find the algebraic solution for: 6>3(4+5d)>-3?

2 > 4 + 5d > - 1

- 2 > 5d > - 5

d > - 1 , - 2/5 > d

{ d : - 1 < d < - 2/5 , d Є R }

1st set:
6 > 3(4 + 5d)
2 > 4 + 5d
- 2 > 5d
- 2/5 > d

2nd set:
3(4 + 5d) > - 3
4 + 5d > - 1
5d > - 5
d > - 1

Answer: - 2/5 > d > - 1

6>12+15d>-3
-6>15d>-15

-2/5>d>-1

6 > 3(4 + 5d) > -3
6/3 > 4 + 5d > -3/3
2 - 4 > 5d > -1 - 4
-2 > 5d > -5
-2/5 > d > -5/5
-2/5 (-0.4) < d < -1

6 > 3(4 + 5d) > -3

Divide everything by 3:
6/3 > 3(4 + 5d)/3 > -3/3

2 > 4 + 5d > -1

Subtract 4 from all 3 parts:
2 - 4 > 4 - 4 + 5d > -1 - 4

-2 > 5d > -5

Divide everything by 5
-2/5 > 5d/5 > -5/5

-2/5 > d > -1

But, this doesn't look right, so reverse it (so the smallest value is on the left):
-1 < d < -2/5

Check your answer:
Pick a value < -1, between -1 and -2/5, and > -2/5:
For your answer to be correct, the values you choose for < -1 and > -2/5 should result in a false answer and the value between -1 and -2/5 should result in a true answer.

d = -2
6 > 3(4 + 5(-2)) > 3
6 > 3(4 - 10) > 3
6 > 3(-6) > 3
6 > -18 > 3
FALSE

d = -1/2
6 > 3(4 + 5(-1/2)) > 3
6 > 3(4 - 5/2) > 3
6 > 3(3/2) > 3
6 > 9/2 > 3
TRUE

d = 0
6 > 3(4 + 5(0)) > 3
6 > 3(4) > 3
6 > 12 > 3
FALSE

So, the answer checks!

-1 < d < -2/5

split it up
6>3(4+5d) and -3<3(4+5d)
2>4+5d and -1<4+5d
-2>5d and -5<5d

d<(-2/5) and d>-1

this means that domain d is
(-∞,-2/5) U (-1, ∞)

(-1,-2/5)