Equations and Inequalities — Free Algebra 2 Review Games.

x-3=7 gives x=10; x-3=-7 gives x=-4.

|2x|=8 means 2x=8 or 2x=-8, so x=4 or x=-4.

Subtract 2 from all parts: 1 < x < 7.

Absolute value less than means between: -5 < x < 5.

4x = 12, x = 3.

3x+1=10 gives x=3; 3x+1=-10 gives x=-11/3.

Subtract 1: -3 <= 3x <= 9. Divide by 3: -1 <= x <= 3.

Greater than means outside: x-2>5 or x-2<-5, so x>7 or x<-3.

ax = c - b, x = (c-b)/a.

Case 1: 2x-1=x+3, x=4. Case 2: 2x-1=-(x+3), 3x=-2, x=-2/3.

\(x^2 - 4 = 0\), \(x^2 = 4\), \(x = 2\) or \(x = -2\).

-7 < 3x-2 < 7, -5 < 3x < 9, -5/3 < x < 3.

Multiply both sides by 3/(pi*r^2): h = 3V/(pi*r^2).

x < -3 from the first, x > 4 from the second.

Absolute value can never equal a negative number, so no solution.

Split into two cases: x + 5 = 2 gives x = -3, and x + 5 = -2 gives x = -7. Verify: |-3 + 5| = |2| = 2 and |-7 + 5| = |-2| = 2. Choice A is wrong because x = 3 gives |3 + 5| = 8, not 2.

Solve each inequality: x - 1 > 3 gives x > 4, and x - 1 < 8 gives x < 9. The AND condition requires both to hold simultaneously, so the solution is 4 < x < 9. Choice D describes an OR condition, which would include nearly all real numbers since any x > 4 satisfies the first part alone.

To isolate b, divide both sides by h: A/h = bh/h, which simplifies to b = A/h. Choice A (b = Ah) incorrectly multiplies rather than divides. Choice C (b = A - h) confuses multiplication with addition — you cannot undo multiplication by subtracting.

The absolute value of a number equals 0 only when that number is 0. Since |0| = 0, the unique solution is x = 0. Choice A is wrong because |1| = 1 and |-1| = 1, neither of which equals 0. Choice B is wrong because x = 0 is a perfectly valid solution.

Subtract 3 from all three parts: -2 - 3 < x + 3 - 3 ≤ 7 - 3, giving -5 < x ≤ 4. The strict inequality on the left stays strict, and the non-strict inequality on the right stays non-strict. Choice B incorrectly adds 3 instead of subtracting it. Choice D leaves the inequality unchanged without performing the operation.

Split into two cases: x + 1 = 4 gives x = 3, and x + 1 = -4 gives x = -5. Verify: |3 + 1| = 4 and |-5 + 1| = |-4| = 4. Choice A (x = 5 or x = -3) is the solution to |x| = 4 — it ignores the +1 inside the absolute value and shifts the answers incorrectly.

Subtract b from both sides: x + b - b = a - b, giving x = a - b. Choice B (x = a + b) moves b to the wrong side by adding instead of subtracting. Choice C (x = b - a) reverses the order of subtraction, giving the negative of the correct answer.

Case 1: 4x - 3 = 9 gives 4x = 12, so x = 3. Case 2: 4x - 3 = -9 gives 4x = -6, so x = -3/2. Verify: |4(3) - 3| = 9 and |4(-3/2) - 3| = |-9| = 9. Choice B is wrong because x = 3/2 gives |4(3/2) - 3| = |3| = 3, not 9.

Subtract 2l from both sides: P - 2l = 2w. Then divide both sides by 2: w = (P - 2l)/2. Choice A is wrong because it skips the division by 2. Choice D incorrectly divides P by the product 2l rather than first subtracting 2l.

Solve each part: 2x - 3 > 1 gives 2x > 4, so x > 2. And 3x + 2 ≤ 14 gives 3x ≤ 12, so x ≤ 4. The AND condition means both must hold simultaneously: 2 < x ≤ 4. Choice B uses OR instead of AND, which would produce a much larger solution set. Choice D incorrectly makes x = 2 included and x = 4 excluded.

For |expr| ≥ c (c > 0), the solution is expr ≥ c or expr ≤ -c. Here: x + 4 ≥ 3 gives x ≥ -1, and x + 4 ≤ -3 gives x ≤ -7. The solution is their union. Choice B is the solution to |x + 4| ≤ 3 — it describes a bounded interval, which applies to the less-than case, not the greater-than case.

Add n to both sides: mx = p + n. Then divide both sides by m: x = (p + n)/m. Choice A uses (p - n)/m, which comes from subtracting n instead of adding it. Choice B (x = p + n - m) treats dividing by m as subtracting m, which is not a valid algebraic operation.

Subtract 1 from all three parts: -6 < 2x ≤ 8. Divide all parts by 2: -3 < x ≤ 4. Choice A comes from only dividing by 2 without first subtracting 1. Choice D incorrectly swaps which endpoint is strict and which is non-strict.

An absolute value equals 0 only when the expression inside equals 0. Set 3x + 6 = 0: 3x = -6, so x = -2. There is exactly one solution. Choice B is wrong because x = 2 gives |3(2) + 6| = 12, not 0. Choice C is wrong because x = -2 is a valid solution — do not confuse |expr| = 0 with |expr| = negative, which has no solution.

Solve each inequality: x + 3 < -2 gives x < -5, and x - 1 > 4 gives x > 5. Because this is an OR condition, the solution is the union of both sets: x < -5 or x > 5. Choice B describes the region between the two boundaries, which is actually excluded from this OR solution. Choices C and D each capture only half the answer.

For |A| = |B|, either A = B or A = -B. Case 1: x - 5 = x + 1 simplifies to -5 = 1, a contradiction with no solution. Case 2: x - 5 = -(x + 1) gives x - 5 = -x - 1, so 2x = 4, and x = 2. Verify: |2 - 5| = 3 and |2 + 1| = 3. Choice A (x = -2) gives |-7| = 7 and |-1| = 1, which are not equal.

Multiply both sides by 2: 2A = bh. Then divide both sides by h: b = 2A/h. Choice A (b = A/(2h)) inverts the operation on the 1/2 — it divides by 2 instead of multiplying. Choice D omits the factor of 2 entirely, as if the formula were A = bh rather than A = (1/2)bh.

Write -3 ≤ 5 - 2x ≤ 3. Subtract 5 from all parts: -8 ≤ -2x ≤ -2. Divide by -2 and flip both inequality signs: 4 ≥ x ≥ 1, equivalently 1 ≤ x ≤ 4. Choice A is the solution to |5 - 2x| ≥ 3 — it applies the wrong direction of the original inequality. Choice C results from forgetting to flip the signs when dividing by the negative number -2.

