Question 741884
Very often you do not have to solve for x or y.
Your boundary lines will be straight lines, and to graph each line, you will need at most 2 points. Finding and plotting those 2 points may be easier than you think.
 
The book uses
{{{x >= 0}}} because the number for x cannot be negative
{{{y >= 0}}} because the number for y cannot be negative
Those are two of the constraint.
{{{1.2x + 2y}}} is the surface area that the plants will cover, in square feet, which could be up to 600 square feet. That gives you the inequality
{{{1.2x + 2y <= 600}}}
The total cost, in $, will be {{{2.50x +2.50y}}} and must not be more than $1000, and that gives you the inequality
{{{2.50x + 2.50y <= 1000}}}
 
THE BOUNDARIES OF THE FEASIBLE REGION:
{{{x=0}}} and {{{y=0}}} are the y an x axes respectively (no calculations needed to graph those lines), and the space where {{{x>=0}}} and {{{y>=0}}} is the first quadrant including the pieces of the axes that border the quadrant.
The boundary line {{{1.2x+2y=600}}} is a straight line that passes through A(0,300) and P(500,0)
because {{{x=0}}} means {{{1.2*0+2y=600}}} --> {{{2y=600}}} --> {{{y=300}}} and
{{{y=0}}} means {{{1.2x+2*0=600}}} --> {{{1.2x=600}}} --> {{{x=500}}}
Another boundary is {{{2.50x+2.50y=1000}}} --> {{{x+y=400}}} (dividing everything by 2.50). 
Making {{{x=0}}} first and then {{{y=0}}} (as for the other line), you can easily find that {{{x+y=400}}} passes through Q(0,400) and C(400,0).
 
The graph of the boundaries looks like this:
{{{drawing(400,400,-50,620,-50,620,
grid(0),
red(line(400,0,0,400)),
red(line(0,300,500,0)),
locate(5,25,O),locate(405,25,C),
locate(5,325,A),locate(255,170,B),
locate(505,25,P),locate(5,425,Q)
)}}} The feasible region is the quadrilateral ABCO.
How do we know that the feasible region is the part below both slanted lines?
We know that
{{{1.2x+2y=600}}} (line AP), and
{{{x+y=400}}} (line QC) are boundary lines.
For {{{x+y<=400}}} the boundary line is {{{x+y=400}}}.
The points to one side of that line have {{{x+y<400}}}, and the points to the other side have {{{x+y>400}}}.
What side satisfies {{{x+y<400}}}?
We can check with a test point that is not on the boundary line.
An easy choice for a test point is O(0,0). Is it part of the solution to {{{x+y<=400}}}?
When {{{x=0}}} and {{{y=0}}}, {{{x+y=0<400}}}, so O(0,0) satisfies {{{x+y<400}}} and so do all the points on the same side of QC.
Similarly, you can find that the side of {{{1.2x+2y=600}}} that is the solution you {{{1.2x+2y<=600}}} is the side that contains O(0,0).
The only hard-work calculation you could have to do is finding the coordinates of point B, if you need to, but you may not be expected to do that for this problem.
If you had to, the coordinates of point B are the solution to
{{{system(1.2x+2y=600,x+y=400)}}}.
 
NOTE:
You may think that you have to solve for y to get
{{{x+y<=400}}} --> {{{y<=400-x}}} to figure out that the points that satisfy
{{{x+y<=400}}} are those below the line {{{x+y=400}}} <--> {{{y=400-x}}} but you do not need to do that much work (unless your teacher wants you to do it that way).