Equation I: $$5x - 7 = 5x + 3$$ simplifies to $$-7 = 3$$ (no solution). Equation II: $$4(x + 1) = 4x + 4$$ simplifies to $$4x + 4 = 4x + 4$$ or $$4 = 4$$ (infinitely many solutions). Equation III: $$3x + 2 = 7x - 10$$ gives $$-4x = -12$$, so $$x = 3$$ (one solution). Choice A misidentifies equation I as having one solution. Choice B misidentifies equation II as having one solution. Choice D incorrectly includes equation I.

Simplifying the left side: $$\frac{1}{2}(4x + 6) = 2x + 3$$. The equation becomes $$2x + 3 = 2x + n$$. Subtracting $$2x$$ from both sides gives $$3 = n$$. When $$n = 3$$, we have $$3 = 3$$, which is always true, indicating infinitely many solutions. Choice A misinterprets the identity as a unique solution. Choice C would be correct if we had $$3 = n$$ where $$n \neq 3$$. Choice D incorrectly distributes $$\frac{1}{2} \cdot 6 = 6$$ instead of $$3$$.

Simplifying the left side: $$2(3x - 1) - 6x = 6x - 2 - 6x = -2$$. The equation becomes $$-2 = k - 2$$, which gives $$k = 0$$. When $$k = 0$$, we get $$-2 = -2$$ (infinitely many solutions). For no solution, we need $$k \neq 0$$, giving us $$-2 = k - 2$$ where $$k - 2 \neq -2$$. Choice A gives infinitely many solutions. Choice B also gives infinitely many solutions. Choice D gives one specific case but misses that any $$k \neq 0$$ works.

Expanding the left side: $$2(x - 3) + 4x = 2x - 6 + 4x = 6x - 6$$. The equation becomes $$6x - 6 = 6x - 8$$. Subtracting $$6x$$ from both sides gives $$-6 = -8$$, which is false. This means no solution exists. Choice A incorrectly solves $$6x - 6 = 6x + 8$$. Choice C mistakes this for an identity. Choice D comes from solving $$2x - 6 = 6x - 8$$ (forgetting to combine like terms).

When you encounter an equation where the same expression appears on both sides, you're dealing with a special case that requires careful analysis of how many solutions exist.

Let's solve this equation step by step. Starting with $$\frac{2x + 6}{3} = \frac{2x}{3} + 2$$, you can split the left side using the distributive property: $$\frac{2x}{3} + \frac{6}{3} = \frac{2x}{3} + 2$$. This simplifies to $$\frac{2x}{3} + 2 = \frac{2x}{3} + 2$$.

Now subtract $$\frac{2x}{3}$$ from both sides: $$2 = 2$$. Since this statement is always true regardless of the value of $$x$$, the equation has infinitely many solutions. Any real number you substitute for $$x$$ will make the original equation true.

Choice A incorrectly assumes there's a unique solution and mistakes the process of eliminating variables. Choice B makes an error by comparing $$\frac{6}{3}$$ and $$2$$ as if they're unequal, when $$\frac{6}{3} = 2$$ exactly. Choice C misunderstands what happens when variable terms cancel—this cancellation is precisely what reveals the equation's true nature.

The correct answer is D because the equation simplifies to the identity $$2 = 2$$.

Study tip: When solving linear equations, pay attention to what happens after you eliminate variables. If you get a true statement like $$2 = 2$$, you have infinitely many solutions. If you get a false statement like $$2 = 3$$, you have no solution.

This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 6(x+1)-3=6x+3, distribute to get 6x+6-3=6x+3, simplify to 6x+3=6x+3, subtract 6x to get 3=3. This results in 3=3, an identity that is always true, so the equation has infinitely many solutions. A common error is thinking an identity like 3=3 means x=0 or one solution, or mistakenly claiming it has no solutions like a contradiction. The strategy is to (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes include stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.

To find when the equation has infinitely many solutions, we need to simplify and get an identity of the form $$a = a$$. Expanding the left side: $$3(x + 2) - 5 = 3x + 6 - 5 = 3x + 1$$. So the equation becomes $$3x + 1 = 3x + k$$. For infinitely many solutions, we need $$1 = k$$, so $$k = 1$$. Choice B gives $$k = 6$$ (incorrect transformation). Choice C gives $$k = 11$$ (adding instead of subtracting). Choice D gives $$k = -1$$ (sign error in simplification).

When you encounter linear equations that simplify to unusual forms, you need to carefully analyze what the result tells you about solutions.

Let's work through Pedro's simplification. Starting with $$5(x - 2) + 10 = 5x + c$$, we distribute: $$5x - 10 + 10 = 5x + c$$, which becomes $$5x = 5x + c$$. Subtracting $$5x$$ from both sides gives us $$0 = c$$, exactly what Pedro found.

This result is crucial because it tells us the original equation's behavior depends entirely on the value of $$c$$. When $$c = 0$$, we get $$0 = 0$$, which is always true regardless of what $$x$$ equals—meaning infinitely many solutions exist. When $$c \neq 0$$ (say $$c = 3$$), we get $$0 = 3$$, which is never true—meaning no solutions exist.

Answer A incorrectly reverses the relationship, claiming no solutions when $$c = 0$$ and infinitely many when $$c \neq 0$$. Answer B suggests there's exactly one solution when $$c = 0$$, but $$0 = 0$$ is satisfied by every possible $$x$$-value, not just one. Answer C claims there's always exactly one solution ($$x = 0$$), but this ignores how the value of $$c$$ fundamentally changes the equation's solvability.

Remember this pattern: when simplifying leads to a statement about constants (like $$0 = c$$), the number of solutions depends on whether that statement is true or false, not on finding a specific $$x$$-value.

Expanding the right side: $$4(x + 2) = 4x + 8$$. The equation becomes $$4x + 8 = 4x + 8$$. Subtracting $$4x$$ from both sides gives $$8 = 8$$, which is always true, confirming infinitely many solutions. Choice B incorrectly expands $$4(x + 2)$$ as $$4x + 8$$ but then writes $$4x = 4x + 8$$. Choice C divides by $$4$$ correctly but doesn't show the final identity clearly. Choice D incorrectly expands $$4(x + 2)$$ as $$4x + 2$$.

This question tests classifying linear equations by solution count: one (x=a), infinitely many (identity a=a), or none (contradiction a≠b). Solving reveals type: distribute and collect terms (2(x+3)=2x+6 → 2x+6=2x+6 → 6=6 identity), result shows classification (x=5 means one solution 5, 3=3 means all x work infinitely many, 4=7 means no x works no solution); type depends on whether variables cancel: if x remains with unique value (one solution), if variables cancel to truth (infinite), if cancel to falsehood (none). For the equation 2x+8=2(x+4), distribute to get 2x+8=2x+8, subtract 2x to get 8=8. This simplifies to 8=8, indicating infinitely many solutions since it's an identity. A common error is thinking it's one solution like x=4 by incorrectly solving without noticing the identity, or claiming no solution like 8=4. Strategy: (1) distribute parentheses, (2) collect like terms (combine x's, combine constants), (3) move variables to one side, (4) observe result (x=number → one, number=number → infinite, number=different → none), (5) verify (substitute x back if one solution, try multiple x values if claiming infinite). Mistakes: stopping too early (not simplifying fully), misinterpreting 0=0 as x=0, claiming 5=8 has solution, distributing incorrectly changing the type.