First solve |2x + 1| ≤ 5: -5 ≤ 2x + 1 ≤ 5 gives -6 ≤ 2x ≤ 4, so -3 ≤ x ≤ 2. Then intersect with x > 0. Because x > 0 is strict, x = 0 is excluded, giving 0 < x ≤ 2. Choice A ignores the constraint x > 0 entirely. Choice B (0 ≤ x ≤ 2) incorrectly includes x = 0, which is ruled out by the strict inequality x > 0.

Case 1: x + 1 = 2x - 3 gives x = 4. Check: |4 + 1| = 5 and 2(4) - 3 = 5. Valid. Case 2: -(x + 1) = 2x - 3 gives -x - 1 = 2x - 3, so 3x = 2 and x = 2/3. Check: |2/3 + 1| = 5/3, but 2(2/3) - 3 = -5/3. Since 5/3 does not equal -5/3, x = 2/3 is extraneous. Only x = 4 is valid. Always check solutions when the right side of the equation contains the variable.

When k > 0, the equation |2x - 4| = k yields two solutions. When k < 0, there is no solution because absolute value is never negative. When k = 0, the equation becomes |2x - 4| = 0, so 2x - 4 = 0, giving exactly one solution: x = 2. Choice A (k = 2) produces two solutions, x = 3 and x = 1. Choice B (k = -1) produces no solution.

Subtract c from both sides: (x - a)/b = d - c. Multiply both sides by b: x - a = b(d - c). Add a to both sides: x = a + b(d - c). Choice B incorrectly subtracts b(d - c) rather than adding it, which would come from a sign error when undoing the minus a. Choice D divides (d - c) by b instead of multiplying, reversing the step of clearing the denominator.

Subtract 4: 2|x - 3| > 6. Divide by 2: |x - 3| > 3. For |expr| > c (c > 0), the solution is expr > c or expr < -c. So: x - 3 > 3 gives x > 6, and x - 3 < -3 gives x < 0. The solution is x < 0 or x > 6. Choice A (0 < x < 6) is the solution to |x - 3| < 3 — it uses the wrong inequality direction. Choice D only captures half the answer.

An absolute value equals 0 only when the expression inside is 0. Set x² - 5x + 6 = 0 and factor: (x - 2)(x - 3) = 0, giving x = 2 or x = 3. Both are valid since |0| = 0. Choice B (x = -2 or x = -3) comes from incorrect factoring — the signs inside the factors must be negative to produce roots of +2 and +3, not +2 and +3 to produce roots of -2 and -3.

Rewrite |2x - k| < m as -m < 2x - k < m, then add k and divide by 2: (k - m)/2 < x < (k + m)/2. Set the endpoints equal to the given boundaries: (k - m)/2 = 1 and (k + m)/2 = 5, so k - m = 2 and k + m = 10. Adding these equations: 2k = 12, giving k = 6. Substituting back: m = 4. Choice B (k = 6, m = 2) would give the interval (6 - 2)/2 < x < (6 + 2)/2, or 2 < x < 4, which does not match.

The absolute value of a number equals its distance from zero on the number line, which is always non-negative. Therefore |−13| = 13. Choosing −13 is a common error — it is the original number, not its absolute value.

The equation |x| = 11 means x is exactly 11 units from zero, which occurs at both x = 11 and x = −11. Absolute value equations of the form |x| = k with k > 0 always produce two solutions: x = k and x = −k.

Divide both sides by 2π: r = C/(2π). Choice A multiplies instead of divides. Choice C incorrectly subtracts 2π, confusing addition and multiplication operations. The variable r is being multiplied by 2π, so the inverse operation is division.

'At least −3' translates to x ≥ −3, and 'at most 10' translates to x ≤ 10. Since both conditions must hold simultaneously, they are joined by 'and.' Choice A uses 'or,' which would include all real numbers. Choice D describes the region outside [−3, 10], the complement of the correct answer.

The absolute value of any real expression is always greater than or equal to zero, so it can never equal a negative number. |4x − 2| = −6 has no solution. Choice A has exactly one solution (x = 3). Choices B and D each have two solutions.

Subtract 5x from both sides: y = 12 − 5x. The term 5x moves to the right side with a sign change. Choice B adds 5x instead of subtracting it. Choices C and D place 5x on the wrong side or use the wrong sign.

The correct first step is to subtract 1 from all three parts: −8 − 1 ≤ 3x ≤ 10 − 1, giving −9 ≤ 3x ≤ 9. Then divide by 3 to get −3 ≤ x ≤ 3. Choice A is incorrect because dividing 3x + 1 by 3 gives x + 1/3, not x + 1. Choice C adds instead of subtracts, moving in the wrong direction.

An absolute value equals zero only when the expression inside equals zero. Setting 6x − 6 = 0 gives x = 1, the only solution. This differs from |x| = k for k > 0, which produces two solutions because both k and −k have absolute value k.

Set 2x + 3 = 11: 2x = 8, so x = 4. Set 2x + 3 = −11: 2x = −14, so x = −7. Choice B comes from a sign error when solving each case. Choice C omits the negative case entirely. Always write both equations before solving.

Add 2 to all three parts: −4 + 2 < 2x ≤ 6 + 2, giving −2 < 2x ≤ 8. Divide all parts by 2: −1 < x ≤ 4. Choice A adds 1 instead of 2. Choice B correctly adjusts the left bound but halves the right bound again after it was already adjusted. Always apply the same operation to all three parts simultaneously.

Multiply both sides by 3: 3V = πr²h. Then divide both sides by πr²: h = 3V/(πr²). Choice A forgets to multiply by 3 first, missing the reciprocal of 1/3. Choice B multiplies by 3πr² instead of dividing. Choice C incorrectly treats the multiplication relationship as addition.

Divide both sides by 3 first: |x − 2| = 4. Then set x − 2 = 4, giving x = 6, and x − 2 = −4, giving x = −2. Choice A omits the negative case (x = −2) and replaces it with the center value 2. Choice D arises from solving |x − 2| = 3 instead of 4, a common arithmetic error.

Solve each part: 3x − 1 > 8 gives x > 3, and 2x + 4 ≤ 0 gives x ≤ −2. For an 'or' compound inequality the solution is the union of both sets: x > 3 or x ≤ −2. Choice A describes the intersection −2 ≤ x ≤ 3, which would be correct for an 'and' inequality — a common mix-up between union and intersection.

Factor x from the left side: x(a − b) = c. Divide both sides by (a − b): x = c/(a − b). Factoring is the essential technique when the variable appears in multiple terms. Choice A incorrectly divides only the first term by a, then subtracts b rather than treating a − b as a single divisor.

Rewrite as −3 < x + 5 < 3. Subtract 5 from all parts: −3 − 5 < x < 3 − 5, giving −8 < x < −2. Choice D is the solution to |x + 5| > 3, the 'greater than' version, which reverses the direction. Choice C incorrectly adds 5 instead of subtracting.

