Solving Linear Equations — Free Algebra 1 Review Games.

2x = 8, x = 4.

5x = 10, x = 2.

x/3 = 3, x = 9.

3x - 2x = 1 + 7, x = 8.

4x + 4 = 20, 4x = 16, x = 4.

6x - 2 = 4x + 6, 2x = 8, x = 4.

5 - 2x - 6 = -9, -2x - 1 = -9, -2x = -8, x = 4.

2y = 12 - 3x, y = (12 - 3x)/2.

0.5x = 2.5, x = 5.

Both sides are identical, so every real number is a solution.

Cross multiply: 5(x+2) = 3(2x-1), 5x+10 = 6x-3, 13 = x. Wait: 10+3 = x, x = 13. Let me recheck: 5x+10=6x-3, -x=-13, x=13. Actually checking: plug 13: (15)/3=5, (25)/5=5. But the answer says -13. Let me recompute: 5(x+2)=3(2x-1) => 5x+10=6x-3 => 10+3=6x-5x => x=13.

2x-3=7 gives x=5; 2x-3=-7 gives x=-2.

A = P + Prt, A - P = Prt, r = (A-P)/(Pt).

6x+3-2x+8 = 3x+19, 4x+11 = 3x+19, x = 8. Hmm, let me recheck: 4x+11=3x+19 => x=8. Check: 3(17)-2(4)=51-8=43, 3(8)+19=43. Yes x=8.

3x - 6 = 3x + 1 gives -6 = 1, which is false. No solution.

Add 4 to both sides: 7x = 21. Divide by 7: x = 3. A common error is dividing 17 by 7 before isolating the variable term, which gives the incorrect answer 13/7.

Subtract 6 from both sides: -3x = -6. Divide both sides by -3: x = 2. A common error is forgetting that dividing a negative by a negative yields a positive, giving the incorrect answer x = -2.

Add 1 to both sides: x/4 = 4. Multiply both sides by 4: x = 16. Choosing x = 12 is a common error — it comes from solving x/4 = 3 (skipping the step of adding 1 first) and then multiplying by 4.

Subtract 4x from both sides: 2x + 2 = 10. Subtract 2: 2x = 8. Divide by 2: x = 4. Choosing x = 6 results from failing to move the 4x term and instead only dividing 12 by 2 incorrectly.

Divide both sides by 3: x - 2 = 3. Add 2 to both sides: x = 5. Choosing x = 3 is a common error — it solves x - 2 = 3 as though the answer is 3, forgetting the final step of adding 2.

Subtract 7 from both sides: -2x = -6. Divide by -2: x = 3. Choosing x = -3 is a sign error — dividing -6 by -2 gives a positive result, not negative.

Subtract mx from both sides: b = y - mx. Choice A adds mx instead of subtracting it. Choice C incorrectly treats the product mx as a sum, subtracting only m and then dividing by x separately.

Distribute: 3x + 12 - 2x = 15. Combine like terms: x + 12 = 15. Subtract 12: x = 3. Choosing x = 27 results from adding 12 to 15 instead of subtracting — a direction-of-operation error.

Distribute: 4x - 2x - 10 = 8. Combine: 2x - 10 = 8. Add 10: 2x = 18. Divide: x = 9. Choosing x = -1 comes from distributing incorrectly as 4x - 2x + 10 = 8, dropping the negative sign on the constant term.

Subtract 3x from both sides: 2x + 3 = -7. Subtract 3: 2x = -10. Divide by 2: x = -5. Choosing x = 5 is a sign error — the constant on the right side is -7, not +7, so the result after subtracting 3 must be negative.

Subtract b from both sides: ax = c - b. Divide both sides by a: x = (c - b)/a. Choice A adds b instead of subtracting it. Choice C inverts the fraction, placing a in the numerator rather than the denominator.

Distribute the left side: 2x + 2 = 2x + 2. Both sides are identical, so the equation is true for every real value of x — this is called an identity. It has infinitely many solutions, not just one.

Subtract 0.4 from both sides: 0.2x = 0.8. Divide by 0.2: x = 4. Choosing x = 0.8 results from stopping after subtraction without completing the division step. Choosing x = 8 comes from dividing 1.2 by 0.2 before subtracting 0.4.

Multiply every term by the LCD of 6: 3x + 2x = 30. Combine: 5x = 30. Divide: x = 6. Choosing x = 10 is a common error — it solves only x/2 = 5 while ignoring the x/3 term entirely.

Distribute: 7 - 3x - 3 = 2x - 6. Combine the left side: 4 - 3x = 2x - 6. Add 3x to both sides: 4 = 5x - 6. Add 6: 10 = 5x. Divide: x = 2. Choosing x = -2 typically results from a sign error when combining the constants on the left side.

Divide both sides by lw: h = V/(lw). Choice A multiplies by lw instead of dividing by it. Choice D inverts the correct answer, placing lw in the numerator and V in the denominator.

Distribute: 2x - 6 = 5x + 3. Subtract 2x from both sides: -6 = 3x + 3. Subtract 3: -9 = 3x. Divide: x = -3. Choosing x = 3 is a sign error — the left side grows more slowly than the right, forcing x into negative territory.

Multiply both sides by 2: 3x + 1 = 14. Subtract 1: 3x = 13. Divide by 3: x = 13/3. Choice D (14/3) is a common order-of-operations error — dividing 14 by 3 before subtracting 1, which skips the subtraction step.

Multiply every term by 4: (2x - 1) - 2(x + 3) = 4. Distribute: 2x - 1 - 2x - 6 = 4. Simplify: -7 = 4. Since -7 never equals 4, the equation has no solution. The variable terms cancel completely, leaving a false statement.

Subtract 2l from both sides: P - 2l = 2w. Divide both sides by 2: w = (P - 2l)/2. Choice A skips the division step entirely. Choice D incorrectly adds l after dividing P by 2, reversing the sign of the l term.

Distribute both sides: 8x - 12 = 6x + 2 + 2x. Combine like terms on the right: 8x - 12 = 8x + 2. Subtract 8x from both sides: -12 = 2. This is a false statement, so the equation has no solution.

Multiply every term by the LCD of 12: 4(x + 1) + 3(x - 2) = 2(x + 3). Expand: 4x + 4 + 3x - 6 = 2x + 6. Combine: 7x - 2 = 2x + 6. Subtract 2x and add 2: 5x = 8. Divide: x = 8/5. Choice D (5/8) is the reciprocal — a classic error from inverting the final division step.

Subtract 3x from both sides: k = -7. When k = -7, the equation becomes -7 = -7, which is always true, giving infinitely many solutions. For any other value of k, the equation produces a false statement and has no solution. Choice A (k = 7) would give 7 = -7, which is false.

