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
1/4 x>-8
1/4 x>-8
|x-2|≤8
1≤ ((2+3x)/2)<7
-x^2+3ax-5a+1<0
Thank you
1 answer