Case 1: 2x − 4 = 2x + 6 → −4 = 6, which is never true. Case 2: −(2x − 4) = 2x + 6 → −2x + 4 = 2x + 6 → −2 = 4x → x = −1/2. Verify: |2(−1/2) − 4| = |−5| = 5 and 2(−1/2) + 6 = 5. The solution checks out. When one case yields a contradiction, do not discard the problem — always check the other case.

Subtract 3x from both sides: −2y = 10 − 3x. Divide by −2: y = (10 − 3x)/(−2) = (3x − 10)/2. Choice C results from dividing by positive 2 instead of −2, missing the sign change in the numerator. Choice A omits the division by 2 entirely and only halves the constant.

Case 1: 3x − 2 = x + 4 → 2x = 6 → x = 3. Verify: |7| = 7 = 3 + 4. Valid. Case 2: 3x − 2 = −(x + 4) → 4x = −2 → x = −1/2. Verify: |−7/2| = 7/2 = −1/2 + 4. Valid. Both solutions work because x + 4 is positive for both values, which is required for the right-hand side of an absolute value equation.

Subtract 3: 2|x − 1| ≤ 6. Divide by 2: |x − 1| ≤ 3. Rewrite: −3 ≤ x − 1 ≤ 3. Add 1: −2 ≤ x ≤ 4. Choice D comes from skipping the step of adding 1 at the end, or using 0 as the center instead of 1. Choice B only captures the upper bound without finding the lower bound from the inequality.

Multiply both sides by (cx + d): ax + b = kcx + kd. Rearrange: ax − kcx = kd − b → x(a − kc) = kd − b → x = (kd − b)/(a − kc). Choice A has the numerator negated: b − kd = −(kd − b). Choice D omits division by (a − kc), leaving out the required denominator.

Solve |x − 4| ≥ 2: x − 4 ≥ 2 or x − 4 ≤ −2, giving x ≥ 6 or x ≤ 2. Intersect with x < 7: the branch x ≥ 6 is restricted to 6 ≤ x < 7, while x ≤ 2 is unaffected since all those values already satisfy x < 7. The final answer is x ≤ 2 or 6 ≤ x < 7. Choice B ignores the additional constraint x < 7.

The right side 2x must be positive, requiring x > 0. Case 1 (x ≥ 3): x − 3 < 2x → −3 < x, always true for x ≥ 3. Case 2 (0 < x < 3): −(x − 3) < 2x → 3 < 3x → x > 1, giving 1 < x < 3. The union of both cases is x > 1. Choice C captures only Case 2. Choice D is too broad and includes values like x = 0.5 that do not satisfy the original inequality.

Multiply through by bd: d(x − a) + b(x + c) = 0 → dx − ad + bx + bc = 0 → x(b + d) = ad − bc → x = (ad − bc)/(b + d). Choice A incorrectly adds bc rather than obtaining ad − bc when rearranging. Choice D omits division by (b + d), which is always required when factoring x from multiple terms.

Case 1 (2x + 3 ≥ 0): 2x + 3 = x + c → x = c − 3, valid when c − 3 ≥ −3/2, i.e., c ≥ 3/2. Case 2 (2x + 3 < 0): −(2x + 3) = x + c → x = (−3 − c)/3, valid only when c > 3/2. For c > 3/2 both cases produce distinct valid solutions. For c < 3/2 neither case is valid and there is no solution. At c = 3/2 exactly, only Case 1 gives a valid solution (x = −3/2), so this is the unique value producing exactly one solution.

Solve |2x − 1| ≤ 3: −3 ≤ 2x − 1 ≤ 3 → −1 ≤ x ≤ 2. Solve |x + 2| > 1: x > −1 or x < −3. Intersect: the part x < −3 does not overlap with −1 ≤ x ≤ 2, so we keep −1 < x ≤ 2. The integers in this interval are 0, 1, and 2. Choice B incorrectly includes x = −1, but |(−1) + 2| = 1, which fails the strict inequality |x + 2| > 1.

The absolute value equation |x| = 5 means the distance from x to 0 on the number line is 5. This gives two cases: x = 5 or x = -5. Choices A and B each capture only one solution. Choice D is wrong because |0| = 0, not 5.

'At most 5' means x cannot exceed 5, so x may equal 5 or be less than 5. This is written x <= 5. 'At least 5' would be x >= 5, 'more than 5' is x > 5, and 'less than 5' (strictly) is x < 5.

An absolute value expression always evaluates to a value greater than or equal to zero. It can never equal a negative number, so |x + 3| = -2 has no solutions. Choices A and B are tempting traps that ignore the sign of the right-hand side.

To isolate r, divide both sides by 2π: r = C / (2π). Choice A incorrectly multiplies C by 2π instead of dividing. Choice C inverts the fraction. Choice D subtracts rather than divides, which is the wrong inverse operation for multiplication.

Absolute value measures distance from zero, so |x| = 9 means x is exactly 9 units from 0 in either direction. This yields x = 9 or x = -9. Choice D incorrectly squares the value. Choices A and B each omit one of the two valid solutions.

A compound 'and' inequality requires both conditions to hold simultaneously. x > -1 excludes -1 itself (open boundary), and x <= 4 includes 4 (closed boundary). The intersection is -1 < x <= 4. Choice C is the complement of this interval and would result from an 'or' with the opposite inequalities.

To isolate b, subtract both a and c from both sides: b = P - a - c. Choice A incorrectly adds a and c rather than subtracting them. Choice C subtracts only a but adds c. Choice D subtracts P instead of isolating it, giving the wrong sign entirely.

Set up two cases. Case 1: 2x + 1 = 7, so 2x = 6 and x = 3. Case 2: 2x + 1 = -7, so 2x = -8 and x = -4. Verify: |2(3)+1| = |7| = 7 and |2(-4)+1| = |-7| = 7. Both check out. Choice A results from incorrect sign handling when setting up the negative case.

Multiply both sides by 2 to clear the fraction: 2A = bh. Then divide both sides by h: b = 2A / h. Choice B divides by 2h directly without first multiplying by 2, effectively only undoing one operation. Choice C multiplies by h rather than dividing. Choice D inverts the relationship entirely.

Subtract 1 from all three parts: -4 < 2x <= 8. Divide all three parts by 2: -2 < x <= 4. Choice B results from subtracting 2 instead of 1 in the first step. Choice C subtracts 1 on the right side but not the left. Choice D reverses which endpoint is strict versus closed.

Subtract 32 from both sides: F - 32 = (9/5)C. Multiply both sides by the reciprocal 5/9: C = (5/9)(F - 32). Choice A adds 32 rather than subtracting it. Choice C uses 9/5 instead of its reciprocal 5/9. Choice D distributes the 5/9 incorrectly, subtracting 32 as if it were not also multiplied by 5/9.