Distribute: 5x + 10 - 6x + 3 = x + 4. Combine like terms on the left: -x + 13 = x + 4. Add x to both sides: 13 = 2x + 4. Subtract 4: 9 = 2x. Divide: x = 9/2. A common error is distributing -3(2x - 1) as -6x - 3, which incorrectly changes the sign of the constant term from +3 to -3.

Multiply both sides by b: x - a = bc. Add a to both sides: x = a + bc. Choice A subtracts a instead of adding it — a sign error on the final step. Choice D divides by b again rather than multiplying, which is the inverse of the correct operation.

Subtract 6 from both sides: 3x = 9. Divide both sides by 3: x = 3. A very common error is dividing 15 by 3 before subtracting 6, which gives the incorrect answer x = 5. Always undo addition and subtraction before multiplication and division.

Add 10 to both sides: 5x = 10. Divide both sides by 5: x = 2. Choosing x = 0 is a classic error — students sometimes assume x must be 0 because the right-hand side is 0, but 0 is the value of the entire expression 5x - 10, not the value of x.

To isolate x, apply the inverse (opposite) of addition, which is subtraction. Subtracting 7 from both sides gives x = 8. Adding 7 would increase the constant and move further from isolating x. Multiplication and division are not applicable here because 7 is added to x, not multiplied.

Divide both sides by -4: x = 28 divided by (-4) = -7. The most common error is forgetting the negative sign and writing x = 7. When dividing a positive number by a negative number, the quotient is always negative.

Multiply both sides by 5 to undo the division: x = 4 times 5 = 20. A common mistake is dividing 4 by 5 instead of multiplying, which gives the incorrect x = 4/5. The operation that undoes division by 5 is multiplication by 5.

Subtract 3 from both sides: 2x = 8. Divide both sides by 2: x = 4. Choosing x = 7 is a common error from adding 3 to 11 instead of subtracting, giving 2x = 14 and then x = 7. Always subtract a positive constant to move it to the other side.

To isolate r in d = rt, divide both sides by t: d/t = r. Choosing r = dt is wrong — multiplying both sides by t would give dt = rt squared, not isolating r. Literal equations are solved by applying inverse operations to isolate the target variable.

Add 1 to both sides: 4x = 12. Divide by 4: x = 3. Choosing x = 12 is a common error where a student correctly sets up 4x = 12 but then stops without dividing both sides by 4, mistakenly treating the equation as already solved.

Subtract 2x from both sides: 3x - 4 = 11. Add 4 to both sides: 3x = 15. Divide by 3: x = 5. The distractor x = 7/3 comes from a sign error: subtracting 4 from 11 (getting 7) instead of adding 4 to both sides, giving 3x = 7. Always move constants by applying inverse operations correctly.

Distribute on the left: 3x - 6 = 2x + 1. Subtract 2x from both sides: x - 6 = 1. Add 6: x = 7. A common error is treating the distributed term as +6 instead of -6, writing 3x + 6 = 2x + 1, which leads to the incorrect x = -5. Remember that 3 times (-2) = -6.

To isolate m in F = ma, divide both sides by a: F/a = m, so m = F/a. Choosing m = Fa is incorrect — multiplying both sides by a would give Fa = ma squared, not isolating m. Division undoes multiplication, so divide by the variable that accompanies m.

Subtract 1.2 from both sides: 0.3x = 2.4. Divide both sides by 0.3: x = 8. A common error is dividing 3.6 by 0.3 directly without first subtracting 1.2, giving 12, which is wrong. When dividing decimals, note that 2.4 divided by 0.3 equals 24 divided by 3 = 8.

Subtract x from both sides: x + 7 = -3. Subtract 7 from both sides: x = -10. Choosing x = 10 is a common sign error — ignoring the negative result. Verify by substituting: 2(-10) + 7 = -13 and (-10) - 3 = -13. Both sides are equal, confirming x = -10.

Distribute the left side: 4x + 4 = 4x + 5. Subtract 4x from both sides: 4 = 5. This is a false statement that can never be true, so the equation has no solution. This differs from infinitely many solutions, which occurs when both sides simplify to the same expression (like 4 = 4), making the equation true for every value of x.

Multiply both sides by 2 to clear the fraction: 5x - 3 = 12. Add 3 to both sides: 5x = 15. Divide by 5: x = 3. Choosing x = 9 is a common error where students subtract the coefficient (5) from 12 + 3 instead of dividing. Choosing x = 9/2 can result from dividing by the original denominator (2) instead of the coefficient (5).

To isolate b, subtract mx from both sides of y = mx + b: y - mx = b. Choosing b = y + mx is wrong because that adds mx instead of subtracting it. Choosing b = (y - m)/x is incorrect because x is only multiplied by m, not by the entire right-hand side, so dividing by x would be invalid here.

Distribute both sides: the left gives 4x - 4 and the right gives 4x - 4. The equation becomes 4x - 4 = 4x - 4, which is true for every value of x. This means there are infinitely many solutions. Contrast this with no solution, which occurs when simplification yields a contradiction such as 4 = 5.

Multiply every term by the LCD of 8: 2(x + 3) - (x - 1) = 8. Distribute carefully: 2x + 6 - x + 1 = 8. Combine like terms: x + 7 = 8. Subtract 7: x = 1. A critical sign error is treating -(x - 1) as -x - 1 instead of -x + 1, which shifts the answer to x = -1.

Subtract x from both sides: 2x - 2y = 6y. Add 2y to both sides: 2x = 8y. Divide both sides by 2: x = 4y. Choosing x = 2y is a common error from dividing 8y by 4 instead of 2, possibly from incorrectly reading the coefficient of x after collecting terms.

Distribute the left side: 2kx + 6 = 6x + 6. Subtract 6 from both sides: 2kx = 6x. For infinitely many solutions, the equation must be true for all values of x, so the coefficients of x must match: 2k = 6, giving k = 3. When k = 1, the equation becomes 2x = 6x, which has only one solution (x = 0).

Multiply every term by the LCD of 6: 2(2x - 3) + 3(x + 1) = 5x - 1. Distribute: 4x - 6 + 3x + 3 = 5x - 1. Combine: 7x - 3 = 5x - 1. Subtract 5x: 2x - 3 = -1. Add 3: 2x = 2. Divide: x = 1. Verify: (-1)/3 + 2/2 = -1/3 + 1 = 2/3, and (5 - 1)/6 = 4/6 = 2/3. Both sides match.

