Answer:
Step-by-step explanation:
Given equations are
[tex]3x+6y\leq 12[/tex]
[tex]-x+y\geq 0[/tex]
Now the inequalities intersect at x=[tex]\frac{4}{3}[/tex]
y=[tex]\frac{4}{3}[/tex]
and we have to satisfy this also
[tex]x\geq 0[/tex]
[tex]y\geq 0[/tex]
Above all conditions is shown in graph
and the shaded area is the required area
For [tex]3x+6y\leq 12[/tex] put (0,0)
it satisfy the point i.e. (0,0) lie on that side which satisfy the inequality similarly check for
[tex]-x+y\geq 0[/tex]
put (4,0) it do not satisfy the inequality therefore it lies on the side which do not satisfy the inequality