Divide both sides by -2: |x + 4| = 5. Now set up two cases: x + 4 = 5 gives x = 1, and x + 4 = -5 gives x = -9. Verify: -2|1+4| = -2(5) = -10 and -2|-9+4| = -2(5) = -10. Choice D solves |x| = 5 without accounting for the +4 shift inside the absolute value.

Solve each inequality separately: 2x - 3 > 1 gives 2x > 4, so x > 2. And x + 4 < 9 gives x < 5. The 'and' requires both conditions simultaneously, so the intersection is 2 < x < 5. Choice A uses only the first inequality. Choice C would be the result for an 'or' compound inequality. Choice D incorrectly intersects x > 2 and x > 5.

To isolate T, divide both sides by nR: T = PV / (nR). Choice A incorrectly subtracts nR rather than dividing. Choice B multiplies all terms together, which is the opposite of solving for T. Choice D inverts the fraction, which would solve for 1/T rather than T.

Multiply both sides by 2/n to clear the coefficient: 2S/n = a + l. Subtract l from both sides: a = 2S/n - l. Choice A adds l rather than subtracting it. Choice C divides S by 2n, which is equivalent to multiplying by 1/(2n) rather than 2/n. Choice D multiplies by 2n instead of 2/n.

When two absolute values are equal, set up two cases. Case 1: 3x - 6 = x + 2, giving 2x = 8, so x = 4. Case 2: 3x - 6 = -(x + 2), giving 3x - 6 = -x - 2, so 4x = 4 and x = 1. Verify: |3(4)-6| = 6 = |4+2| and |3(1)-6| = 3 = |1+2|. Both solutions are valid. Choice A omits the second case.

Solve each part: x - 2 <= 5 gives x <= 7, and 3x > 9 gives x > 3. For an 'or' inequality, take the union. Every number x <= 3 satisfies x <= 7, and every number x > 7 satisfies x > 3. Together these two sets cover all real numbers with no gaps. Choice A is the intersection, which applies to 'and', not 'or'.

Subtract 2l from both sides: P - 2l = 2w. Divide both sides by 2: w = (P - 2l) / 2, which simplifies to P/2 - l. Choice A correctly subtracts 2l but forgets to divide by 2. Choice B subtracts only l before dividing instead of 2l. Choice D makes an arithmetic error, treating 2l/2 as l/2 rather than l.

Case 1: x + 4 = 3x - 2, so 6 = 2x, giving x = 3. Case 2: x + 4 = -(3x - 2), so x + 4 = -3x + 2, giving 4x = -2, so x = -1/2. Verify: |3+4| = 7 = |9-2|; |-1/2+4| = 3.5 = |3(-1/2)-2|. Both solutions check out. Forgetting to set up Case 2 leads to the single-answer trap in Choice B.

Case 1 (x >= 3): x - 3 = kx yields x = 3/(1-k). This requires k < 1 and the solution x >= 3, which holds when k >= 0. Case 2 (x < 3): 3 - x = kx yields x = 3/(1+k). This requires k > -1 and x < 3, which holds when k > 0. Both cases are simultaneously valid only when 0 < k < 1. At k = 0, only x = 3 satisfies the equation. At k >= 1, Case 1 yields no valid solution, leaving only one solution total.

Multiply all three parts by 4: -4 <= 2x - 3 <= 8. Add 3 to all parts: -1 <= 2x <= 11. Divide all parts by 2: -1/2 <= x <= 11/2. Choice B results from adding 3 to -4 as -4+3 = -1 then dividing incorrectly. Choice C mistakenly calculates 11/2 as 5. Choice D makes both of these errors simultaneously.

Isolate 1/d_o by subtracting 1/d_i from both sides: 1/d_o = 1/f - 1/d_i = (d_i - f) / (f times d_i). Take the reciprocal of both sides: d_o = (f times d_i) / (d_i - f). Choice A subtracts d_i directly from f, treating the terms as if they were not reciprocals. Choice C makes the same linear-thinking error. This problem requires combining fractions before taking the reciprocal.

First solve |2x - 5| <= 7: write -7 <= 2x - 5 <= 7, add 5 to get -2 <= 2x <= 12, then divide by 2 to get -1 <= x <= 6. Now intersect with x >= -1: the overlap is -1 <= x <= 6. Choice B ignores the lower bound from the absolute value. Choice C applies the absolute value bound to x directly instead of 2x - 5. Choice D incorrectly uses a strict inequality at x = -1, but both conditions allow equality there.

|2x+1| > 3 gives two cases: 2x + 1 > 3 (so x > 1) or 2x + 1 < -3 (so x < -2). Now intersect each branch with x <= 2: x > 1 combined with x <= 2 gives 1 < x <= 2; x < -2 combined with x <= 2 gives x < -2 (already satisfied). The union of both branches is 1 < x <= 2 or x < -2. Choice B is correct for the absolute value step alone but ignores the constraint x <= 2.

Factor out x on the left side: x(1/a + 1/b) = 1. Combine the fractions in the parentheses: 1/a + 1/b = (b + a) / (ab). So x times (a + b)/(ab) = 1, which gives x = ab / (a + b). Choice B is the coefficient of x rather than x itself. Choices C and D ignore the fractional nature of the coefficients entirely.

Absolute value measures distance from zero on the number line. The two numbers that are exactly 7 units from zero are 7 and -7, so |x| = 7 has exactly two solutions. Choice B misses the negative case: x = -7 also satisfies the equation because |-7| = 7.

To isolate r, divide both sides by t: r = d/t. The variable r is multiplied by t, so division undoes that operation. Choice B inverts the fraction incorrectly, and Choice C subtracts t instead of dividing by it.

An 'and' compound inequality requires both conditions to be satisfied simultaneously. A number must be both greater than 1 and less than 6, which means it must lie strictly between 1 and 6. Choice A describes an 'or' inequality, which would produce two outward rays.

Add 3 to both sides: 5x < 20. Then divide by 5: x < 4. Choice D results from dividing 17 by 5 without first adding 3 — a common error of skipping the addition step.

An absolute value equals 0 only when the expression inside equals 0. Setting x - 5 = 0 gives x = 5. Unlike |x| = k for k > 0 where two solutions exist, here both cases (x - 5 = 0 and -(x - 5) = 0) lead to the same equation, yielding exactly one solution.

|x| >= 4 means the distance from 0 on the number line is at least 4. This means x is 4 or more units from zero in either direction: x <= -4 or x >= 4. Choice A (and D, which states the same thing) describes |x| <= 4, where the distance is at most 4.

To isolate h, divide both sides by b: h = A/b. The variable h is multiplied by b, so we divide to undo that. Choice B multiplies both sides by b rather than dividing, which would increase the product instead of isolating h.

An 'or' compound inequality is satisfied whenever at least one condition holds, which can create two separate outward regions. An 'and' inequality requires both conditions simultaneously, typically producing a single bounded interval (or empty set), not two outward rays.