Factor x from the left side: x(a - b) = c. Divide both sides by (a - b): x = c/(a - b). The condition that a is not equal to b ensures the denominator is nonzero. Choosing c/(a + b) is wrong because the factored expression uses subtraction (a - b), not addition.

Distribute all terms: 6x - 3 - 5x - 10 = 2x - 8 - 3. Simplify both sides: x - 13 = 2x - 11. Subtract x from both sides: -13 = x - 11. Add 11: x = -2. A common sign error is distributing -5(x + 2) as -5x + 10 instead of -5x - 10, which shifts the answer to x = 2.

Subtract 3x from both sides: 2y = 12 - 3x. Divide both sides by 2: y = (12 - 3x)/2. Choosing y = 12 - 3x is a common error where the student subtracts 3x correctly but forgets to divide the entire right side by 2 — both the 12 and the 3x term must be divided.

Substitute x = 7: 5(7 - 2) = 3(7) + k. Simplify: 5(5) = 21 + k, so 25 = 21 + k. Subtract 21: k = 4. Choosing k = 14 is a frequent error from computing 5 times 7 = 35 without first simplifying the parentheses: 5(7 - 2) = 5(5) = 25, not 5(7) = 35.

Subtract 15 from both sides: x = 22 - 15 = 7. Choice A adds 15 instead of subtracting, giving the incorrect result of 37. The inverse operation of adding 15 is subtracting 15.

Divide both sides by 4: x = -20/4 = -5. Choice B ignores the negative sign on -20. Choices C and D result from multiplying by 4 instead of dividing. A positive divided into a negative yields a negative result.

Multiply both sides by 3: x = 9 times 3 = 27. Choice A divides 9 by 3 rather than multiplying by 3. The inverse operation of dividing by 3 is multiplying by 3, not dividing again.

Subtract 7 from both sides: -x = 5. Multiply both sides by -1: x = -5. Choice A forgets to account for the negative sign on x. When you isolate -x, you must still multiply by -1 to solve for positive x.

Add 3 to both sides: 5x = 15. Divide by 5: x = 3. Choice C adds 3 correctly to get 5x = 15 but then forgets to divide by 5, stopping at 15. Choice B divides 12 by 5 before adding 3, reversing the correct order of steps.

Subtract 2 from both sides: y/4 = 3. Multiply both sides by 4: y = 12. Choice A stops after the first step without completing the multiplication by 4. Choice D multiplies the incorrect intermediate value of 7 by 4.

Divide both sides by -3: x = 15 / (-3) = -5. Choice B ignores the negative coefficient, treating -3 as 3. Dividing a positive number by a negative number always yields a negative result.

Distribute the left side: 3x - 12 = 2x + 1. Subtract 2x from both sides: x - 12 = 1. Add 12 to both sides: x = 13. Choice A gives 11, which results from adding 1 and 12 before accounting for the x terms properly. Always distribute before combining like terms.

Subtract 3x from both sides: 2x - 2 = 8. Add 2 to both sides: 2x = 10. Divide by 2: x = 5. Choice A comes from subtracting 2 from 8 to get 6 and then dividing by 2, which skips properly collecting the x terms first.

To isolate r, divide both sides by t: r = d/t. Choice D multiplies d and t rather than dividing. This is the distance-rate-time formula; to find rate, divide distance by time. The variable r is multiplied by t, so the inverse operation is division.

Distribute both sides: 6x + 2 = 4x - 8. Subtract 4x: 2x + 2 = -8. Subtract 2: 2x = -10. Divide by 2: x = -5. Choice B results from treating the -8 as positive, ignoring the negative sign introduced when distributing 4 into (x - 2).

Simplify the left side: (3x + 6)/3 = x + 2. The equation becomes x + 2 = x + 5. Subtract x from both sides: 2 = 5. This is a false statement, so there is no solution. A common error is failing to simplify the fraction completely before trying to solve.

Multiply both sides by 2 to clear the fraction: 2A = bh. Divide both sides by b: h = 2A/b. Choice B divides A by 2b, which is equivalent to dividing by b first and then by 2 — the reverse of the correct steps. This is the triangle area formula solved for height.

Subtract 4x from both sides: 3 = 3x - 9. Add 9 to both sides: 12 = 3x. Divide by 3: x = 4. Choice A results from incorrectly subtracting 7x from both sides to get -3x = -12 and then dropping the negative, yielding -4 instead of 4.

Distribute: 2x + 10 - 3x + 3 = 4. Combine like terms: -x + 13 = 4. Subtract 13: -x = -9. Multiply by -1: x = 9. A common error is distributing -3 into (x - 1) as -3x - 3 instead of -3x + 3, which would incorrectly give x = 13.

Divide both sides by rt: P = I/(rt). This is the simple interest formula where I is interest, P is principal, r is rate, and t is time. Choice A subtracts rt, which is incorrect because P is multiplied by rt in the original equation, not added to it. The inverse of multiplication is division.

Subtract 32 from both sides: F - 32 = (9/5)C. Multiply both sides by 5/9: C = (5/9)(F - 32). Choice B uses 9/5 instead of its reciprocal 5/9. Choice C adds 32 instead of subtracting it. This is the Fahrenheit-to-Celsius conversion formula.

Let w = width, so length = w + 2. Perimeter = 2(length + width) = 2((w + 2) + w) = 2(2w + 2) = 4w + 4. Set equal to 36: 4w + 4 = 36. Subtract 4: 4w = 32. Divide by 4: w = 8. Choice D gives 9 by solving 4w = 36, which omits the constant 4 from the perimeter expression.

Distribute the left side: 3x + 3k = 3x - 6. Subtract 3x from both sides: 3k = -6. Divide by 3: k = -2. When k = -2 the equation becomes 3x - 6 = 3x - 6, which is always true, yielding infinitely many solutions. Choice B, k = 2, leads to 3x + 6 = 3x - 6, or 6 = -6, which is a contradiction with no solution.

Multiply every term by 4 to clear denominators: (3x - 1) + 2(x + 3) = 20. Distribute: 3x - 1 + 2x + 6 = 20. Combine like terms: 5x + 5 = 20. Subtract 5: 5x = 15. Divide by 5: x = 3. A common error is multiplying only the fraction terms by 4 while forgetting to also multiply the right side of the equation.

Subtract cx from both sides and subtract b from both sides: 2ax - cx = d - b. Factor out x: x(2a - c) = d - b. Divide both sides by (2a - c): x = (d - b)/(2a - c). Choice A incorrectly adds b and d in the numerator. Choice D reverses the denominator sign, making it the negative of the correct answer since c - 2a = -(2a - c).

