📝 Solve equations with variables divided by a constant (28 MCQs)

📖 Explanation: The student is incorrect. The negative sign in the divisor directly affects the sign of the quotient. Since $\frac{x}{-5} = 12$, multiplying both sides by -5 yields $x = -60$. This tests the understanding that the multiplicative inverse of -5 is -1/5, and the sign must be preserved. Confusing magnitude with value is a common error.

📖 Explanation: To isolate x, you must undo the operations in reverse order. The variable is multiplied by 3 and divided by 4. So, first multiply both sides by 4 to cancel the division, then divide by 3 to cancel the multiplication. Option B would give $\frac{3x}{16}$, and Option C is applying the coefficient itself, not its inverse.

📖 Explanation: This is a direct application of the speed formula. Speed = distance/time, so $\frac{d}{5} = 48$. Multiplying both sides by 5 gives $d = 240$ miles. Option B incorrectly treats speed as 5 times distance, and Option C incorrectly divides 48 by 5. This models a real-world scenario requiring correct equation setup.

📖 Explanation: The student likely divided 9 by 3 instead of multiplying both sides by 3. To solve $\frac{x}{3} = 9$, you must multiply both sides by 3 to get $x = 27$. The student’s answer $x=3$ suggests they performed the inverse operation incorrectly, confusing division with multiplication. This is a classic error.

📖 Explanation: The solution to the equation is a single value, not a range. Solving $\frac{x}{4} = -3$ gives $x = -12$. This is represented as a single point on the number line at -12. Option B comes from multiplying -3 by 4 incorrectly (or dividing), Option C is the positive root, and Option D is an inequality. This tests representation of exact solutions graphically.

📖 Explanation: The first equation has a coefficient of 2 in the numerator. Solving the first: multiply by 3 → $2x = 30$ → $x=15$. Second: multiply by 3 → $x=15$. Wait, they are actually equal. Let's check: Second gives x=15, first gives x=15. So they are equal. The correct answer is C. This compares two equations with different coefficients but same solution. It tests proportional reasoning and equivalence.

📖 Explanation: First, eliminate denominators by multiplying through by 4: $10x = x + 36$ → $9x = 36$ → $x=4$. Check: LHS = $5(4)/2 = 10$, RHS = $4/4 + 9 = 1+9=10$. This is a multi-step equation involving variables on both sides and constants in denominators. It requires careful distribution and combining like terms. Option B, C, D do not satisfy the equation.

📖 Explanation: Perimeter = 2(length + width) = 56. So $2(\frac{3w}{4} + w) = 56$ → $2(\frac{7w}{4}) = 56$ → $\frac{7w}{2} = 56$ → $7w = 112$ → $w = 16$ cm. This requires setting up an equation from a geometry context and solving with a variable divided by a constant. Common error: forgetting to multiply by 2 for both dimensions.

📖 Explanation: The student's method is correct: $\frac{x}{6} = \frac{1}{2}$ → cross-multiply: $2x = 6$ → $x=3$. Alternatively, multiply both sides by 6: $x = 6 \times \frac{1}{2} = 3$. The student used a valid proportional reasoning. Option C and D come from misapplying inverse operations. This question checks understanding of equivalent methods and their validity.

📖 Explanation: Student A is correct: $\frac{x}{-7} = -5$. Multiply both sides by -7: $x = (-5) \times (-7) = 35$. The quotient of two negatives is positive. Student B incorrectly assumed the product of two negatives is negative. This tests sign rules in division equations. A common misconception is mishandling signs when multiplying by a negative constant.

📖 Explanation: This is the fundamental rule for solving equations of the form variable divided by a constant. To isolate x, multiply both sides by the constant $a$, giving $x = ab$. This is a direct recall of the inverse operation. Options B, C, D represent common errors: dividing instead of multiplying, adding, or subtracting. This is the simplest form of the concept.

📖 Explanation: The equation $\frac{x}{8} = 3.5$ means a total quantity $x$ is divided into 8 equal parts, each of size 3.5. So, total weight = 8 × 3.5 = 28 oz. Option A correctly describes this. Option B misinterprets the divisor and quotient. This requires translating a mathematical equation into a contextual story, testing modeling skills.

📖 Explanation: First: $x/3=6$ → $x=18$. Second: $2x/3=6$ → multiply by 3: $2x=18$ → $x=9$. The solution is indeed halved because the numerator is doubled, so to keep the quotient same, x must be halved. This is consistent. The student's reasoning is correct. This tests the relationship between coefficients and solutions, and whether proportional changes are understood.

📖 Explanation: This is a tricky conceptual question. Since $x = 12k$, for any integer $k$, $x$ is an integer. Therefore, all given values of $k$ (5,3,4,6) are possible. However, if the problem intended that $x$ must be a multiple of something else, it's not stated. The correct approach is to recognize that no option is invalid, but since the question forces a choice, the least obvious or most likely to be overlooked is k=5, but mathematically all work. This tests critical reading and number theory basics.

