Question 548268
I will first look at the equations (I rewrote (3) a bit... if you don't see what I did let me know):
{{{y >= 2x - 8}}}          (1)
{{{y <= 10 - x}}}          (2)
{{{y <= (1/2)*x + 4}}}     (3)

(2)-(3) leaves:
{{{0<=6-(3/2)*x}}}
{{{(3/2)*x<=6}}}
{{{3x<=12}}}
{{{x<=4}}}


Equation (1) can be rewritten as 
{{{-y<=-2x+8}}}           (4)

Equation (4) +equation (2) gives
{{{0<=18-3x}}}
{{{3x<=18}}}
{{{x<=6}}}   (Note: we already know x<=4... so this is no surprise...)

Equation (4)+ Equation(3) gives
{{{0<=(-3/2)*x+12}}}
{{{(3/2)*x<=12}}}
{{{3x<=24}}}
{{{x<=8}}} (Again... no surprise... but it's good to check as we could have gotten a different value)

So we know now that {{{1<=x<=4}}}.(So the maximum value of x is 4)   


We have to do the same thing for y by eliminating x
Just a reminder:
{{{y >= 2x - 8}}}          (1)  which is really just {{{-y<=-2x+8}}}           (4)
{{{y <= -x+10}}}           (2)
{{{y <= (1/2)*x + 4}}}     (3)

Let's first compare (4) and (2) 
We need to eliminate x so I will multiply equation (2) by 2 to get:
{{{2y<=-2x+20}}}       (5)
So adding (4) and (5) we get 
{{{y<= 28}}}

Now let's compare (1) and (3) 
We will first multiply equation (3) by -4 to get ((WHEN MULTIPLYING OR DIVIDING BY A NEGATIVE NUMBER, YOU MUST CHANGE INEQUALITY SIGN))
{{{-4y>=-2x-16}}}        (6) 
Now adding (6) to (1) we get
{{{-3y>=-24}}}
{{{y<=8}}}   

One more comparison...  (2) and (3) 
We will multiply (3) by -2 (and switch the sign) to get:
{{{-2y>=-x-8}}}        (7)
Now we want to compare (7) with (2) but first we must rearrange (2) so the inequality signs match 
{{{y <= -x+10}}}           (2)
{{{-y>=x-10}}}            (8)
And finally adding (7) and (8) we get
{{{-3y>=-18}}}
{{{y<=6}}}     (This is a surprise!)    
So if all these must be true we have {{{2<=y<=6}}} and the maximum bound on y is 6.  


To maximize p we must sub in both our maximum x and maximum y into the formula p=7x-4y... I will let you do this part:) 


Phew!    Hopefully this helps!