Distribute on both sides: 5 - 6x + 8 = 3x + 3 - 4x. Combine like terms: 13 - 6x = -x + 3. Add 6x to both sides: 13 = 5x + 3. Subtract 3: 10 = 5x. Divide by 5: x = 2. A frequent error is distributing -2 into (3x - 4) as -6x - 8 instead of -6x + 8, because multiplying two negatives gives a positive.

Expand the right side: 5x - k = 5x + 10. Subtract 5x from both sides: -k = 10. Solve: k = -10. Substituting back, the equation becomes 5x + 10 = 5x + 10, which is always true, giving infinitely many solutions. Choice A, k = 10, gives 5x - 10 = 5x + 10, or -10 = 10, which is a contradiction with no solution.

Factor the right side by pulling out a: a^2 + ab = a(a + b). The equation becomes (a + b)x = a(a + b). Divide both sides by (a + b): x = a. The key step is recognizing the common factor (a + b) on both sides. Choice C is incorrect because substituting x = a + b would require (a + b)^2 = a^2 + ab, which does not hold in general.

Distribute: 8x - 12 - 2x - 10 = 6x - 6 + 2. Combine like terms on each side: 6x - 22 = 6x - 4. Subtract 6x from both sides: -22 = -4. This is a false statement, so the equation has no solution. Had the result been -22 = -22, there would be infinitely many solutions, which is why Choice D is a plausible but incorrect answer.

Subtract 5 from both sides: 3x = 9. Then divide both sides by 3: x = 3. Choice B (x = 6) comes from subtracting 5 from only the coefficient 3 instead of the constant 14, and choice D results from treating the equation as 3x = -14 + 5.

Divide both sides by -4: x = 20 / (-4) = -5. Choice A (x = 5) is the most common error — students divide 20 by 4 correctly but ignore the negative sign on the coefficient.

Multiply both sides by 3: x = 9 times 3 = 27. Choice A (x = 3) results from dividing 9 by 3 instead of multiplying — the inverse operation needed to isolate x is multiplication, not division.

A literal equation contains two or more variables. A = lw involves three variables (A, l, and w). The other three choices each contain only the single variable x, making them standard one-variable linear equations.

Add 8 to both sides: 2x = 8. Divide both sides by 2: x = 4. Choice A (x = -4) results from subtracting 8 instead of adding it, while choice B (x = 8) skips the division step entirely.

Subtract 7 from both sides: x = -2 - 7 = -9. Choice A (x = 9) results from adding 7 to -2 instead of subtracting, and ignoring the negative sign on -2.

Subtract 3x from both sides: 2x = 8. Divide both sides by 2: x = 4. Choice C (x = 8) is the result of ignoring the 3x term and dividing 8 by 1 instead of 2.

Subtract x from both sides: 2y = 10 - x. Divide both sides by 2: y = (10 - x)/2. Choice A forgets the final division by 2. Choice C incorrectly subtracts 2x rather than x from both sides.

Distribute: 3x - 12 = 2x + 1. Subtract 2x from both sides: x - 12 = 1. Add 12: x = 13. Choice B (x = 11) results from distributing as 3x - 4 instead of 3x - 12, forgetting to multiply 4 by 3.

Let n represent the number. Write the equation: n + 5 = 3n - 7. Subtract n from both sides: 5 = 2n - 7. Add 7: 12 = 2n. Divide by 2: n = 6. Choice D (12) is the value of 2n before dividing, a common step-stopping error.

Subtract 1 from both sides: (2x)/3 = 6. Multiply both sides by 3: 2x = 18. Divide by 2: x = 9. Choice A (x = 6) comes from stopping after multiplying by 3 and forgetting the final division by 2.

Distribute the left side: 3x + 6 = 3x + 6. Both sides are identical for any value of x, making this an identity with infinitely many solutions. Contrast this with a contradiction such as 3x + 5 = 3x + 6, where the variable cancels and leaves 5 = 6 — a false statement meaning no solution.

Divide both sides by t: r = d/t. Choice A (r = d - t) is a common error — because r and t are multiplied in the original equation, you undo multiplication with division, not subtraction.

Subtract 2l from both sides: P - 2l = 2b. Divide both sides by 2: b = (P - 2l)/2. Choice A omits the final division by 2. Choice B subtracts only l from P before dividing, rather than subtracting the full 2l term.

Distribute the right side: 2x + 6 = 2x + 2k. The 2x terms cancel, leaving 6 = 2k. Solve: k = 3. For any other value of k, the equation becomes a false statement (such as 6 = 4 when k = 2), meaning no solution.

Her total earnings are 12h dollars. Spending $30 reduces that amount, giving 12h - 30 = 90. Solving: 12h = 120, so h = 10 hours. Choice A would mean she gained $30 on top of her earnings rather than spending it.

Distribute the negative sign: -x - 3 = 2x - 6. Add x to both sides: -3 = 3x - 6. Add 6 to both sides: 3 = 3x. Divide: x = 1. A critical error is treating -(x + 3) as -x + 3, which fails to distribute the negative to the constant term.

Simplify the inner bracket first: 3(x + 1) - 4 = 3x + 3 - 4 = 3x - 1. Multiply by 2: 6x - 2. Right side: 3x + 6 + 1 = 3x + 7. Set equal: 6x - 2 = 3x + 7. Subtract 3x: 3x = 9. Divide: x = 3. Choice D (x = 9) comes from skipping the step of combining like terms on the right side.

Substitute x = 5: 3(2(5) - 1) = 4(5) + k. Simplify inside parentheses: 3(10 - 1) = 20 + k, so 3(9) = 20 + k, giving 27 = 20 + k. Therefore k = 7. Choice C (k = 10) results from an order-of-operations error computing 3 times 2 times 5 = 30, then 30 - 1 = 29, ignoring proper grouping.

Subtract 3x from both sides: 2y = 4x - 3x - 7 = x - 7. Divide both sides by 2: y = (x - 7)/2. Choice A forgets to divide by 2. Choice C leaves the 4x unchanged, failing to subtract the 3x from the right side.

Multiply every term by 12 (the LCM of 3, 4, and 2): 8x - 3 = 6x + 9. Subtract 6x from both sides: 2x - 3 = 9. Add 3: 2x = 12. Divide: x = 6. Choice B (x = 3) results from multiplying through by 6 instead of 12, which fails to clear the denominator of 4.

Subtract 4: 2kx = 4. When k = 0, the left side is 0 regardless of x, giving 0 = 4 — a false statement with no solution. For any nonzero k, divide by 2k to get x = 2/k, which is exactly one solution. Choices B, C, and D all produce a unique solution.

