(8) Sketch solutions of 4x + 8y ≥ 32

First we sketch the graph of 4x + 8y = 32,
which has an x intercept of 8 and a y intercept
of 4. Now take (0,0) as a test point. Plugging
this into the inequality gives 0x + 0y ≥ 32,
or 0 ≥ 32 which is FALSE. Thus we shade
in the side of the line not containing (0,0).

(10) Sketch Solutions of 6x ≥ 4y
(which is the same as 6x - 4y ≥ 0 ).

First we sketch the graph of 6x = 4y. Right
away we see that the x and y intercepts are both
0, and that's not going to help us with the graph
(we need TWO points to draw a line through).
Therefore, we rewrite 6x = 4y as

4y = 6x
y = 6/4 x
y= 3/2 x + 0 (form y=mx+b)

Now it's clear that the line has slope 3/2 and
y-intercept 0, and that's easy to sketch.

But our much-loved test point (0,0) is actually on
the line this time, so we have to use another one,
say (1,0). Plugging (x,y)=(1,0) into 6x ≥ 4y
gives a true statement, so we shade in the side of
the line containing (1,0).

The line 3x + 4y = 12 has x-intercept 4 and y-intercept 3. The line y=-3 is a horizontal line with y-intercept -3. These lines are sketched on the right.

The test point (0,0) makes both inequalities true, so we shade in that side of the two lines.

By just looking at the graph we see a corner point of (8,-3).

Line 6x + 3y = 24 has x and y intercepts 4 and 8, respectively. It's sketched on the right.

Line 3x + 6y = 30 has x and y intercepts 10 and 5, respectively. It's also sketched.

Lines x=0 and y=0 are just the y and x axes.

Test point (1,1) satisfies all the inequalities so we shade in the region which contains that point.