Set up two cases. Case 1: 2x - 6 = 10 gives 2x = 16, so x = 8. Case 2: 2x - 6 = -10 gives 2x = -4, so x = -2. Both solutions are valid. Choice D uses x = 2, which results from the error 2x = 4 instead of 2x = -4 in the second case.

Multiply both sides by 2: 2A = (b1 + b2)h. Divide both sides by h: 2A/h = b1 + b2. Subtract b2: b1 = 2A/h - b2. Choice A adds b2 instead of subtracting it. Choice C uses A/(2h), which would come from dividing by 2h before multiplying by 2, missing the factor of 2 in the numerator.

First inequality: x > 3. Second inequality: x < 5. Both must hold simultaneously, giving 3 < x < 5. Choice B says 'or' instead of 'and' — x > 3 OR x < 5 would be satisfied by virtually all real numbers since every number is less than 5 or greater than 3.

|x + 3| < 5 means the expression is between -5 and 5: -5 < x + 3 < 5. Subtract 3 throughout: -8 < x < 2. This is a single connected interval. Choice D would be the solution to |x + 3| > 5, which is the opposite inequality.

Multiply both sides by 2: 2KE = mv^2. Divide both sides by v^2: m = 2KE/v^2. Choice B omits multiplying by 2 first, producing KE/(2v^2) — the numerator is missing the factor of 2 that comes from clearing the (1/2).

First: 4x > 12, so x > 3. Second: 2x < 4, so x < 2. An 'or' inequality is satisfied by either region: x > 3 or x < 2. Choice D (2 < x < 3) is actually the gap between the solution regions — numbers in that gap satisfy neither original inequality.

Add 2 to all three parts: 3 < 3x < 12. Divide all three parts by 3: 1 < x < 4. Choice C correctly adds 2 but forgets to divide by 3 in the final step. Choice D only divides the right side, leaving the left incorrectly as 1.

Case 1: x/3 - 2 = 4 gives x/3 = 6, so x = 18. Case 2: x/3 - 2 = -4 gives x/3 = -2, so x = -6. Choice D uses x = 6 in place of x = -6, which is a sign error in Case 2: -2 times 3 is -6, not 6.

Expand: 2x - 2a + b = 0. Move constants: 2x = 2a - b. Divide by 2: x = a - b/2. Choice B adds b/2 instead of subtracting it — a sign error from treating +b as -b when moving it to the right side.

Case 1 (expressions equal): x + 2 = 2x - 1, so 3 = x. Case 2 (expressions opposite in sign): x + 2 = -(2x - 1) gives x + 2 = -2x + 1, so 3x = -1, meaning x = -1/3. Both solutions check out. Choice D uses x = 1/3, which comes from the error 3x = 1 instead of 3x = -1 in Case 2.

Add 4: 2|3x - 3| > 12. Divide by 2: |3x - 3| > 6. Split into two cases: 3x - 3 > 6 gives x > 3, and 3x - 3 < -6 gives 3x < -3, so x < -1. The solution is x > 3 or x < -1. Choice A describes the solution to |3x - 3| < 6. Choice D has x < 1, which is a division error: 3x < -3 divided by 3 gives x < -1, not x < 1.

Subtract nx from both sides and add 4: mx - nx = 10. Factor out x: x(m - n) = 10. Divide by (m - n): x = 10/(m - n). Choice A has 2 in the numerator, which comes from using 6 - 4 = 2 rather than 6 + 4 = 10. Choice D uses m + n, a sign error from incorrectly moving nx to the left side.

The absolute value of any real number expression is always greater than or equal to 0, so it can never equal a negative number. Therefore |x + 4| = -2 has no solution. Choice A has exactly one solution (x = 3), Choice B has two solutions (x = 2 and x = -3), and Choice D has two solutions (x = 4 and x = 2).

First solve |2x - 4| < 6: this gives -6 < 2x - 4 < 6, then -2 < 2x < 10, so -1 < x < 5. Now intersect with x >= 1: the overlap is 1 <= x < 5. Choice A ignores the condition x >= 1. Intersection tightens the left boundary from x > -1 to x >= 1 while preserving the right boundary x < 5.

Take the reciprocal of both sides: x - 1 = 1/k. Add 1 to both sides: x = 1/k + 1. Choice C results from the error of treating the equation as x - 1 = k before solving, adding 1 to get x = k + 1, which ignores that k is in the denominator. Choice D subtracts 1 instead of adding it.

Multiply all parts by 4 (positive, so inequality directions are preserved): -8 < 3x + 1 <= 12. Subtract 1 throughout: -9 < 3x <= 11. Divide by 3: -3 < x <= 11/3. Choice C correctly multiplies by 4 and subtracts 1, but forgets to divide by 3 in the last step. Choice B incorrectly reverses which endpoint is strict.

Since both sides are non-negative, squaring preserves the inequality: \((x - 3)^2 > (x + 1)^2\). Expanding: \(x^2 - 6x + 9 > x^2 + 2x + 1\). The \(x^2\) terms cancel: \(-6x + 9 > 2x + 1\), so \(8 > 8x\), giving \(x < 1\). Choice A reverses the direction. Choice C describes a pattern for an absolute value inequality of the form \(|x - a| > \text{constant}\), but does not match this problem.

Set up two cases: x - 1 = 6 gives x = 7, and x - 1 = -6 gives x = -5. Both are valid solutions. Choice D incorrectly uses the constant 6 directly without accounting for the -1 inside the absolute value.

The word AND means both conditions must hold simultaneously, so we take the intersection. The result is all numbers strictly between 2 and 9, written 2 < x < 9. Choice A uses OR logic, which would include nearly all real numbers.

Subtract 2l from both sides: P - 2l = 2w. Then divide both sides by 2: w = (P - 2l) / 2. Choice A forgets to divide by 2 after isolating the 2w term.

An absolute value equals zero only when the expression inside is exactly zero. Setting 2x + 4 = 0 gives x = -2, the unique solution. This contrasts with |2x + 4| = k for k > 0, which always yields two solutions.

OR means the union of both solution sets: every number at or below -3, plus every number at or above 7. Choice A describes the AND solution (-3 <= x <= 7). Choice D is wrong because a number like x = 2 satisfies neither condition.

Divide both sides by r (assuming r is not equal to 0): t = d / r. This is the distance-rate-time formula rearranged. Choice B inverts the variables incorrectly, and Choice A subtracts instead of divides.

The compact form -4 < x < 6 is shorthand for x > -4 AND x < 6, meaning both conditions are satisfied simultaneously. Choice A uses OR, which covers almost all real numbers. Choice B is the complement, describing numbers outside the interval.

Case 1: 2x - 3 = 11 gives 2x = 14, so x = 7. Case 2: 2x - 3 = -11 gives 2x = -8, so x = -4. Both solutions check out. Choice C has sign errors from misapplying the negative case.

