📝 Solve equations with variables on both sides (15 MCQs)

📖 Explanation: The first step should aim to collect variable terms on one side. Subtracting $3x$ gives $4x - 9 = 15$, which is a valid and efficient first move. Adding 9 is also valid but usually done after variable collection. Dividing or multiplying prematurely complicates the equation without reducing the variable terms.

📖 Explanation: The student subtracts $2x$ from both sides to maintain equality. This is a direct application of the Subtraction Property of Equality, which states that if you subtract the same quantity from both sides of an equation, the equality remains true.

📖 Explanation: First distribute: $4x - 12 = 2x + 6$. Subtract $2x$: $2x - 12 = 6$. Add 12: $2x = 18$. Divide by 2: $x = 9$. Substituting back: $4(6)=24$ and $2(9)+6=24$, so correct. Common mistakes include forgetting to distribute the 4 to both terms.

📖 Explanation: The solution is perfectly correct. The student moved variable terms to the left and constants to the right: $6x - 4x = 13 - 5$, simplifying to $2x=8$, then $x=4$. The error-analysis distractor is that some may think the signs are wrong, but they are correctly applied.

📖 Explanation: Subtracting $8x$ from both sides gives $-7 = 3$, which is false. Therefore, no value of $x$ can satisfy this equation. This is a contradiction, so the equation has no solution. This tests understanding of identity vs. contradiction.

📖 Explanation: Distribute RHS: $3x + 9 = 3x + 9$. Subtracting $3x$ gives $9=9$, a true statement for all $x$. Thus the equation is an identity and has infinitely many solutions. This tests the student's ability to recognize an identity.

📖 Explanation: The first taxi cost is $5 + 2m$, second is $3m$. Setting them equal: $5 + 2m = 3m$. Subtract $2m$: $5 = m$, so at 5 miles both cost \$15. This models a real-life scenario requiring equation setup with variables on both sides.

📖 Explanation: Set equations equal: $2x + 1 = -x + 7$. Add $x$: $3x + 1 = 7$. Subtract 1: $3x = 6$. Divide: $x = 2$. The intersection point is (2,5). This connects algebraic solving to graphical interpretation of intersection.

📖 Explanation: All steps are correct. $2(3x-4) = 6x -8$. Then $6x - 5x = x$, so $x - 8 = 3$, then $x=11$. Checking: $2(33-4)=58$ and $5(11)+3=58$. The distractors are common error points, but here the solution is flawless.

📖 Explanation: Multiply by 6 (LCM of 3 and 2): $4x + 30 = 3x + 42$. Subtract $3x$: $x + 30 = 42$. Subtract 30: $x = 12$. This involves fractions and variable on both sides, requiring careful multiplication to clear denominators.

📖 Explanation: Method A: $4x-2x = -8-6$ gives $2x=-14$, $x=-7$. Method B: $4x+14=2x$ gives $2x=-14$, $x=-7$. Both work, but Method A avoids extra steps (no need to move 8 first). Efficiency means fewer steps and less chance of sign errors.

📖 Explanation: For a unique solution, the coefficients of $x$ must differ: $a \neq c$. If $a = c$, then the equation becomes $b = d$ (identity if true, or contradiction if false). So uniqueness requires the variable terms to not cancel out completely.

📖 Explanation: Perimeter formula: $2L + 2W = 46$. So $2(3x+2)+2(x+5)=46$. Simplify: $6x+4+2x+10=46$ → $8x+14=46$ → $8x=32$ → $x=4$. This requires setting up a linear equation from geometry and solving it.

📖 Explanation: If we subtract $5x$ from both sides, we get $-3 = 2$, which is false. Therefore, the equation has no solution, not $x=0$. This is a classic error where students think cancelling variables leaves zero, but they ignore the constant contradiction.

📖 Explanation: For infinitely many solutions, both sides must be identical for all $x$. Comparing coefficients: $2x = kx$ gives $k=2$, and constants $5=5$ match. So $k=2$ makes the equation $2x+5=2x+5$, true for all $x$. This requires deep understanding of identity conditions.