Week 2: Student Response to Discussion Two

8:44amJul 27 at 8:44am

Manage Discussion Entry

33. -12 < 12 + 4x < 0

I will be solving for an "and" compound inequality. A "and" compound inequality has to meet two conditions at the same time. Meaning 12+4x must be less than 0 and also greater than -12.

-12 < 12 + 4x < 0

-24 < 4x < -12 I subtracted -12 from three parts of this

inequality.

-6 < x < -3 Next, I divided 4 from three parts.

Values that make x greater than -6 and less than -3 render this inequality true.

33. 12 - x > 15 or 7x - 13 > 1

With this equation I'll be solving for an "or" compound inequality. Meaning, there will be two conditions that will need to be met, but of both them will not be true at the same time.

12 - x > 15

- x > 15 - 12 I subtracted 12 from both sides.

- x > 3 Then subtracted 15 from 12. Note: the

x < 3 negative variable. To remove this

negative variable, -1 must be divided

by each side and the inequality symbol

must be flipped.

7x - 13 > 1 This is the next part of my "or" compound

inequality.

7x > 1 +13 I added 13 to each side.

7x > 14

x > 2 I divided 7 to each side.

Complete set written algebraically: x < 3 or x > 2