Divide both sides by P: A / P = 1 + r. Subtract 1: r = (A / P) - 1. Choice B subtracts P directly without first dividing, which is algebraically incorrect. Choice A forgets to subtract 1 in the final step.

First inequality: 2x + 1 > 9 gives x > 4. Second: x - 4 < -6 gives x < -2. Since the connector is OR, take the union: x > 4 or x < -2. Choice B gives the intersection (AND), which here is empty because no number is simultaneously greater than 4 and less than -2.

Case 1: 3x + 6 = 15 gives 3x = 9, so x = 3. Case 2: 3x + 6 = -15 gives 3x = -21, so x = -7. Choice B correctly solves Case 1 but sets Case 2 equal to +15 again instead of -15.

Subtract 1 from all three parts: -6 <= 2x <= 6. Divide all parts by 2: -3 <= x <= 3. The inequality signs remain as <= throughout because no step reverses the inequality. Choice A incorrectly subtracts 1 from -5 to get -4 instead of -6.

Subtract mx from both sides: b = y - mx. Choice B incorrectly adds mx instead of subtracting. Choice D incorrectly separates m from x and divides by x, treating mx as if it were m plus x.

Case 1: x - 4 = 2x - 10 gives x = 6. Check: |6 - 4| = 2 and 2(6) - 10 = 2. Valid. Case 2: -(x - 4) = 2x - 10 gives x = 14/3. Check: 2(14/3) - 10 = -2/3, which is negative. Since an absolute value cannot equal a negative number, x = 14/3 is extraneous and rejected.

From 3x - 2 >= 7: 3x >= 9, so x >= 3. From x + 1 <= 6: x <= 5. The AND connector requires both conditions, giving the intersection: 3 <= x <= 5. Choice A uses OR logic and would cover nearly all real numbers.

Subtract cx from both sides: ax - cx = b. Factor: x(a - c) = b. Divide by (a - c): x = b / (a - c). The condition a is not equal to c ensures we never divide by zero. Choice A sums instead of subtracts in the denominator.

Case 1: 5 - x = 3 gives -x = -2, so x = 2. Case 2: 5 - x = -3 gives -x = -8, so x = 8. Both are valid. Choice C has a sign error on the first case; Choice B has a sign error on the second case.

Subtract 32 from both sides: F - 32 = (9/5)C. Multiply both sides by 5/9: C = (5/9)(F - 32). Choice D applies the factor 5/9 only to F and subtracts 32 separately, which is incorrect because the entire quantity (F - 32) must be multiplied by 5/9.

Case 1: 3x + 1 = x - 5 gives 2x = -6, so x = -3. Case 2: 3x + 1 = -(x - 5) gives 3x + 1 = -x + 5, then 4x = 4, so x = 1. Both solutions are valid. Choice D has sign errors from mishandling the negative case in Case 2.

Add 5 and divide by 2: |x + 3| >= 4. For |u| >= k with k > 0, the solution is u >= k OR u <= -k. So x + 3 >= 4 gives x >= 1, and x + 3 <= -4 gives x <= -7. These form a union. Choice A is the solution to |x + 3| <= 4, the less-than version.

Since |2x - 5| >= 0 for all real x, the expression |2x - 5| + 3 >= 3 for all real x. It can never be less than 0, so no value of x satisfies the inequality. Choice B is the solution to the related inequality |2x - 5| < 3, which is a solvable but different problem.

Add c to both sides: (b/a)x = c + d. Multiply both sides by a/b: x = (a/b)(c + d) = a(c + d) / b. Choice D swaps a and b in the fraction, and Choice B places both a and b in the denominator instead of moving a to the numerator.

For x > 8 AND x < 2, a number must simultaneously be greater than 8 and less than 2, which is impossible since those intervals do not overlap. Choice A (5 < x < 10) has infinitely many solutions. Choice D (OR) covers all real numbers because every number is either at least 0 or at most 1.

Since both sides are non-negative, square both sides: \((x - 5)^2 < (x + 3)^2\). Expanding gives \(x^2 - 10x + 25 < x^2 + 6x + 9\). Cancel \(x^2\) and simplify: \(16 < 16x\), so \(x > 1\). Geometrically, this asks when x is closer to 5 than to -3 on the number line, which occurs to the right of their midpoint: \((-3 + 5) / 2 = 1\).

The property |a| = a holds only when a >= 0. So |3x - 6| = 3x - 6 requires 3x - 6 >= 0, meaning x >= 2. For x < 2, the right side is negative while the left side is non-negative, making the equation false. The student's error was assuming |a| = a holds even when a is negative.

Absolute value measures the distance from zero on the number line and is always non-negative. |−7| = 7 because −7 is 7 units from zero. The answer −7 is wrong because absolute value is never negative; 49 confuses absolute value with squaring; 1/7 is the reciprocal and unrelated.

The equation |x| = 9 asks which numbers are exactly 9 units from zero. Both 9 and −9 satisfy this, so there are two solutions: x = 9 or x = −9. Choosing only one value misses the negative case. x = 81 confuses absolute value with squaring (9 squared is 81).

Solving x + 3 < 7 gives x < 4. Testing each option: 4 + 3 = 7, which is not strictly less than 7; 5 + 3 = 8 and 6 + 3 = 9, both greater than 7; only 3 + 3 = 6 < 7. x = 4 is a common trap because it satisfies x + 3 ≤ 7 but not the strict inequality.

To isolate r, divide both sides of d = rt by t, giving r = d/t. The choice r = d − t incorrectly subtracts t rather than dividing; r = dt multiplies when division is needed; r = t/d inverts the correct fraction.

Subtract 2L from both sides: P − 2L = 2W. Then divide both sides by 2: W = (P − 2L)/2. Choice A forgets to divide by 2 after subtracting. Choice C uses addition instead of subtraction, giving the wrong sign. Choice D multiplies P by 2 rather than dividing.

An absolute value equals zero only when the expression inside is exactly zero. Setting x − 3 = 0 gives x = 3, the only solution. Unlike the typical two-solution case, there is no positive value to assign ± here. Choosing x = −3 comes from incorrectly treating it like |x| = 3; x = 0 confuses the constant inside with the solution.

For an OR compound inequality, a value is a solution if it satisfies at least one condition. x = 4 satisfies x > 3, so it is in the solution set. x = 1 satisfies neither condition. x = −1 fails x < −1 because the inequality is strict (not ≤). x = 3 fails x > 3 for the same reason — a strict inequality does not include the boundary value.

Multiply both sides by 2 to clear the fraction: 2A = bh. Then divide both sides by b: h = 2A/b. Choice A forgets the multiplication by 2 step. Choice C multiplies A by b instead of dividing. Choice D inverts the relationship, placing b in the numerator.

Split into two cases. Case 1: 2x − 3 = 7 gives 2x = 10, so x = 5. Case 2: 2x − 3 = −7 gives 2x = −4, so x = −2. Both values satisfy the original equation. The distractor x = 2 or x = −5 comes from solving 2x = 3 ± 7 but dividing incorrectly, or from confusing the coefficient of x.

