y = 12

When ,

x = 0 , y = 3

x = 4 , y = 0

Thus , the graph of 3x + 4y = 12 is given in the figure

Let a point (0,0) in first part and putting x = 0 and y = 0 in the given inequality , we can see that

3(0) + 4(0) ≤ 12

0 ≤ 12 which is true

Thus , the shaded region below the line 3x + 4y = 12 is the solution of given inequation including every point on the line