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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the given inequality graphically in two- dimensional plane $3x+4y \leq 12$

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

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Toolbox:
  • 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
 
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