41 and 46 2 x 4 3 x 9 x3 25 2 x2 13 1 u 2 12 4 2x 3

# 41 and 46

|-2| x = 4 |3| x =9 |x/3 + 2/5| = 2 |x/2 - 1/3| = 1 |u - 2| = -1/2 4 - |2x| =3 5 -|1/2 x| = 3 |x^2 - 9| = 0 |x^2 - 2x| = 3 |x^2 + x| = 12 |x^2 + x 1| = 1 |3x - 2/2x - 3| = 2 |2x + 1/3x + 4| = 1 |x^2 + 3x| = |x^2 - 2| solve each inequality. Express your answer using set notation or interval |2x| |-3| |-x - 2| greaterthanorequalto 1 -|2x - 1| |2x|

Solution

41. I x -2 I + 2 < 3 so that I x - 2 I < 3 -2 or,  I x - 2 I < 1 Thus, - 1 < x - 2 < 1 or, - 1 + 2 < x < 1 + 2 or,1< x< 3.Thus x ( 1, 3)

46. I 3x + 4 I 2.Then either (3x + 4) -2 or, (3x + 4) 2 . Now, if (3x + 4) - 2 , then 3x -2 -4 or, 3x -6 or, x -2. Further, if (3x + 4) 2, then 3x 2 -4 or 3x -2 or, x -2/3 . Thus x ( - , -2] U [ -2/3, )