Section 51
(8) Sketch solutions of 4x + 8y ≥ 32 First we sketch the graph of 4x + 8y = 32, 
(10) Sketch Solutions of 6x ≥ 4y First we sketch the graph of 6x = 4y. Right 4y = 6x Now it's clear that the line has slope 3/2 and But our muchloved test point (0,0) is actually on 
The line 3x + 4y = 12 has xintercept 4 and yintercept 3. The line y=3 is a horizontal line with yintercept 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. 
Three of the corner points can be read off the graph. They are (0,0), (0,5), and (4,0). The other corner point is at the intersection of the two diagonal lines. To find it we just solve the system
6x + 3y

= 24 
3x + 6y

= 30 
Multiply the first equation by 2 and add it to the second and you will get 9x = 18, so x=2. Now plug that back into either one of the equations to get y=4. The final corner point is (2,4).