aČ + 4 < 4a + 9
Add -4a - 9 to both sides:
aČ - 4a - 5 < 0
Factor the left side:
(a + 1)(a - 5) < 0
The zeros of the left side are -1 and 5, and
thus are the critical numbers:
We plot them on a number line:
-------------o-----------------------o------------
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
We test a value of "a" in the region to the left of -1, say -2,
by substituting it for "a" in
(a + 1)(a - 5) < 0
(-2 + 1)(-2 - 5) < 0
(-1)(-7) < 0
7 < 0
This is false, so we cannot shade the region to the left of -1
Next we test a value of "a" in the region between -1 and 5, say 0,
by substituting it for "a" in
(a + 1)(a - 5) < 0
(0 + 1)(0 - 5) < 0
(1)(-5) < 0
-5 < 0
This is true, so we shade the region between -1 and 5:
-------------o=======================o------------
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
Next we test a value of "a" in the region to the right of 5, say 6,
by substituting it for "a" in
(a + 1)(a - 5) < 0
(6 + 1)(6 - 5) < 0
(7)(1) < 0
7 < 0
This is false, so we cannot shade the region to the right of 5:
So the graph of the solution is
-------------o=======================o------------
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
In set builder notation the solution set is {a | -1 < a < 5}
In interval notation the solution set is (-1, 5)
The correct choice is e)
Edwin