Apply inverse operations to all three parts simultaneously. Subtract 1 throughout: −4 < 4x < 12. Divide by 4: −1 < x < 3. Choice C shows the result before dividing by 4 — a common stopping-too-early error. Choice D results from dividing by 2 instead of 4. Choice B comes from subtracting incorrectly in the first step.

Subtract 3x from both sides: −2y = 12 − 3x. Divide by −2: y = (12 − 3x)/(−2) = (3x − 12)/2. Choice A forgets to divide by −2. Choice C keeps the wrong sign — dividing 12 − 3x by positive 2 ignores that −2y was divided. Choice D has an incorrect sign on the constant term.

Case 1: 4x − 8 = 12 gives 4x = 20, so x = 5. Case 2: 4x − 8 = −12 gives 4x = −4, so x = −1. Both solutions check out in the original equation. Choice C results from reading 4x = 20 as x = 4 rather than dividing by 4. Choice D reverses the signs in both solutions.

Subtract a from both sides: P − a = bt. Divide both sides by b: t = (P − a)/b. Choice A incorrectly applies the division only to P rather than the entire expression (P − a). Choice C multiplies by b instead of dividing. Choice D adds a to P rather than subtracting, producing the wrong sign.

For |expression| > k with k > 0, the solution is expression > k OR expression < −k. So x + 2 > 5 gives x > 3, and x + 2 < −5 gives x < −7. The complete solution is x > 3 or x < −7. Choice B is the solution to |x + 2| < 5 (the less-than case). Choice D is impossible since no number is both greater than 3 and less than −7 at the same time.

Solve each part separately. 2x − 1 > 5 gives 2x > 6, so x > 3. x + 4 < 2 gives x < −2. Because this is an OR compound inequality, the solution is the union: x > 3 or x < −2. Choice B is the intersection (AND), which would require both conditions to hold simultaneously. Choices C and D each omit one branch of the solution.

For |expression| ≤ k, rewrite as −k ≤ expression ≤ k. So −4 ≤ 2x − 6 ≤ 4. Add 6 to all parts: 2 ≤ 2x ≤ 10. Divide by 2: 1 ≤ x ≤ 5. Choice B is the solution to |2x − 6| ≥ 4 — the greater-than version flips to an OR inequality. Choice D results from adding 6 correctly but forgetting to divide by 2.

Add y to both sides: (2x)/3 = y + z. Multiply both sides by 3: 2x = 3(y + z). Divide both sides by 2: x = 3(y + z)/2. Choice D inverts the final step — it multiplies by 2 and divides by 3 instead of the reverse. Choice B divides by 6, which treats the coefficient as 2 times 3 rather than the fraction 2/3. Choice C forgets to divide by 2 in the last step.

Isolate the absolute value first. Subtract 1: 3|2x − 4| = 12. Divide by 3: |2x − 4| = 4. Case 1: 2x − 4 = 4 gives 2x = 8, so x = 4. Case 2: 2x − 4 = −4 gives 2x = 0, so x = 0. Choice B skips the division by 3 step, solving |2x − 4| = 12 instead. Choice D treats the ±4 as the final answer without solving for x.

Solve the absolute value inequality first: −3 ≤ x + 1 ≤ 3, so −4 ≤ x ≤ 2. Next, apply the AND condition x > 0 by intersecting: the overlap of −4 ≤ x ≤ 2 and x > 0 is 0 < x ≤ 2. Choice B gives only the absolute value solution without applying the second condition. Choice C incorrectly includes x = 0, which violates the strict inequality x > 0.

Subtract 1/b from both sides: (1/a)x = c − 1/b. Multiply both sides by a: x = a(c − 1/b), which expands to ac − a/b. Choice B has the wrong sign — it adds a/b rather than subtracting. Choice C divides by a rather than multiplying, reversing the operation. Choice D incorrectly distributes, treating the equation as if c and 1/b had been divided by a separately.

Square both sides (valid since both sides are non-negative): (2x + 1)² > (x − 2)². Expand: 4x² + 4x + 1 > x² − 4x + 4. Simplify: 3x² + 8x − 3 > 0. Using the quadratic formula, the roots are x = 1/3 and x = −3. Since the parabola opens upward, the quadratic is positive outside the roots: x < −3 or x > 1/3. Choice C is the solution to the less-than version. Choice B captures only the right branch.

Solve each part. 3(x − 2) > 9 simplifies to 3x − 6 > 9, giving x > 5. For −2x + 1 ≥ −5: subtract 1 to get −2x ≥ −6, then divide by −2 and flip the inequality: x ≤ 3. The AND condition requires x > 5 and x ≤ 3 simultaneously — no number satisfies both, so the solution is the empty set. A key error in Choice B is forgetting to flip the inequality when dividing by −2.

Multiply both sides by 2 to clear the fraction: 2S = n(a + l). Divide both sides by (a + l): n = 2S/(a + l). Choice B inverts the steps — multiplying S by (a + l) moves the factor to the wrong side. Choices C and D subtract (a + l) from both sides, but the variable n is being multiplied by (a + l), so division is the correct inverse operation.

When c > 0, the equation has two solutions: x = 4 + c and x = 4 − c. When c < 0, there is no solution because absolute value is never negative. When c = 0, the equation |x − 4| = 0 forces x − 4 = 0, giving exactly one solution: x = 4. So c = 0 is the only value producing a single solution. For example, c = 4 gives x = 8 and x = 0 — two distinct solutions.

Subtract 6 from both sides: −2|x − 1| ≤ −10. Divide both sides by −2 and flip the inequality sign: |x − 1| ≥ 5. For an absolute value greater-than-or-equal inequality: x − 1 ≥ 5 gives x ≥ 6, and x − 1 ≤ −5 gives x ≤ −4. The solution is x ≥ 6 or x ≤ −4. Choice B results from forgetting to flip the inequality when dividing by −2, which incorrectly treats ≤ as ≥ for the absolute value step.

Dividing both sides of A = lw by l gives w = A/l. Choice B multiplies instead of divides. Choice C subtracts l, which would only apply if the formula used addition. Choice D inverts the wrong variable.

Absolute value measures distance from zero. Both 9 and -9 are exactly 9 units from zero, so both satisfy the equation. For any equation |x| = c with c > 0, there are always exactly two solutions: x = c and x = -c.

'At least -3' means x ≥ -3, and 'at most 5' means x ≤ 5. Combined with 'and', this gives -3 ≤ x ≤ 5. Choice A describes values outside the interval. Choice C requires x to be simultaneously at most -3 and at least 5, which is impossible. Choice D excludes the endpoints -3 and 5 with strict inequalities.

Dividing both sides of C = 2πr by 2π isolates r: r = C / (2π). Choice A subtracts 2π instead of dividing. Choice B multiplies both sides by 2π rather than dividing. Choice D inverts the roles of C and π incorrectly.

