Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
0 votes

Solve the given inequality graphically in two- dimensional plane $3x+4y \leq 12$

$\begin{array}{1 1} 1 \\ 0 \\ equal \\ not\; equal \end{array} $

Can you answer this question?

1 Answer

0 votes
  • To represent the solution of linear inequality of one or two variable in a plane if the inequality involves $'\geq'$ or $' \leq$ we draw the graph of the line as a thick line to indicate the line is included in the solution set.
  • If the inequality involves $'>'$ as $'<'$ we draw the graph of the line using is not included in the solution set.
  • To solve an inequality $ax+by > c \qquad a \neq 0, b \neq 0 ( or \;> )$
  • We consider the corresponding equation $ax+by =c$ which represents a straight line This line divides the plane into two half planes I and II
  • We take any point in I half plane and check if it satisfies the given inequality will be one half plane (called solution region ) Containing the point satisfying the inequality
Step 1:
The given inequality is $ 3x+4y \leq 12$
Consider the equation $3x +4y=12$
The point $(4,0) $ and $(0,3)$ Satisfy the equation
Step 2:
The graphical representation of the line is given below the line divides the x-y plane into two half planes I and II
Select a point not on the line (0,0)
We see that $3(0) +4(0) \leq 12$
or $0 \leq 12$ is true.
Step 3:
Therefore the half plane II is not the solution .
The half plane I containing the point (0,0) is the solution region is the shaded half plane including the points on the line.
answered Jul 29, 2014 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App