📖 Explanation: This is a straightforward application of the multiplication property of equality. To isolate $x$, multiply both sides of the equation by the constant 5, since $x$ is divided by 5. Thus, $x = 12 \times 5 = 60$. The other options represent common arithmetic errors like adding, subtracting, or dividing instead of multiplying.

📖 Explanation: This question tests the ability to translate a verbal statement into a mathematical equation. The phrase 'a number divided by -4' means the variable $x$ is the numerator and -4 is the denominator, written as $\frac{x}{-4}$. The word 'is' translates to the equals sign, and '9' is the result. Option B is the reciprocal, which is a common mistranslation.

📖 Explanation: The area of a rectangle is given by $A = l \times w$. Substituting the known values gives $72 = 9w$, which is equivalent to $w = \frac{72}{9}$. Solving this gives $w = 8$ meters. This is a real-world application where the variable is divided by a constant in the context of the problem's formula.

📖 Explanation: The student's answer of 5 indicates they performed the operation $15 \div 3$. However, the equation $\frac{x}{3} = 15$ requires the inverse operation. Since $x$ is divided by 3, we must multiply both sides by 3 to isolate $x$, yielding $x = 45$. The error is a fundamental misconception of inverse operations.

📖 Explanation: The graph represents the equation $y = \frac{x}{4}$. To find $x$ when $y = 3$, we set up the equation $3 = \frac{x}{4}$. Multiplying both sides by 4 gives $x = 12$. This requires interpreting a graphical relationship and solving the resulting one-step equation. The other options are common arithmetic mistakes from misreading the graph or solving incorrectly.

📖 Explanation: This is a two-step problem. First, subtract 3 from both sides to isolate the term with the variable: $\frac{x}{2} = 8$. Then, multiply both sides by 2 to solve for $x$, giving $x = 16$. Options B, C, and D result from performing the operations in an incorrect order, such as multiplying before subtracting or subtracting after multiplying incorrectly.

📖 Explanation: This is a direct application of the speed formula $\text{speed} = \frac{\text{distance}}{\text{time}}$. Substituting the given speed and time, we get $65 = \frac{d}{4}$. To solve for $d$, multiply both sides by 4: $d = 65 \times 4 = 260$ km. This connects a real-world formula to the algebraic skill of solving equations with a variable divided by a constant.

📖 Explanation: Solve both equations: For $x$, multiply by 3: $x = 18$. For $y$, multiply by 6: $y = 18$. Thus $x = y$. This question requires the student to solve both equations and then compare their values. It goes beyond simple computation to demand relational reasoning between two different equations.

📖 Explanation: The student is incorrect. The equation has $x$ divided by -7. To isolate $x$, we must perform the inverse operation, which is multiplication, not division. Since the variable is divided by -7, we multiply both sides by -7 to get $x = 21 \times -7 = -147$. This is a classic error where students confuse the inverse operation needed for a division with the operation needed for a multiplication.

📖 Explanation: The variable $x$ is currently divided by 3. The goal is to isolate $x$. The inverse operation of division is multiplication. Therefore, the first and most logical step is to multiply both sides of the equation by 3. This cancels the division on the left side, leaving $x$ alone. The other options suggest incorrect or irrelevant operations that would not help in isolating the variable.

📖 Explanation: This is a generalization of the concept. To solve $\frac{x}{a} = b$, multiply both sides by $a$ (the constant divisor) to isolate $x$. This yields $x = a \times b$. This abstraction requires students to apply the algebraic rule without concrete numbers, testing their ability to handle symbolic representations and understand the underlying principle rather than just memorizing a procedure.

📖 Explanation: The formula relates perimeter to side length. Substituting $P = 36$ gives $36 = 4s$. This is equivalent to $s = \frac{36}{4}$, which simplifies to $s = 9$. This is a modeling question where the equation naturally arises from a geometric context, and solving it requires dividing the constant (the multiplier 4) to isolate the variable, reinforcing the inverse operation concept.

📖 Explanation: To solve, multiply both sides by -3: $x = 12 \times (-3) = -36$. A common error is to forget the negative sign, leading to 36. Another error is to divide instead of multiply, leading to -4 or 4. The correct answer requires careful attention to the sign of the constant and the application of the multiplication property of equality with negative numbers.

📖 Explanation: The equation $y = \frac{x}{2}$ is a linear function. We need to find the x-coordinate where the y-value is -3. This is exactly solving $-3 = \frac{x}{2}$. Multiply both sides by 2 to get $x = -6$. On the graph, this corresponds to the point (-6, -3). This question blends graph interpretation with equation solving, requiring the student to connect the algebraic solution to a point on a graph.