Perimeter = 2(length) + 2(width): 2(3x + 5) + 2(x - 2) = 46. Distribute: 6x + 10 + 2x - 4 = 46. Combine like terms: 8x + 6 = 46. Subtract 6: 8x = 40. Divide: x = 5. Choice A (x = 4) results from combining 10 and -4 as -6 instead of +6.

Multiply both sides by 2: 2S = n(a + l). Divide both sides by (a + l): n = 2S/(a + l). Choice A subtracts (a + l) instead of dividing by it. Choice C divides by an extra factor of 2, which would only be correct if the original equation had no denominator.

Distribute: 0.3x + 0.4x - 0.8 = 0.5x + 0.8. Combine like terms on the left: 0.7x - 0.8 = 0.5x + 0.8. Subtract 0.5x: 0.2x - 0.8 = 0.8. Add 0.8: 0.2x = 1.6. Divide: x = 8. Choice B (x = 6) results from computing 0.7x - 0.5x as 0.3x instead of 0.2x.

Subtract 5 from both sides: 3x = 12. Divide by 3: x = 4. A common error is choosing x = 12 by stopping after 3x = 12 and not dividing by 3.

Add 7 to both sides: 2x = 10. Divide by 2: x = 5. Choosing x = -2 results from subtracting 7 instead of adding, giving 2x = -4 incorrectly.

Distribute: 4x + 12 = 20. Subtract 12: 4x = 8. Divide by 4: x = 2. Choosing x = 5 is a common error where students divide 20 by 4 first and forget to subtract 3 afterward, leaving x + 3 = 5 unsolved.

Multiply both sides by 4: x = 24. Choosing x = 1.5 results from dividing 6 by 4 instead of multiplying, which is the inverse operation error.

Combine like terms on the left: 2x = 10. Divide by 2: x = 5. Choosing x = 2 is an error where the student only used the coefficient 5 (from 5x = 10) without combining like terms first.

Subtract x from both sides: 9 = 2x - 1. Add 1: 10 = 2x. Divide by 2: x = 5. Choosing x = 4 results from subtracting instead of adding the constant (9 - 1 = 8, then 8/2 = 4).

Dividing both sides by the nonzero value a gives x = b/a, a unique value. Infinitely many solutions occur when both sides are identical expressions (like 0x = 0); no solution occurs when a = 0 and b is not zero.

Distribute: 3x - 6 = 2x + 5. Subtract 2x: x - 6 = 5. Add 6: x = 11. Choosing x = 7 results from a distribution error where 3(x - 2) is treated as 3x - 2 instead of 3x - 6.

Subtract 3x from both sides: 2x + 2 = -8. Subtract 2: 2x = -10. Divide by 2: x = -5. Choosing x = 5 is an error where the student ignores the negative sign, treating 2x = 10.

Multiply both sides by 2: 2A = bh. Divide both sides by b: h = 2A/b. Choosing h = A/(2b) is a common error where the student divides by b first and then divides by 2 again instead of multiplying by 2.

Subtract 0.5x from both sides: 2 = 0.25x - 3. Add 3: 5 = 0.25x. Multiply by 4: x = 20. Choosing x = 4 results from moving 0.75x to the left incorrectly: 0.5x + 0.75x = 5, giving 1.25x = 5 and x = 4.

Translate to: 3n - 8 = 16. Add 8: 3n = 24. Divide by 3: n = 8. Choosing 4 is an error from setting up 3n = 16 - 8 = 8 (misreading 'decreased by 8' as dividing), then n = 8/3, rounding incorrectly.

Subtract 32: F - 32 = (9/5)C. Multiply both sides by 5/9: C = (5/9)(F - 32). Choosing C = (9/5)(F - 32) is wrong because the student flipped the fraction when multiplying instead of using the reciprocal of 9/5.

Distribute both sides: 8x - 4 = 3x + 6. Subtract 3x: 5x - 4 = 6. Add 4: 5x = 10. Divide by 5: x = 2. Choosing x = 10 is an error where the student finds 5x = 10 but forgets to divide by 5.

Subtract 3x from both sides: k = 9. For the equation to be an identity (true for all x), the constants must be equal, so k = 9. If k were any other value, the equation would have no solution because 3x cancels and leaves a false statement.

Distribute: 2x + 10 - 3 = 4x - 1. Simplify left side: 2x + 7 = 4x - 1. Add 1: 2x + 8 = 4x. Subtract 2x: 8 = 2x. Divide: x = 4. Choosing x = 3 results from subtracting the constants incorrectly: 7 - 1 = 6, then 6/2 = 3.

For 3x + 5 = 3x - 2: subtract 3x to get 5 = -2, which is always false, so no solution exists. The first choice simplifies to 4 = 4 (infinitely many solutions). The third gives x = 2, and the fourth gives x = 0 — both single solutions.

Subtract b from both sides: ax = c - b. Divide by a: x = (c - b)/a. Choosing x = c/a - b is wrong because division by a must apply to the entire numerator (c - b), not just c.

Distribute left side: 6x - 12 - 2x - 2 = 4x - 14. Distribute right side: 4x - 12 + 6 = 4x - 6. The equation becomes 4x - 14 = 4x - 6. Subtract 4x: -14 = -6, which is always false, so the equation has no solution.

Distribute: A = P + Prt. Subtract P: A - P = Prt. Divide both sides by Pt: r = (A - P)/(Pt). Choosing r = A/(Pt) - 1 is wrong because the -1 would need to be divided by t as well, making it (A - P)/(Pt), not A/(Pt) - 1.

The LCD of 4, 3, and 6 is 12. Multiply every term by 12: 9x - 4x = 10. Combine: 5x = 10. Divide by 5: x = 2. Choosing x = 5 results from incorrectly multiplying only the right side by 12 and forgetting to combine coefficients on the left.

Let w = width and l = 2w + 3. Use P = 2l + 2w: 2(2w + 3) + 2w = 54. Simplify: 4w + 6 + 2w = 54, then 6w = 48, so w = 8 cm. Choosing 48 cm is an error from computing 6w = 48 and reporting the intermediate value instead of dividing by 6.

Subtract v₀ from both sides: v - v₀ = at. Divide both sides by a: t = (v - v₀)/a. Choosing t = v/(v₀ + a) incorrectly treats v₀ and a as combined in the denominator rather than isolating the at term first.

Distribute the left side: 6x - 2a = 6x - 10. Subtract 6x from both sides: -2a = -10. Divide by -2: a = 5. For infinitely many solutions, the equation must be an identity, so the constants must match. If a equals any other value, the equation becomes a false statement with no solution.

