Answer to Question #22415 in Trigonometry for Brenden Fields

Question

Solve, write your answer in interval notation and graph the solution set.

14c. |2x-1| + 7 greater than or equal to 1

14d. |5x+3| less than or equal to -1

Answer 1

14a. |5y-2| < 13
-13 < 5y -2 < 13
-13+2 < 5y < 13+2
-11 < 5y < 15
-11/5 < y < 15/5
-2.2 < y < 3

So y in (-2.2, 3)


Answer: (-2.2, 3)
-------------------o=============================o-------------------->
-2.2 3
the ends are notincluded, and the solution is shown with====





14b. |x+1|>= 5

Consider two cases:

1) x+1<0.Then |x+1|=-x-1, and so we have the inequality

-x-1 >=5
-x >=5+1=6
x <= -6
x in(-infinity, -6]


1) x+1>=0.Then |x+1|=x+1, and so we have the inequality

x+1 >= 5
x >=5-1=4
x >= 4
x in [4,+infinity)

Thus the solution has the following form:
y in (-infinity, -6] U [4, +infinity)


Answer: (-infinity, -6] U [4, +infinity)

================*---------------------*==============================>
-6 4
the ends are included, and the solution is shown with====