The problem asks to match each of the six given inequalities to the graph that represents its solution set.

AlgebraInequalitiesLinear InequalitiesSolving InequalitiesGraphing Inequalities
2025/3/25

1. Problem Description

The problem asks to match each of the six given inequalities to the graph that represents its solution set.

2. Solution Steps

1. $6x \le 3x$

Subtract 3x3x from both sides:
6x3x06x - 3x \le 0
3x03x \le 0
x0x \le 0
This corresponds to graph F, a closed circle at 0 and an arrow pointing to the left.

2. $\frac{1}{4} x > -\frac{1}{2}$

Multiply both sides by 4:
x>2x > -2
This corresponds to graph A, an open circle at -2 and an arrow pointing to the right.

3. $5x + 4 \ge 7x$

Subtract 5x5x from both sides:
42x4 \ge 2x
Divide by 2:
2x2 \ge x or x2x \le 2
This corresponds to graph C, a closed circle at 2 and an arrow pointing to the left.

4. $8x - 2 < -4(x - 1)$

8x2<4x+48x - 2 < -4x + 4
Add 4x4x to both sides:
12x2<412x - 2 < 4
Add 2 to both sides:
12x<612x < 6
Divide by 12:
x<612x < \frac{6}{12}
x<12x < \frac{1}{2}
This corresponds to graph D, an open circle at 1/2 and an arrow pointing to the left.

5. $\frac{4x - 1}{3} > -1$

Multiply both sides by 3:
4x1>34x - 1 > -3
Add 1 to both sides:
4x>24x > -2
Divide by 4:
x>24x > -\frac{2}{4}
x>12x > -\frac{1}{2}
This corresponds to graph B, an open circle at -1/2 and an arrow pointing to the right.

6. $\frac{12}{5} - \frac{x}{5} \le x$

Add x5\frac{x}{5} to both sides:
125x+x5\frac{12}{5} \le x + \frac{x}{5}
1256x5\frac{12}{5} \le \frac{6x}{5}
Multiply both sides by 5:
126x12 \le 6x
Divide by 6:
2x2 \le x or x2x \ge 2
This corresponds to graph C. However, graph C was already matched to inequality

3. There is an error. Re-examine inequality 3:

5x+47x    42x    2x5x+4 \ge 7x \implies 4 \ge 2x \implies 2 \ge x.
Thus, x2x \le 2. This matches graph C, a closed circle at 2, and the line extending to the left.
Re-examine inequality 6: 125x5x\frac{12}{5} - \frac{x}{5} \le x becomes 12x5x    126x    2x12 - x \le 5x \implies 12 \le 6x \implies 2 \le x or x2x \ge 2. Thus, x is greater than or equal to

2. This must be graph E, which has a solid circle at 2 and a line going to the right.

3. Final Answer

1. F

2. A

3. C

4. D

5. B

6. E

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