The LCD of 3, 4, and 6 is 12. Multiply every term by 12: 4(2x - 1) - 3(x + 2) = 2(x - 3). Expand: 8x - 4 - 3x - 6 = 2x - 6. Simplify: 5x - 10 = 2x - 6. Subtract 2x: 3x - 10 = -6. Add 10: 3x = 4. Divide: x = 4/3. Choosing x = 4 is an error from finding 3x = 4 and forgetting to divide by 3.

Subtract 5 from both sides to get 2x = 6, then divide by 2 to get x = 3. A common error is adding 5 instead of subtracting, which gives 2x = 16 and x = 8 — the goal is to isolate the variable by undoing addition with subtraction.

Add 3 to both sides to get 4x = 16, then divide by 4 to get x = 4. The distractor x = 2.5 results from subtracting 3 instead of adding it, giving 4x = 10 and x = 2.5 — this reverses the required inverse operation.

Subtract 2 from both sides to get x/3 = 5, then multiply both sides by 3 to get x = 15. The distractor x = 5 comes from skipping the final step of multiplying by 3, leaving the answer as x/3 = 5 and reading x = 5 incorrectly.

A literal equation contains two or more variables. A = lw (area equals length times width) has three variables, making it a literal equation. The other three choices each contain only one variable and are standard linear equations solved for a specific numerical value.

Subtract 3x from both sides to get 2x + 4 = 10, then subtract 4 to get 2x = 6, and divide by 2 to get x = 3. A common mistake is subtracting the constants incorrectly — for instance, using 10 - 4 = 6 but then dividing incorrectly — leading to answers like x = 7.

Subtract 9 from both sides to get -3x = -9, then divide both sides by -3 to get x = 3. The distractor x = -3 results from a sign error: dividing -9 by positive 3 instead of by -3, which ignores the negative coefficient.

Divide both sides by 3: x + 2 = 7, then subtract 2 to get x = 5. Alternatively, distribute first: 3x + 6 = 21, then 3x = 15 and x = 5. The distractor x = 7 comes from solving x + 2 = 7 but then stopping there and writing x = 7 rather than x = 5.

Subtract 2x from both sides to collect all variable terms on the left: 3x = 12. Then divide by 3 to get x = 4. The distractor x = 12 comes from reading the constant directly without first subtracting 2x from both sides to combine variable terms.

Distribute both sides: 3x + 6 = 2x + 10. Subtract 2x from both sides: x + 6 = 10. Subtract 6: x = 4. A common error is distributing incorrectly on one side — for example, writing 3x + 2 = 2x + 5 — which leads to wrong answers like x = 3 or x = 8.

To isolate h, divide both sides of V = lwh by lw: h = V/(lw). The distractor h = V - lw incorrectly applies subtraction instead of division — it confuses additive inverse operations with the multiplicative inverse needed to undo multiplication.

Distribute the negative sign: 2x - x - 4 = 3x - 12, giving x - 4 = 3x - 12. Subtract x from both sides: -4 = 2x - 12. Add 12: 8 = 2x, so x = 4. A frequent error is failing to distribute the negative sign, treating -(x + 4) as -x + 4, which leads to x - 4 becoming x + 4 and gives x = -4.

Subtract 0.4x from both sides: 2 = 0.4x - 2. Add 2: 4 = 0.4x. Divide by 0.4: x = 10. The distractor x = 5 results from dividing 4 by 0.8 instead of 0.4 — a mistake that comes from using the wrong coefficient when isolating x.

Let the three consecutive integers be n, n + 1, and n + 2. Their sum is 3n + 3 = 57, so 3n = 54 and n = 18. The smallest integer is 18. The distractor 19 is the middle integer — a common error when students solve for n correctly but then misidentify which integer the variable n represents.

Subtract b from both sides: y - b = mx. Divide by m: x = (y - b)/m. The distractor x = y/m - b looks similar but is not equivalent. For example, if y = 10, m = 2, and b = 4, the correct answer gives (10 - 4)/2 = 3, while y/m - b gives 10/2 - 4 = 1. The subtraction must happen before the division.

Distribute the -2: 5x - 2x - 6 = x + 6, giving 3x - 6 = x + 6. Subtract x: 2x - 6 = 6. Add 6: 2x = 12. Divide by 2: x = 6. The distractor x = 3 comes from a collection error that produces 2x = 6 instead of 2x = 12, which can happen when the -6 is accidentally moved to the right side as +6.

Expanding 3(x + 2) = 3x + 6 gives 3x + 6 = 3x + 6, which is always true — an identity with infinitely many solutions. The other choices each simplify to false statements: 5 = -5, -3 = 4, and 5 = 6 respectively, meaning they have no solution.

Multiply both sides by 3 to clear the fraction: 2x - 1 = 15. Add 1: 2x = 16. Divide by 2: x = 8. The distractor x = 3 comes from skipping the multiplication by 3 and solving 2x - 1 = 5 instead. The distractor x = 16 results from correctly reaching 2x = 16 but forgetting to divide by 2.

Distribute all terms: 10x - 15 - 8x - 2 = 3x - 21. Combine like terms on the left: 2x - 17 = 3x - 21. Subtract 2x: -17 = x - 21. Add 21: x = 4. A sign error when distributing -2(4x + 1) — writing -8x + 2 instead of -8x - 2 — shifts the constant sum and leads to x = -4.

Divide both sides by 2: P/2 = l + w. Subtract l: w = P/2 - l. The distractor (P - l)/2 is a common error — subtracting l before dividing by 2 gives (P - l)/2, which is not equivalent. For example, if P = 20 and l = 6, the correct answer gives 10 - 6 = 4, while (20 - 6)/2 gives 7.

Multiply every term by the LCD of 6: 2(2x - 1) + 3(x + 1) = 3x + 5. Expand: 4x - 2 + 3x + 3 = 3x + 5. Combine the left side: 7x + 1 = 3x + 5. Subtract 3x: 4x = 4. Divide: x = 1. Errors in multiplying each fraction by the correct factor of the LCD commonly produce x = 3 or x = -1.

Multiply both sides by 2: 2A = (b₁ + b₂)h. Divide both sides by h: 2A/h = b₁ + b₂. Subtract b₂: b₁ = 2A/h - b₂. The distractor A/(2h) - b₂ arises from dividing by 2h in a single step without first multiplying by 2, which incorrectly halves the result.

Let a = number of adult tickets, so 50 - a = number of child tickets. The revenue equation is 12a + 7(50 - a) = 450. Distribute: 12a + 350 - 7a = 450. Combine: 5a = 100, so a = 20. The distractor 25 assumes an equal split of the 50 tickets and ignores the price constraint entirely.

