Consider the first inequality $3x+4y \leq 60$ ---------(1)
Let corresponding the equation see $3x +4y =60$
The points $(20,0)$ and $(0,15)$ satisfy the equation.
The graph of the equation is drawn as shown .
The line divides the xy- plane into two half planes.
Consider the points $(0,0)$
We see that $ 3(0) +4(0) \leq 60$
$0 < 60$ is true.
Thus the inequality (1) represents region below the line containing the point (0,0) (including the line 3x+4y=60)
The second inequality is $x +3y \leq 30$
Consider the equation $x+3y =30$
The points (30,0) and (0,15) satisfy the equation .
The graph of this line is drawn as shown .The line divides the x-y into two half - planes.
Consider point (0,0). We see that , Thus the inequality represent the region below the line $x+3y =30$ containing the point (0,0) (including the line x+3y=30)
The third inequality is $x \geq 0$
It represent the region to the right of the y axis including y axis.
The fourth inequality is $y \leq 0$
It represented the region above the x-axis including the x axis
Step 5 :
The given system of linear inequalities represents the common shaded region including all the lines.