help solving linear inequalities symbolically?

Question

equation:
(2/3) (1-2x) - (3/2x) + (5/6x) "greater than or equal to" (2x-1) / (3) + 1

would i have to distribute the 2/3 first and mulitply by 3 then solve?
would the interval notation be (-infinity, -1/3] ?
thanks for the help.

Answer 1

2/3 (1 - 2x) - (3/2 x) + (5/6 x) >= (2x - 1)/3 + 1
Indeed, distribute & multiply (I would say, by 6)
4 - 8x - 9x + 5x >= 4x - 2 + 6
4 - 12x >= 4x + 4
0 >= 16x
x <= 0

interval: (-oo, 0]

(The previous answer is almost correct; the only problem is that the + 1 at the right hand side should be multiplied with 6, just like the rest.)

Answer 2

My way is different from the others,and i think easier.

(2/3) (1-2x) - (3/2x) + (5/6x) = 2/3 - 4/3x - 3/2x + 5/6x
(2x-1) / (3) + 1 = (2x+2)/3

2/3 - 4/3x - 3/2x + 5/6x ≥ (2x+2) /3
( 4 - 8x - 9x + 5x) / 6 ≥ (2x+2 ) /3
(4 -12 x) / 6 ≥ (2x+2 ) /3
3 * (4 -12 x) 6 * (2x +2)
12 - 36x ≥ 12x -12
-36x - 12x ≥ 12 - 12
- 48x ≥ 0
multiply this inequalities by ' -1 '
48x ≤ 0
x ≤ 0/48
x ≤ 0
so duchprof is right as well.Good Luck.i hope i could help u.

Answer 3

2/3-(4/3)x-(3/2)x+(5/6)x >= (2/3)x-1/3+1
2/3 +((-8-9+5)/6)x >= (2/3)x+2/3
-(12/6)x >= (2/3)x
(-6/3)x >= (2/3)x
(-8/3)x>=0

So the answer is (-infinity,0]
You can chech that 0 is a solution too

Answer 4

taking LCD on both sides
2*2(1-2x)-3*3x+5x>=2(2x-1)+2
4-8x-9x+5x>=4x
-12x>=-4
12x<=4
x<=1/3
x=(-infinity, 1/3]