Expand the left side: 2kx - 6 = 4x + 8. For no solution, the x-coefficients must be equal (so x cancels) while the constants remain unequal. Setting 2k = 4 gives k = 2. Substituting back yields 4x - 6 = 4x + 8, which simplifies to -6 = 8 — a false statement, confirming no solution. At k = 7, a unique solution exists.

Multiply every term by the LCD of 15: 9x - 10(x - 1) = 1. Distribute: 9x - 10x + 10 = 1. Combine: -x + 10 = 1. Subtract 10: -x = -9. Divide by -1: x = 9. The distractor x = -9 results from a sign error in the final step — dividing -9 by positive 1 instead of by -1.

Supplementary angles sum to 180 degrees: (4x + 10) + (2x + 20) = 180. Combine like terms: 6x + 30 = 180. Subtract 30: 6x = 150. Divide by 6: x = 25. Substituting back gives 4(25) + 10 = 110 degrees and 2(25) + 20 = 70 degrees. The larger angle is 110 degrees. The distractor 25 degrees is the value of x — not an angle measure.

Subtract 7 from both sides: 3x = 9. Divide both sides by 3: x = 3. Choice B results from adding 7 to both sides instead of subtracting, giving 3x = 23 and x = 23/3. Choice C is wrong because after correctly finding 3x = 9, you must divide by 3 — x cannot equal 9.

Distribute: 2x + 10 = 14. Subtract 10 from both sides: 2x = 4. Divide by 2: x = 2. Choice B (x = 7) comes from dividing 14 by 2 to get x + 5 = 7 and then failing to subtract 5. Choice C (x = 12) results from adding 10 to both sides instead of subtracting, giving 2x = 24.

Subtract 2x from both sides: 2x - 3 = 7. Add 3 to both sides: 2x = 10. Divide by 2: x = 5. Choice B (x = 2) comes from subtracting the constants instead of adding: 7 - 3 = 4, so 2x = 4. Choice C (x = 5/3) comes from adding x-terms instead of subtracting: 6x = 10. Choice D (x = 10) comes from finding 2x = 10 but not dividing by 2.

Subtract 8 from both sides: -2x = -8. Divide both sides by -2: x = 4. Choice B (x = -4) results from incorrectly treating -2x = 8 when moving 8, which gives x = -4. Choice C (x = 16) comes from multiplying 8 by 2 instead of dividing.

Multiply both sides by 3: x = 21. Choice B (x = 7/3) comes from dividing 7 by 3 instead of multiplying by 3. Choices C and D come from adding or subtracting 3 from 7, which are not the correct inverse operations when the variable is divided by 3.

Subtract 5 from both sides: -x = -2. Divide both sides by -1: x = 2. Choice B (x = -2) is a common sign error — forgetting to negate both sides when dividing by -1. Choice C (x = 8) comes from adding 3 and 5 instead of subtracting 3 from 5.

When all variable terms cancel and the remaining statement is always true (such as 4 = 4), the equation is an identity — it is true for every value of the variable, giving infinitely many solutions. Choice A describes a contradiction, which occurs when the result is always false (like 4 = 5). Choice B is wrong because the constant in the simplified statement does not represent a specific value of x.

Distribute: 3x - 12 = 2x + 1. Subtract 2x: x - 12 = 1. Add 12: x = 13. Choice B (x = -11) results from moving the constant incorrectly: computing x = 1 - 12 = -11. Choice D (x = 5) comes from a distribution error where 3(x - 4) is written as 3x - 4 instead of 3x - 12, giving 3x - 4 = 2x + 1 and x = 5.

Multiply both sides by 4: x + 3 = 8. Subtract 3: x = 5. Choice B (x = 11) comes from adding 3 instead of subtracting after reaching x + 3 = 8. Choice C (x = 8) is the result of stopping after multiplying by 4, without completing the final step of subtracting 3.

Subtract 4x from both sides: 5 = -3. This is always false regardless of x, so the equation has no solution. Choice B (x = -8) is incorrect — once 4x is eliminated, no x-term remains to solve for. Choice D is wrong because the constant equation 5 = -3 is a contradiction, not an identity; for infinitely many solutions, the result must be a true statement like 5 = 5.

Divide both sides by t: r = d/t. Choice B (r = dt) results from multiplying by t rather than dividing. Choice D (r = t/d) inverts the ratio. When a variable is multiplied by another quantity, the inverse operation — division — isolates it.

Distribute: 10x + 5 = 9x + 12 - 1 = 9x + 11. Subtract 9x: x + 5 = 11. Subtract 5: x = 6. Choice D (x = 10) results from a partial distribution error — writing 5(2x + 1) as 10x + 1 instead of 10x + 5, giving 10x + 1 = 9x + 11 and x = 10. Choice B is a sign error from incorrectly combining variable terms.

Divide both sides by lw: h = V/(lw). Choice B (h = Vlw) results from multiplying by lw rather than dividing. Choice D (h = lw/V) inverts the ratio. When a variable is multiplied by a group of terms, isolating it requires dividing by that entire group.

Add 1.5 to both sides: 0.5x = 4. Divide both sides by 0.5 (or multiply by 2): x = 8. Choice B (x = 2) comes from incorrectly computing 2.5 - 1.5 = 1 instead of adding 1.5, giving 0.5x = 1. Choice C (x = 4) comes from correctly finding 0.5x = 4 but not completing the division step.

Let x be the smaller number. Then x + (x + 12) = 50, so 2x + 12 = 50, 2x = 38, and x = 19. Choice B (31) is the larger number, not the smaller. Choice C (25) comes from taking half of 50 while ignoring the difference condition. Choice D (13) comes from subtracting 12 from half of 50, a common setup error.

Let w = width and l = w + 4. Using P = 2l + 2w: 2(w + 4) + 2w = 44, so 4w + 8 = 44, 4w = 36, and w = 9 cm. Choice B (13 cm) is the length, not the width. Choice C (11 cm) comes from omitting the subtraction of 8: treating 4w = 44 instead of 4w = 36.

Distribute: 6 - 6x + 8 = 4x + 2, giving 14 - 6x = 4x + 2. Add 6x and subtract 2: 12 = 10x, so x = 6/5. Choice B (x = -2/5) results from a sign error: writing -2(3x - 4) as -6x - 8 instead of -6x + 8, which gives 5x - 5 = 12 and x = 17/5. Choice D (x = 0) comes from distributing -2 only to 3x and not to -4, giving 6 - 6x - 4 = 4x + 2 and 0 = 10x.

