Question #22415

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

Expert's answer

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====

-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====

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