Module 4: Lesson 8
 
Self-Check 3

 

Questions 5.a., 5.c., 6.b., and 6.d. on page 485 of the textbook

  1.  


    1.  
      This is a number line with lowest value -6 and highest value 3. The following intervals are marked: x < -6, -6 < x < 3, and x > 3.

       
        x < −6 −6 −6 < x < 3 3 x > 3
      (x + 6) − 0 + + +
      (x – 3) − − − 0 +
      (x + 6)(x – 3) + 0 − 0 +

      The solution is −6 ≤ x ≤ 3.



    1.  
      This is a number line with lowest value 0 and highest value 6. The following intervals are marked: x < - three over four, - three over four < x < 6, and x > 6.
       
        6 x > 6
      (4x – 3) − 0 + + +
      (x – 6) − − − 0 +
      (4x – 3)(x – 6) + 0 − 0 +

      The solution is .
  1.  


    1. Case 1: positive × positive > 0

      Case 2: negative × negative > 0

      Case 1:

      (x + 12) > 0 ∴ x > −12
      (x + 1) > 0 ∴ x > −1

      The two red arrows meet up at x > −1, which is a solution.

       
      This is a number line with lowest value -12 and highest value 0. Open circles are shown at -12 and -1. A red line extends from -12 to the right and another red line extends from -1 to the right.

      Case 2:

      (x + 12) < 0 ∴ x < −12
      (x + 1) < 0 ∴ x < −1

      The two red arrows meet up at x < −12, which is a solution.

       
      This is a number line with lowest value -12 and highest value 0. Open circles are shown at -12 and -1. A red line extends from -12 to the left and another red line extends from -1 to the left.

      Therefore, the solution is x < −12 or x > −1.



    1. Case 1: positive × positive ≥ 0
      Case 2: negative × negative ≥ 0

      Case 1:

      (2x − 1) ≥ 0 ∴

      (x + 8) ≥ 0 ∴ x ≥ −8

      The red arrows meet up at , which is a solution.

       
      This is a number line with lowest value -8 and highest value 1. Solid circles are shown at -8 and one over two. A red line extends from -8 to the right and another red line extends from one over two to the right.

      Case 2:

      (2x − 1) ≤ 0 ∴
      (x + 8) ≤ 0 ∴ x ≤ −8

      The red arrows meet up at x ≤ −8, which is a solution.

       
      This is a number line with lowest value -8 and highest value 1. Solid circles are shown at -8 and 1 over 2. A red line extends from -8 to the left and another red line extends from 1 over 2 to the left.

      Therefore, the solution is .