# Hi, I need some help with some inequalitites. Thanks!!

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 don't 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 don't 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 don't 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
-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
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

## 1. The intersection between (-9,0) and [-4,∞) is [-4,0).

2. The union between (-9,0) and [-4,10] is (-9,10].

4. To solve the linear inequality 21x - 21 > 3(6x - 2):

First, distribute the 3 on the right side: 21x - 21 > 18x - 6.

Next, combine like terms: 3x > 15.

Divide both sides by 3 to isolate x: x > 5.

The solution set in interval notation is (5, ∞) and the graph would be an open circle at 5 with an arrow pointing to the right.

5. To solve the inequality 5(4x+7)-4x<4(8+4x)-6:

First, distribute on both sides: 20x + 35 - 4x < 32 + 16x - 6.

Next, combine like terms: 16x + 35 < 16x + 26.

Since the variables cancel out, there is no solution to this inequality.

6. To solve the inequality 6(x+4) ≥ 5(x-3)+x:

First, distribute on both sides: 6x + 24 ≥ 5x - 15 + x.

Next, combine like terms: 6x + 24 ≥ 6x - 15.

The variables cancel out, leaving the true statement 24 ≥ -15.

Since this is true for all x, the solution set is (-∞, ∞) and the graph would be a line that extends in both directions.

10. To solve the compound inequality -24 ≤ -5x+1 < -9:

First, isolate the variable between the inequalities: -25 ≤ -5x < -10.

Next, divide by -5, remembering to reverse the inequalities: 5 ≥ x > 2.

The solution set in interval notation is [2, 5) and the graph would be a solid line between 2 and 5.

14. To solve the compound inequality 3 ≤ (8/5x)-5 < 11:

First, add 5 to all parts of the inequality: 8 ≤ (8/5x) < 16.

Next, multiply by 5 to clear the fraction: 40 ≤ 8x < 80.

Finally, divide by 8: 5 ≤ x < 10.

The solution set in interval notation is [5, 10) and the graph would be a solid line between 5 and 10.

20. To solve the absolute value inequality |x+9| - 4 ≤ 0:

First, add 4 on both sides: |x+9| ≤ 4.

Next, split the inequality into two cases: x+9 ≤ 4 and x+9 ≥ -4.

For the first case, subtract 9 from both sides: x ≤ -5.

For the second case, subtract 9 from both sides: x ≥ -13.

The solution set in interval notation is [-13, -5] and the graph would be a closed interval between -13 and -5 inclusive.

21. To solve the absolute value inequality |(10y+30)/3| < 10:

First, split the inequality into two cases: (10y+30)/3 < 10 and -(10y+30)/3 < 10.

For the first case, multiply both sides by 3: 10y + 30 < 30.

Next, subtract 30 from both sides: 10y < 0.

Divide by 10: y < 0.

For the second case, multiply both sides by -3: -(10y+30) < -30.

Distribute the negative sign: -10y - 30 < -30.

Add 30 to both sides: -10y < 0.

Divide by -10 (remembering to flip the inequality sign): y > 0.

The solution set in interval notation is (0,∞) and the graph would be an open circle at 0 with an arrow pointing to the right.