# Solve the system of inequalities graphically $x +y \leq 5;x+y \geq 4$

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 $x+y \leq 6$---------(1)
Consider the equation $x+y =6$
The points (6,0) and (0,6) satisfy the equation.
The graph of the line $x+y =6$
It is drawn as shown which divides the XY palne into half plane . Consider the point (0,0)
We see that
$0+0 \leq 6$ is true.
The inequality $x+y \leq 6$ represents the half plane below the line $x+y =6$ Containing the point (0,0) (including the line)
Step 2:
The Second inequality is $x +y \leq 4$-----(2)
Consider the equation $x+y =4$
The points $(0,4)$ and $(4,0)$ satisfy the equation .
The graph of the line is drawn as shown in the diagram. It divides the XY- plane into two half planes.
Consider the point (0,0)
We see that $0+0 \geq 4$ is false
Thus the inequality $x+y \geq 4$ represents the half plane above the line not containing the point (0,0) (including the line $x+y =4)$
Step 3:
Hence the solution of the given system of linear inequalities is represented by the common shaded region, including the points on the lines.