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Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
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Solve the following system of inequalities graphically $3x+2y \leq 150; x +4y \leq 80 ; x \leq 15 ; y \geq 0, x \geq 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.
  • 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 $3x +2y \leq 150$------(1)
Consider the equation $3x +2y =150$
The points $(50,0) $ and $(0,75)$ 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,
$3(0) +2(0) \leq 150$
=> $ 0 \leq 150$ is true.
Thus the inequality (1) is represented by the region below the line $ 3x+2y =150$ Containing the point $(0,0) $ (including the line)
Step 2:
The second inequality is $ x+4y \leq 80$
Consider the equation $x+4y=80$
The point $(80,0) $ and $(0,20)$ satisfy the equation.
The graph of the equation is drawn as shown.
The line divides the xy plane into two half parts.
Consider the point (0,0)
We see that $)+ 4(0) \leq 80$
=> $0 \leq 8$ is true.
Thus the inequality (2) is represented by the region below the line $ x+4y =80$ Containing the point $(0,0) $ (including the line)
Step 3:
The third is $ x \leq 15$
The inequality (3) is represented by the region to the lest of the line $ x =15$
Step 4:
The fourth inequality $ x \leq 0; y \leq 0$ 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 corresponding lines.
answered Aug 1, 2014 by meena.p
 
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