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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the system of inequalities graphically $x+2y \leq 10 ; x+y \geq 1 ; x -y \leq 0 ; x \leq 0 , y\leq 0 $

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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.
  • 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 first inequality is $ x+2y \leq 10$---------(1)
Consider the equation $x+2y =10$
The points $(10,0) $ and $ (0,5)$ satisfy the equation .
The graph of the equation is drawn as shown .
The line divides the xy plane into two half planes.
Consider the point$(0,0)$
We see that, $ 0+ 2(0) \leq 10$
$=> 0 \leq 10$ is true.
Hence the inequality (1) represents the region below the line $ x+2y =10$ containing the point (0,0) (including the line )
Step 2:
The second inequality is $x+y \leq 1$ -----(2)
Consider the equation $x+y =1$
We see that $(1,0)$ and $(0,1)$ satisfy the equation .
The graph of the equation is drawn as shown.
The line divides the XY plane into two regions ( half plane)
Consider the point (0,0)
We see that $ 0+0 \geq 1$
$ => 0 \leq 1$ is false
Thus the inequality (2) is represented by the region above the line $x+y=1$ not including the point (0,0) (including the line y=2x)
Step 3 :
The third inequality is $x-y \leq 0$ -----(3)
Consider the equation $x-y =0$
We see that $(0,0) ,(1,1) $ Satisfy equation.
The graph of the equation is drawn as shown.
The line divides the XY plane into two regions ( half plane)
Consider the point (1,0)
We see that $1-0 \leq 0$
$ 1 \leq 0$ is false
Thus the inequality (3) is represented by the region above the line $x-y=0$ not including the point (0,0) (not including the line )
Step 4:
The fourth inequality is $x \geq 0, y \geq 0$
It represents the first quadrant
Step 5:
Hence the solution of the system of linear inequalities is represented by the common shaded region in the first quadrant including the line.
answered Aug 1, 2014 by meena.p
 
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