Answer:
(a)
if then
and if x<0 then
(b)
That is straightforward from what you showed on theorem 3.25
(c)
Following the same ideas from (a) x>a or -x > a.
Step-by-step explanation:
Remember how we define the absolute value of a number.
(a)
In general
Therefore if you have two cases, if
then
and if x<0 then
(b)
That is straightforward from what you showed on theorem 3.25
(c)
Following the same ideas from (a) x>a or -x > a.
Step 1) Graph the line y = (-1/2)x + 3. This line goes through (0,3) as its y intercept and it also goes through (2,2). You can start at (0,3) and move down one unit and to the right two units to arrive at (2,2). The slope of the boundary line is -1/2, meaning that the rise is -1 and the run is 2. A negative rise indicates we go down instead of up.
Step 2) Make the line in step 1 to be a dashed line. This is because there is no "or equal to" as part of the inequality sign. The dashed line says that points on this line are not part of the solution set.
Step 3) Shade above the dashed line. We shade above due to the "greater than" sign. Points above the dashed line are in the shaded solution set. An alternative is to test a point like (x,y) = (0,0) and you'll find y > (-1/2)x+3 turn into 0 > 3 which is a false statement; therefore, (0,0) is not in the solution set. This is expected as (0,0) is below the dashed line.
The final result is what you see in the attached image below. Points A and B help set up the dashed boundary line. Point C is the test point mentioned above which is not in the blue shaded solution region.
Note: the points shown in the diagram are optional when it comes to showing the final answer to your teacher. All you need really is the boundary line and its proper shaded region.
