# Solve the following system of inequalities graphically $2x +y \geq 6; 3x +4y \leq 12$

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 $2x +y \geq 6$ -----(1)
Consider the equation $2x+ y =6$
The point $(3,0)$ and $(0,6)$ satisfy the equation.
The graph of this is given below, Consider the point (0,0)
we see that, $2(0) +0 \geq 6$
$0 \geq 6$ is false
The inequality $2x+y \geq 6$ represents the half plane not containing (0,0) .
It represents the region above the line $2x+y=6$ including the point on the line.
Step 2:
The second inequality is $3x+4y \leq 12$
the points (4,0) and (0,3) satisfy the equation.
The graph of the line $3x+4y =12$ is given below,
Consider the point (0,0)
we see that, $3(0)+4(0) \leq 12$
$0 \leq 12$ is true.
The inequality $3x+4y \leq 12$ represents the region below the line $3x+4y =12$ (including the points on the line) Containing the point (0,0)
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.