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# Solve the system of inequalities : $x+y \leq 9; y > x;x \geq 0$ Comment
A)
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 $x+y \leq 9$----(1)
Consider the equation $x +y =9$
The points (9,0) and (0,9) satisfy the equation.
The graph of the line is drawn as shown.
The line divides the xy -plane into two half planes .
Consider the point (0,0) We see that
$0+0 \leq 9$
$0 \leq 9$ is true.
Thus the inequality $x+y \leq 9$ represents the region below the line $x+y =9$ containing the point (0,0) including the line .
Step 2:
The second inequality $y > x$ ----(2)
consider the equation $y=x$
The points $(1,1) ,(0,0)$ satisfy the equation.
The graph of the line $y=x$ is drawn using dotted line as shown .
The graph divides the xy plane into two half planes.
consider the point $(1,0)$ we see that
$0 > 1$ is false.
Thus the region represented by the inequality $y >x$ is above the line not containing point (1,0) (excluding the line y=x)
Step 3:
The third inequality $x \geq 0$
The inequality represents the region right of the y-axis (including y axis )
Step 4:
The solution of the system of inequality is represented by the common shaded region including the points on the line $x+y =9$ and $x=0$ and excluding the points on the line $y=x$.