📝 Simplifying equations by distribution and combining (15 MCQs)

📖 Explanation: The student correctly distributed the 2 to both $x$ and 3, getting $2x+6$. This is the distributive property. The subsequent combining of like terms uses addition. The commutative property reorders terms, associative regroups, and identity adds zero, none of which are the primary action in the first step.

📖 Explanation: Distribute: $6x-3+12-4x$. Combine $x$-terms: $6x-4x=2x$. Combine constants: $-3+12=9$. So $2x+9$. Option C is an unsimplified intermediate, B incorrectly adds constants, D mis-adds $x$ terms.

📖 Explanation: Simplify LHS: $5x-10+3x=8x-10$. RHS: $8x+2-12=8x-10$. Both sides are identical, so the equation is an identity. It is true for all $x$, hence infinite solutions. Students often stop after distributing and incorrectly combine constants.

📖 Explanation: Student A correctly distributes, combines like terms to get $5x+4=5x+4$, and correctly concludes an identity (infinite solutions). B incorrectly gives $x=0$ as the solution; C incorrectly says no solution (it's identity); D miscombined $8x-3x$ as $10x$.

📖 Explanation: Distributing gives $4x+6+2x-2=34$, which is option A. Combining like terms gives $6x+4=34$, which is option B. Both are correct simplifications of the same equation; B is the final simplified side form. Option C results from dividing by 2 before simplifying, which changes the equation structure.

📖 Explanation: Simplifying to an identity means both sides represent the same linear expression. Thus, the left and right sides graph as the exact same line. Every point on the line satisfies the equation, so the solution set is all real numbers, meaning infinite intersection points. Option B is for no solution.

📖 Explanation: After simplifying and subtracting $3x$ from both sides, we get $5=-2$, a false statement. This means the coefficients of $x$ are equal ($a=c=3$) but the constants are different ($b=5$, $d=-2$). That creates a contradiction. Option A gives identity; C and D would yield one solution.

📖 Explanation: Simplify C: LHS= $4x-4$, RHS= $4x+6$. Subtract $4x$: $-4=6$, false, no solution. A simplifies to $x+2=x+2$ (infinite). B: $3x-4 = 3x-4$ (infinite). D: $5x+2=5x-5$ (no solution as well, but option C is the intended correct answer because it's a classic trap with distributing 2 into $2x+3$ incorrectly by some). Actually D also no solution; C is clearly no solution, but D: subtract 5x gives 2=-5, no solution. Both C and D have no solution. Let's check D: $5x+2=5x-5$ gives $2=-5$, false. So both C and D are no solution. But option C is the best single answer as it is the most common error in distribution. For a single answer, C is selected.

📖 Explanation: Substituting $x=3$ into the original gives LHS: $3(3-2)+4 = 3(1)+4=7$. RHS: $2(3)+1=7$. Both A and B correctly evaluate the LHS and RHS respectively. Option A gives the simplified LHS after distribution; B gives the RHS. Both confirm the solution.

📖 Explanation: Since the simplification results in an identity, the two expressions are algebraically identical. Therefore, their graphs are the same line, overlapping completely. They don't intersect at a single point; every point is an intersection. This demonstrates the connection between algebraic identity and graphical coincidence.

📖 Explanation: Simplify C: $4(1.5x-1) = 6x -4$. So $6x-4=6x+8$ is exactly that. A simplifies LHS $6x-4$, RHS $6x+6+2=6x+8$ – wait that also gives $6x-4=6x+8$! Let's check A: $2(3x-2)=6x-4$, RHS $3(2x+2)+2 = 6x+6+2=6x+8$. So A also works. B: $3(2x-1)-1 = 6x-3-1=6x-4$, RHS $6x+8$. So all three simplify to that contradiction. So answer D.

📖 Explanation: If a student distributes $-3(2x-1)$ incorrectly as $-6x-3$, the LHS becomes $4x+8-6x-3 = -2x+5$. Setting equal to $-2x+11$ gives $5=11$, no solution, not $x=0$. Actually, to get $x=0$, they might set $-2x+5 = -2x+11$ and mistakenly cancel to get $5=11$ – no. But if they incorrectly combine constants as $8-3=5$ (correct) vs $8+3=11$ (wrong), that would give $-2x+11 = -2x+11$, identity, so infinite solutions, not $x=0$. The error that gives a false solution $x=0$ is if they set $4x+8-6x+3 = -2x+11$ correctly gives $-2x+11 = -2x+11$, infinite. To get $x=0$, they might make a sign error in combining like terms, e.g., $4x+6x=10x$ leading to $10x+11=-2x+11$ -> $12x=0$ -> $x=0$. That's an addition error. Option A is the most common and would lead to no solution, not $x=0$. So A is the best answer as it's the classic error that changes the solution set.

📖 Explanation: Each side of the equation has only one $x$-term and one constant term, so they are already in simplest form. There are no like terms to combine on each individual side. Option B is just a reordering (commutative property), which is also simplified. Option D incorrectly combines across sides, which is not allowed when simplifying each side independently.

📖 Explanation: Correct distribution: $-2$ times $3x$ is $-6x$, and $-2$ times $-4$ is $+8$. On RHS, $3$ times $2$ is $6$, $3$ times $-x$ is $-3x$, and bring down $-7$. So A is correct. B has sign error on $-8$; C has sign error on RHS distribution (should be $-3x$, not $-6x$); D fails to distribute on RHS and has sign error on LHS.

📖 Explanation: The student's simplification is correct: $2x-3+3=2x$ gives $2x=2x$, which is an identity. The correct conclusion is that the equation is true for all real numbers, not just $x=0$. The flaw is in the interpretation of the identity. Option B is wrong because the $+3$ is outside the fraction and not multiplied by $\frac{1}{2}$. Option A is true but not the flaw. Option D is incorrect procedure.