Browse Questions

# Solve the following system of inequalities graphically $x+y \geq 4 ; 2x -y > 0$

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
The first inequality is $x+y \leq 4$ --------(1)
Consider the equation $x+y =4$
The points $(4,0)$ and $(0,4)$ satisfy the equation.
The graph of the line $x+y =4$ is 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.
The inequality does not represented the half plane containing . (0,0) It represented the half plane above the line $x+y =4$ (including the line)
Step 2:
The second inequality is $2x-y >0$
Consider the equation $2x-y =0$
The points $(-2,4)$ and $(0,0)$ satisfy the equation $2x-y =0$
The graph of the line is shown in the diagram .
It divides the XY plane into two half planes.
Consider the point (1,0)
$2(1)-0 >0$
$2> 0$ is true.
Therefore the inequality represents the half plane to the right of the line $2x-y =0$ (excluding the line) Containing the point (1,0)
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.
$x+y =4$ and excluding the points on the line $2x-y=0$