Solve the following equations and inequalities: 1. 2x + 3 = 1x 2. 2a + 3 = 7x − 6 3. 6x − 5 < 8x 4. (3x+5)/2 >= (6x+1)/2

$$\frac{3}{-1} = \frac{-x}{-1}$$ $$x = -3$$

The student's solution is correct.

$$\boxed{x = -3}$$

:

$$2a + 3 = 7x - 6$$

$$2a + 3 + 6 = 7x$$ $$2a + 9 = 7x$$

$$\frac{2a + 9}{7} = x$$ $$x = \frac{2a + 9}{7}$$ $$\boxed{x = \frac{2a + 9}{7}}$$

:

$$6x - 5 < 8x$$

$$-5 < 8x - 6x$$ $$-5 < 2x$$

$$\frac{-5}{2} < \frac{2x}{2}$$ $$-\frac{5}{2} < x$$

This can also be written as $x > -\frac{5}{2}$.
The student made an error by flipping the inequality sign when dividing by a positive number (2), and also by moving the constant to the wrong side initially. When dividing by a positive number, the inequality sign does not flip.

$$\boxed{x > -\frac{5}{2}}$$

:

$$\frac{3x+5}{2} \ge \frac{6x+1}{2}$$

$$2 \times \frac{3x+5}{2} \ge 2 \times \frac{6x+1}{2}$$ $$3x+5 \ge 6x+1$$

$$5 \ge 6x - 3x + 1$$ $$5 \ge 3x + 1$$

$$5 - 1 \ge 3x$$ $$4 \ge 3x$$

$$\frac{4}{3} \ge \frac{3x}{3}$$ $$\frac{4}{3} \ge x$$

This can also be written as $x \le \frac{4}{3}$.
The student made an error in their third line of working, where they incorrectly multiplied terms by 2 again.

$$\boxed{x \le \frac{4}{3}}$$