Subtract 1 from both sides: 3x/4 = 6. Multiply both sides by 4: 3x = 24. Divide by 3: x = 8. Choice B (x = 24) comes from correctly finding 3x = 24 but not dividing by 3. Choice C (x = 6) comes from stopping at 3x/4 = 6 without completing the remaining steps. Choice D (x = 32/3) results from adding 1 to both sides instead of subtracting: 3x/4 = 8, 3x = 32.

Multiply through by the LCD of 12: 3(3x + 1) - 4(x - 2) = 12. Distribute: 9x + 3 - 4x + 8 = 12, giving 5x + 11 = 12. Subtract 11: 5x = 1, so x = 1/5. Choice C (x = 17/5) results from a sign error in distribution: writing -4(x - 2) as -4x - 8 instead of -4x + 8, giving 5x - 5 = 12. Choice B (x = 1) is a common arithmetic slip — setting up 5x = 1 correctly but then writing x = 1 rather than x = 1/5.

Distribute: ax + ab = c. Subtract ab from both sides: ax = c - ab. Divide by a: x = (c - ab)/a. Choice B uses addition instead of subtraction, reversing the sign when moving ab. Choice C divides by ab instead of just a. Choice D divides by b and subtracts only a rather than ab, misidentifying the structure of the equation.

Distribute: 3x + 3c = 3x - 9. Subtract 3x from both sides: 3c = -9. Divide by 3: c = -3. Choice B (c = 3) is a sign error — the equation requires 3c = -9, not 3c = 9. Choice C (c = -9) comes from forgetting to divide by 3. Choice D (c = 0) would give 3x = 3x - 9, or 0 = -9, which is a contradiction with no solution rather than infinitely many.

Simplify inside the brackets: 3x - (x + 4) = 2x - 4. Multiply by 2: 4x - 8 = 5x - 12. Subtract 4x from both sides: -8 = x - 12. Add 12: x = 4. Choice C (x = 20) comes from a sign error when distributing the negative: treating -(x + 4) as -x + 4, making the bracket 2x + 4, then 4x + 8 = 5x - 12 and x = 20. Choice D (x = 8/3) results from ignoring the outer factor of 2 and solving 2x - 4 = 5x - 12 directly.

Subtract 32 from both sides: F - 32 = (9/5)C. Multiply both sides by 5/9 (the reciprocal of 9/5): C = (5/9)(F - 32). Choice B uses 9/5 instead of its reciprocal 5/9, multiplying by the same fraction rather than its inverse. Choice C applies 5/9 only to F and then subtracts 32 separately, failing to distribute the multiplication over the full expression (F - 32). Choice D uses addition instead of subtraction when moving 32.

Set the costs equal: 50 + 35h = 20 + 45h. Subtract 35h and 20 from both sides: 30 = 10h. Divide by 10: h = 3. Choice B (7 hours) comes from adding the service fees and dividing by the rate difference: (50 + 20) / 10 = 7. Choice D (30 hours) comes from finding the fee difference of 30 but forgetting to divide by the hourly rate difference of 10.

Multiply both sides by b: 2x - a = b(c + d). Add a to both sides: 2x = b(c + d) + a. Divide by 2: x = [b(c + d) + a] / 2. Choice C subtracts a instead of adding, reversing the sign. Choice B divides only a by 2 rather than the entire right-hand side, applying the final division incorrectly. Choice D divides by 2b rather than first multiplying by b and then separately dividing by 2.

Subtract 7 from both sides: 2x = 8. Divide both sides by 2: x = 4. A common error is skipping the division step and writing x = 8, which confuses 2x = 8 with x = 8.

Adding 8 to both sides eliminates the constant on the left: 4x - 8 + 8 = 20 + 8, giving 4x = 28. Choice B results from subtracting 8 instead of adding, and choice C skips the isolation step entirely.

Subtract x from both sides: 4 = 2x - 2. Add 2 to both sides: 6 = 2x. Divide by 2: x = 3. Check: 3 + 4 = 7 and 3(3) - 2 = 7. A common error is moving the variable incorrectly and treating -2 as +2, which gives x = 1.

Distribute on both sides: 4x - 12 = 2x + 2. Subtract 2x from both sides: 2x - 12 = 2. Add 12: 2x = 14. Divide by 2: x = 7. Check: 4(4) = 16 and 2(8) = 16. A common error is distributing incorrectly as 4x - 3 = 2x + 1, leading to x = 2.

Subtract 2a from both sides: P - 2a = 2b. Divide both sides by 2: b = (P - 2a) / 2. Choice B is a common error where students forget to divide by 2. Choice D incorrectly divides only P by 2, failing to distribute the division to the entire numerator.

Set up the equation: 4 + 2.50m = 19. Subtract 4 from both sides: 2.50m = 15. Divide by 2.50: m = 6. Check: 4 + 2.50(6) = 19. Choice C (7.6) is a common error from dividing the full fare by the per-mile rate without subtracting the base fare first: 19 / 2.50 = 7.6.

Distribute -3: 5 - 3x - 6 = 2x - 3, which simplifies to -1 - 3x = 2x - 3. Add 3x to both sides: -1 = 5x - 3. Add 3: 2 = 5x. Divide by 5: x = 2/5. Choice C (x = 2) results from distributing -3(x + 2) incorrectly as -3x + 2, preserving the wrong sign on the constant. Choice B (x = 14/5) results from treating -3(x + 2) as -3x + 6.

Distribute all terms: 8x - 4 - 3x - 6 = 2x + 10 - 1. Combine like terms on each side: 5x - 10 = 2x + 9. Subtract 2x: 3x - 10 = 9. Add 10: 3x = 19, so x = 19/3. Choice C (x = 7/3) results from miscomputing -4 - 6 as +2 (sign error), giving 5x + 2 = 2x + 9 and then 3x = 7. Choice B (x = 7) results from expanding 2(x + 5) - 1 as 2x + 11 instead of 2x + 9.

Set the heights equal: 12 - t = 6 - 0.5t. Subtract 0.5t from both sides: 12 - 0.5t = 6. Wait, rearranging correctly: subtract 6 from both sides and add 0.5t: 6 = 0.5t. Divide by 0.5: t = 12. After 12 hours both candles are at 0 inches. Choice B (6 hours) results from ignoring that the second candle also decreases, solving 12 - t = 6 instead of setting both heights equal.

Multiply both sides by 2: 2S = n(a + l). Divide both sides by (a + l): n = 2S / (a + l). Choice B results from dividing by 2 instead of multiplying, which moves the 2 into the denominator. Choice C results from multiplying both sides by (a + l) rather than dividing, and choice D inverts the result entirely.