📝 Solving equations with simplification (14 MCQs)

📖 Explanation: The student made an error in distributing the negative sign. The expression $-4(x-1)$ should be expanded as $-4x + 4$, not $-4x - 4$. The negative sign applies to both terms inside the parentheses, changing the sign of the constant term. This is a common sign error in multi-step equations.

📖 Explanation: Option A correctly applies the distributive property: $4(2x-3)=8x-12$ and $-3(x+1)=-3x-3$. The others have either incorrect distribution of the constant or the variable term. This tests conceptual understanding of the distributive property as the foundational step for simplifying equations.

📖 Explanation: The perimeter of a rectangle is $2l + 2w$. Substituting $l=3x+2$ and $w=2x-1$ gives $2(3x+2)+2(2x-1)=42$. This is an application problem where students must translate a real-world situation into an algebraic equation using the distributive property and combining like terms.

📖 Explanation: Expanding: $5x-15-2x = 4x+6-3x$ simplifies to $3x-15 = x+6$. Solving gives $2x=21$, so $x=10.5$. Wait, recalculating: $3x - x = 6+15$ gives $2x=21$, so $x=10.5$. None of the options match? Let me re-check the expansion: left side $5x-15-2x = 3x-15$. Right side $4x+6-3x = x+6$. So $3x-15 = x+6$ -> $2x=21$ -> $x=10.5$. Actually $10.5$ is not listed. Let me check option A: If $x=-21$, LHS $5(-24)-2(-21) = -120+42=-78$, RHS $4(-21)+3(23)=-84+69=-15$. Not equal. There must be a misprint in options. The correct solution is $x=10.5$. I will correct options to include it. Actually, let me re-solve: $5x-15-2x=4x+6-3x$ -> $3x-15=x+6$ -> $2x=21$ -> $x=10.5$. So the correct answer is not listed. I will adjust the options. Option A: 10.5, B: -10.5, C: 21, D: -21. Correct is A.

📖 Explanation: Expanding: $6x-2+4=6x+3-5$ simplifies to $6x+2=6x-2$. Subtracting $6x$ gives $2=-2$, which is false. Therefore, the equation has no solution. The student's answer $x=0$ is incorrect; it is a common error to assume a variable can be isolated when it cancels out. This is an error analysis question that tests understanding of identity vs. contradiction.

📖 Explanation: Setting the equations equal gives $2x+3 = -x+9$. Adding $x$ to both sides: $3x+3=9$. Subtracting 3: $3x=6$. Dividing by 3: $x=2$. This is a graph-based question that requires interpreting intersection as solving a simplified linear equation. It tests the ability to move between graphical and algebraic representations.

📖 Explanation: Expanding: $8x-12 -6x+10 = 6 - x - 2$. Simplifies to $2x-2 = 4 - x$. Adding $x$: $3x-2=4$. Adding 2: $3x=6$. Dividing: $x=2$. This requires careful distribution of the second term $-2(3x-5)$ which gives $-6x+10$, and the negative sign before the parenthesis on the right. Multi-step reasoning is needed.

📖 Explanation: To clear fractions, multiply every term by the LCD, which is 6. This gives $6 \cdot \frac{3x-1}{2} - 6 \cdot \frac{x+2}{3} = 24$, simplifying to $3(3x-1) - 2(x+2) = 24$. Option A correctly applies this. The others either use wrong multipliers or forget to multiply the right side. This tests conceptual understanding of the clearing-fractions method.

📖 Explanation: Expanding the left: $2x-8+3x = 5x-8$. This is $5x-8 = 5x-8$, which is an identity. Every real number is a solution, not just $x=3$. The student's claim is partially true but incomplete; they missed the infinite solution set. This is an error analysis question requiring deep conceptual understanding of identities.

📖 Explanation: Expanding: $3ax+6-4x = 4x+6-ax$. Simplify: $(3a-4)x+6 = (4-a)x+6$. For infinitely many solutions, the coefficients of $x$ and constants must match on both sides. So $3a-4 = 4-a$ and $6=6$. Solving $3a-4 = 4-a$ gives $4a=8$, so $a=2$. This is an Olympiad-style problem that requires understanding conditions for infinite solutions and manipulating parameters.

📖 Explanation: Simplifying the left: $2x-6+4 = 2x-2$. Right: $kx+6-2x = (k-2)x+6$. For no solution, the $x$-coefficients must be equal but constants different. So $2 = k-2$ gives $k=4$. Then constants: $-2 \neq 6$, so no solution. This tests the conditions for no solution in a linear equation, requiring comparison of coefficients and constants.

📖 Explanation: The cost for the first company is $25 + 0.15m$, and for the second is $35 + 0.10m$. Setting them equal gives $25+0.15m = 35+0.10m$. Solving: $0.05m=10$, so $m=200$ miles. This is a modeling question that requires translating a real-world scenario into a linear equation and then simplifying it.

📖 Explanation: Method A: $4x-8 = 4x-2-6$ -> $4x-8=4x-8$ -> infinite solutions. Method B: Divide by 2: $2(x-2) = (2x-1)-3$ -> $2x-4 = 2x-4$ -> infinite solutions. Both yield the identity. This question compares solution strategies and tests understanding that different valid algebraic manipulations lead to the same conclusion. It encourages flexible thinking.

📖 Explanation: The simplification is correct: $5x-2x-3 = 3x-3$ gives $3x-3=3x-3$, which is an identity. Therefore, every real number satisfies the equation, not just $x=0$. The student made the error of stopping at a true statement and arbitrarily assigning a value. This is a classic error analysis question that requires recognizing identities.