Each inequality should be matched to a graph that represents its solutions as follows;
1. 6x ≤ 3x ↔ graph F.
2. 1/4(x) > -1/2 ↔ graph E.
3. 5x + 4 ≥ 7x ↔ graph B.
4. 8x - 2 < -4(x - 1) ↔ graph D.
5. (4x - 1)/3 > -1 ↔ graph A.
6. 12/5 - x/5 ≤ x ↔ graph C.
Part 1.
Based on the inequalities shown above, we would determine the solution set for each inequality as follows;
6x ≤ 3x
6x - 3x ≤ 0
3x ≤ 0
x ≤ 0/3
x ≤ 0
Therefore, a solid dot would be placed at point 0 on the number line with an arrow that decreases to the left as correctly depicted by graph F.
Part 2.
1/4(x) > -1/2
x > -4/2
x > -2
Therefore, a hollow dot would be placed at point -2 on the number line with an arrow that increases to the right as correctly depicted by graph E.
Part 3.
5x + 4 ≥ 7x
5x - 7x ≤ -4
-2x ≤ -4
x ≤ 4/2
x ≤ 2
Therefore, a solid dot would be placed at point 2 on the number line with an arrow that decreases to the left as correctly depicted by graph B.
Part 4.
8x - 2 < -4(x - 1)
8x - 2 < -4x + 4
8x + 4x < 4 + 2
12x < 6
x < 6/12
x < 1/2
Therefore, a hollow dot would be placed at point 1/2 on the number line with an arrow that decreases to the left as correctly depicted by graph D.
Part 5.
(4x - 1)/3 > -1
4x - 1 > -3
4x > -3 + 1
4x > -2
x > -2/4
x > -1/2
Therefore, a hollow dot would be placed at point -1/2 on the number line with an arrow that increases to the right as correctly depicted by graph A.
Part 6.
12/5 - x/5 ≤ x
12 - x ≤ 5x
5x + x ≥ 12
6x ≥ 12
x ≥ 12/6
x ≥ 2
Therefore, a solid dot would be placed at point 2 on the number line with an arrow that increases to the right as correctly depicted by graph C.