Please find the attached files for a better understanding of the solution presented here.
The system is made up of the following two inequalities.
[tex] y\geq-3x+3 [/tex]..........................................(Inequality 1)
and [tex] y<\frac{3}{2}x-6 [/tex]...........................(Inequality 2)
PART A
The graphs of the given two inequalities are attached in the first and the second graph respectively.
As we can see, graph 1 represents all those values of y which are greater than or atleast equal to [tex] -3x+3 [/tex]. Likewise, graph 2 represents the second inequality.
Now, when we superimpose the two graphs on each other we will get the third graph which has been attached too.
As we can clearly see in the third graph, the common area is the one which has been enclosed by the yellow straight lines and this common area represents the common solution for both the inequalities which represent this system. This area is the solution area of the system.
PART B
In graph 4, we can see that the point (-6,3) falls outside the common solution area. Thus the point (−6, 3) is not included in the solution area for the system.
This can be justified mathematically.
Let us take the first inequality [tex] y\geq-3x+3 [/tex]
If we plug in [tex] x=-6 [/tex], we see that for the inequality to be satisfied, [tex] y\geq -3(-6)+3\geq 18+3\geq 21 [/tex]. But we can clearly see that this is not the case as the given point [tex] (-6,3) [/tex] has 3 as it's y coordinate and we know that [tex] 3\ngeqslant 21 [/tex].
Likewise, let us take the second inequality, [tex] y<\frac{3}{2}x-6 [/tex]. This gives that as per the second inequality we should have:
[tex] y<\frac{3}{2}(-6)-6=3(-3)-6=-9-6=-15 [/tex] but as we know, [tex] y\nleqslant -15 [/tex] as the y in the point [tex] (-6,3) [/tex] is 3 and again, [tex] 3\nleqslant -15 [/tex].
Thus, we have mathematically proven that the given point (−6, 3) is not included in the solution area for the system.
