Since $$x = 4$$ is a solution to $$ax + 1 = 3x - 7$$, we can substitute $$x = 4$$ into the equation: $$a(4) + 1 = 3(4) - 7$$. This gives $$4a + 1 = 12 - 7 = 5$$. Therefore, $$4a = 4$$, so $$a = 1$$. Now we can find $$a^2 - 3a = 1^2 - 3(1) = 1 - 3 = -2$$.

First, expand both sides of the equation $$4x - (x + 3) = 6 + (x - 3)$$. Left side: $$4x - x - 3 = 3x - 3$$. Right side: $$6 + x - 3 = 3 + x$$. So the equation becomes $$3x - 3 = 3 + x$$. Subtracting $$x$$ from both sides: $$2x - 3 = 3$$. Adding 3 to both sides: $$2x = 6$$, so $$x = 3$$. Therefore, $$x^2 + x = 3^2 + 3 = 9 + 3 = 12$$.

To solve $$|2x - 6| = 4x - 10$$, we need to consider two cases based on the definition of absolute value. First, note that for the equation to have real solutions, we need $$4x - 10 \geq 0$$, which means $$x \geq 2.5$$. Case 1: $$2x - 6 \geq 0$$, i.e., $$x \geq 3$$. Then $$|2x - 6| = 2x - 6$$, so the equation becomes $$2x - 6 = 4x - 10$$. Solving: $$-6 + 10 = 4x - 2x$$, so $$4 = 2x$$ and $$x = 2$$. But we need $$x \geq 3$$ for this case, and $$2 < 3$$, so $$x = 2$$ is not valid for Case 1. Case 2: $$2x - 6 < 0$$, i.e., $$x < 3$$. Then $$|2x - 6| = -(2x - 6) = 6 - 2x$$, so the equation becomes $$6 - 2x = 4x - 10$$. Solving: $$6 + 10 = 4x + 2x$$, so $$16 = 6x$$ and $$x = \frac{8}{3} \approx 2.67$$. For this case, we need $$x < 3$$, and indeed $$\frac{8}{3} < 3$$. Also, we need $$x \geq 2.5$$ for the right side to be non-negative: $$\frac{8}{3} = \frac{8}{3} \approx 2.67 > 2.5$$, so this condition is satisfied. Let's verify: $$|2(\frac{8}{3}) - 6| = |\frac{16}{3} - 6| = |\frac{16 - 18}{3}| = |\frac{-2}{3}| = \frac{2}{3}$$. And $$4(\frac{8}{3}) - 10 = \frac{32}{3} - 10 = \frac{32 - 30}{3} = \frac{2}{3}$$. Since both sides equal $$\frac{2}{3}$$, $$x = \frac{8}{3}$$ is indeed a solution. Therefore, the equation has exactly one solution: $$x = \frac{8}{3}$$. Choice A is wrong because we found a solution. Choice C is wrong because we only found one solution, not two. Choice D is wrong because the solution is unique, not infinitely many.

To solve $$\frac{x}{4} + \frac{x}{12} = \frac{x}{6} - 4$$, first find a common denominator. The LCM of 4, 12, and 6 is 12. Multiply the entire equation by 12: $$3x + x = 2x - 48$$, which simplifies to $$4x = 2x - 48$$. Subtracting $$2x$$ from both sides gives $$2x = -48$$, so $$x = -24$$. Therefore, $$\frac{2x}{3} = \frac{2(-24)}{3} = \frac{-48}{3} = -16$$.

When you encounter a system of linear equations and need to find the product of the variables, you'll need to solve for each variable first, then multiply them together.

To solve this system, substitution works well since the second equation can easily be rearranged. From $$x - y = 1$$, you get $$x = y + 1$$. Substituting this into the first equation: $$2(y + 1) + 3y = 12$$, which simplifies to $$2y + 2 + 3y = 12$$, then $$5y = 10$$, so $$y = 2$$.

Substituting back: $$x = 2 + 1 = 3$$. Therefore, $$xy = 3 \times 2 = 6$$.

Let's check why the wrong answers appear. Choice A ($$15$$) likely comes from incorrectly adding the two equations and getting confused about what operation to perform. Choice B ($$9$$) might result from solving incorrectly and getting $$x = 3, y = 3$$, then calculating $$3 \times 3 = 9$$. Choice C ($$12$$) is tempting because it's the constant from the first equation—students sometimes grab familiar numbers without completing the calculation.

The key strategy here is to always verify your solution by substituting both values back into both original equations. With $$x = 3$$ and $$y = 2$$: the first equation gives $$2(3) + 3(2) = 6 + 6 = 12$$ ✓, and the second gives $$3 - 2 = 1$$ ✓. This confirmation step catches calculation errors and builds confidence in your final answer.