The only number with absolute value equal to 0 is 0 itself, so x = 0 is the unique solution. This is a special boundary case: when c > 0, |x| = c has two solutions; when c = 0, there is one solution; when c < 0, there are no solutions.

An 'or' compound inequality (such as x < -2 or x > 3) combines solutions from two separate conditions using union, forming two disjoint intervals. An 'and' inequality takes the intersection, producing a single interval. A three-part inequality and |x| ≤ c also produce single connected intervals.

Multiply both sides by 2: 2A = (b₁ + b₂)h. Divide by h: 2A/h = b₁ + b₂. Subtract b₂: b₁ = (2A/h) - b₂. Choice B subtracts h instead of dividing by it. Choice C divides A by 2h rather than first multiplying A by 2 and then dividing by h. Choice D places the subtraction of b₂ inside the division, which is algebraically incorrect.

Subtract 1 from all three parts: -4 < 2x < 8. Divide all parts by 2: -2 < x < 4. Choice B results from subtracting 2 instead of 1. Choice C correctly subtracts 1 but forgets to divide by 2. Choice D drops the negative lower bound.

Case 1: 3x - 9 = 12 → 3x = 21 → x = 7. Case 2: 3x - 9 = -12 → 3x = -3 → x = -1. Both values check out. Choice B uses x = 1 for Case 2, forgetting to apply the negative. Choice C only solves 3x - 9 = 0. Choice D has an arithmetic error in Case 1.

Divide both sides by πr²: h = V / (πr²). Choice B multiplies both sides by πr² rather than dividing. Choice C subtracts, which would only be valid if the formula used addition. Choice D inverts the equation, effectively solving for 1/h.

Add 2 to all three parts: -3 ≤ 3x ≤ 9. Divide by 3: -1 ≤ x ≤ 3. Choice A adds 2 correctly but does not divide by 3. Choice C divides -3 by 3 incorrectly as -7/3. Choice D discards the negative lower bound entirely.

Rewrite as a three-part inequality: -3 < x - 5 < 3. Add 5 to all parts: 2 < x < 8. Choice A reverses the direction, which would be the solution to |x - 5| > 3. Choice B applies a sign error to the bounds. Choice D is also the 'greater than' case with shifted values.

Dividing both sides by (x - x₁) gives m = (y - y₁) / (x - x₁), the standard slope formula. Choice B swaps numerator and denominator, yielding 1/m instead of m. Choice C subtracts rather than divides. Choice D multiplies rather than divides.

Solve each part: 2x + 3 > 7 → 2x > 4 → x > 2. Also, 3x - 1 < -4 → 3x < -3 → x < -1. Because this is an 'or' inequality, the solution is the union: x > 2 or x < -1. Choice B is the intersection (and), which is empty here. Choice C uses 'and' incorrectly. Choice D describes the complement of the correct solution.

Absolute value represents distance, which is always non-negative. An equation setting an absolute value equal to a negative number, like |x + 4| = -2, has no solution. Choice A has solutions x = 8 and x = -2. Choice B has exactly one solution at x = -1/2. Choice D has solutions x = 3 and x = -3.

Subtract 32 from both sides: F - 32 = (9/5)C. Multiply both sides by 5/9: C = (5/9)(F - 32). Choice A adds 32 instead of subtracting before scaling. Choice C multiplies by 9/5 rather than its reciprocal 5/9. Choice D only partially applies the 5/9 factor, distributing it incorrectly.

Rewrite as a three-part inequality: -3 ≤ 2x + 5 ≤ 3. Subtract 5: -8 ≤ 2x ≤ -2. Divide by 2: -4 ≤ x ≤ -1. Choice B is the solution to the 'greater than or equal to' case, |2x + 5| ≥ 3. Choice C results from ignoring that the bounds are negative. Choice D uses strict inequalities, incorrectly excluding the endpoints.

Solve each part. First: (x/2) + 1 > 3 → x/2 > 2 → x > 4. Second: (x/3) - 1 < -2 → x/3 < -1 → x < -3. The union is x > 4 or x < -3. Choice B uses positive 3 for the second bound, ignoring the negative. Choice C describes the region between the two bounds. Choice D multiplies the first part by 3 instead of 2.

Case 1: 3x - 2 = x + 4 → 2x = 6 → x = 3. Check: |7| = 7 and 3 + 4 = 7. Case 2: 3x - 2 = -(x + 4) → 3x - 2 = -x - 4 → 4x = -2 → x = -1/2. Check: |-7/2| = 7/2 and -1/2 + 4 = 7/2. Both solutions are valid. Choice D arises from not distributing the negative sign in Case 2, giving 4x = -6 incorrectly.

Subtract 7 from all parts: -9 < -3x ≤ 6. Divide by -3 and flip both inequality signs: 3 > x ≥ -2. Rewritten in standard form: -2 ≤ x < 3. The critical step is reversing both inequalities when dividing by a negative number. Choice C loses the equality on the left side. Choice D makes both endpoints closed, ignoring the strict left inequality. Choice B isolates only part of the solution.

Multiply both sides by (x + b): x - a = c(x + b) = cx + bc. Gather x terms on the left: x - cx = a + bc. Factor: x(1 - c) = a + bc. Divide by (1 - c), valid since c ≠ 1: x = (a + bc) / (1 - c). Choice B has a sign error when distributing c. Choice C rearranges the numerator and denominator inconsistently. Choice D forgets to divide by (1 - c).

Add 5 to both sides: 2|x + 3| = 14. Divide by 2: |x + 3| = 7. Case 1: x + 3 = 7 → x = 4. Case 2: x + 3 = -7 → x = -10. Choice D results from solving |x + 3| = 5 instead of 7, a common error from dividing 9 by 2 before adding 5.

The system requires a number that is simultaneously greater than k and less than k - 1. Since k - 1 is always less than k for every real k, no such number exists. The intersection of x > k and x < k - 1 is always empty, regardless of k. Choices A, B, and D incorrectly suggest the system has solutions for some specific values of k.

Multiply all three parts by 3 (positive, so no sign flip): -9 ≤ 2x + 1 < 15. Subtract 1: -10 ≤ 2x < 14. Divide by 2: -5 ≤ x < 7. Choice B results from adding 1 instead of subtracting before dividing. Choice C reverses which endpoint is strict. Choice D correctly subtracts 1 and gets -10 ≤ 2x < 14, but forgets the final step of dividing by 2.

Case 1: 2x - 1 > x + 2 → x > 3. Case 2: -(2x - 1) > x + 2 → -2x + 1 > x + 2 → -1 > 3x → x < -1/3. The union is x < -1/3 or x > 3. Choice A uses -1 instead of -1/3 in Case 2, an error from writing -1 > 3x as x < -1. Choice B lists the correct bounds but reverses their positions. Choice C describes the complement of the correct answer.