Wednesday

October 1, 2014

October 1, 2014

Posted by **Christie** on Saturday, August 18, 2007 at 11:49pm.

Use graphs to find the set.

1.(-9,0) intersection [-4,] I got [-4,0)

2. (-9,0) U [-4,10] I got (-9,10]

Solve the linear inequality. Other than (empty set, use interval notation to express the solution set and graph the solution set on a number line.

4. 21x -21>3(6x-2) I got x<5, graph would be <------)5

5. 5(4x+7)-4x<4(8+4x)-6 I got 0>9, but I dont know how the graph would be.

6. 6(x+4)> or equal to 5(x-3)+x I got 0 > greater than or equal to -39, but dont know how the graph would be.

Solve the compound inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.

10. -24 < or equal to -5x+1 < -9 I am not too sure how to solve this.

14. 3 < or equal to (8/5x)-5<11 one this one am I suppose to multiply each side by 5 to undo the fraction.

Solve the absolute value inequality. Other than empty set, use interval notation to express the solution set and graph the solution set on a number line.

20. |x+9| -4 < or equal to I got [-5 -13] and the graph would be <----)-13 -5---->

21. |10y+30/3| < 10 This one I dont understand what to.

I am not quite familiar with your notation for #1 and #2,

but if you mean the intersection between all values between -9 and zero and the value -4, it would be -4

#4, I got x>5

21x - 21 > 18x - 6

3x>15

x>5

#5 Since the variables disappeared and the statement 0>9 is FALSE, there is no solution to your inequation.

#6 This time the statement is TRUE, so any x will do. Draw a line on your graph with arrows in both directions.

#10

-24 ≤ -5x+1 < -9

add -1 to each part

-25 ≤ -5x ≤ -10

divide by -5, remember that we switch the inequality signs after a division and multiplication of negatives

5 ≥ x ≥ 2

and finally writing this in the common notation of the smaller number on the left

2 ≤ x ≤ 5

draw a solid line between 2 and 5, including the 2 and the 5 (solid dots)

#14

3 ≤ (8/5x)-5 < 11

add 5

8 ≤ (8/5)x < 16

multiply by 5

40 ≤ 8x < 80

divide by 8

5 ≤ x < 10

draw a line from 5 to 10, including the 5 but excluding the 10

#21 I will assume you meant |(10y+30)/3| < 10 or else you would just change 30/3 to 10 and it would be easy

|(10y+30)/3| < 10

(10y+30)/3 < 10 and -(10y+30)/3 < 10

10y+30 < 30 and -10y-30 < 30

y < 0 and -10y < 60

y < 0 and y > -6

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