1/4 x>-8

1/4 x>-8

|x-2|≤8

1≤ ((2+3x)/2)<7

-x^2+3ax-5a+1<0

Thank you

first two are the same,

(1/4)x > -8
x > -32

|x-2| ≤ 8
x-2≤8 AND -x+2≤8
x≤10 AND x > -6

- 6 ≤ x ≤ 10 , that is, x is between -6 and +10

1 ≤ (2x+3x)/2 < 7
times 2

2 ≤ 2x+3 < 14
subtract 3
-1 ≤ 3x < 11
divide by +3
-1/3 ≤ x < 11/3

-x^2+3ax-5a+1<0
x^2 - 3ax + 5a - 1 > 0
x-intercepts of the corresponding parabola:
x = (3a ± √(9a^2 - 4(1)(5a-1))/2
= (3a ± √(9a^2 - 20a + 4) )/2

so x < (3a - √(9a^2 - 20a + 4) )/2 OR x > (3a + √(9a^2 - 20a + 4) )/2