A student solving |3x + 6| − 4 = 8 immediately writes: 3x + 6 = 12 and 3x + 6 = −12. What error did they make?
AThey set up the wrong pair of cases — they should use +8 and −8, not +12 and −12
BThey failed to isolate the absolute value first; they should add 4 to both sides to get |3x + 6| = 12 before splitting
CThey should only write one case, since the equation has a unique solution
DThere is no error; this setup correctly leads to the right answers
The rule is to isolate the absolute value expression before splitting into cases. With |3x + 6| − 4 = 8, the −4 is outside the bars. Splitting immediately — as if the equation were |3x + 6| = 8 — gives the wrong right-hand side. You must first add 4 to both sides to get |3x + 6| = 12, and only then write 3x + 6 = 12 and 3x + 6 = −12. The student's equations happen to use ±12 but for the wrong reason — they derived it by folding in the 4, which is precisely the procedure they should have done explicitly.
Question 2 Multiple Choice
What is the solution to |x − 5| = −3?
Ax = 2 and x = 8
Bx = 8 only
CNo solution — absolute value cannot equal a negative number
Dx = −2 and x = −8
Absolute value measures distance from zero, and distance is never negative. For any value of x, |x − 5| ≥ 0 always. No number can be −3 units from zero, so there is no solution. You don't need to split into cases — you should recognize immediately that the right side is negative and conclude no solution exists. This saves work and builds the mathematical instinct that absolute value equations with negative right-hand sides are always unsolvable.
Question 3 True / False
If the right side of an absolute value equation equals zero, there is no solution.
TTrue
FFalse
Answer: False
When |expression| = 0, there is exactly one solution: the value that makes the expression equal to zero. For example, |x + 2| = 0 has the solution x = −2. Zero is not negative — absolute value can equal zero when the expression inside evaluates to zero. The 'no solution' rule only applies when the right side is strictly negative. Confusing |expression| = 0 with |expression| = negative is a subtle but consequential error.
Question 4 True / False
To correctly solve |5x − 2| + 3 = 11, you must subtract 3 from both sides before splitting into two cases.
TTrue
FFalse
Answer: True
The +3 is outside the absolute value bars, so it must be removed before the two-case split. Subtracting 3 from both sides gives |5x − 2| = 8, which then splits into 5x − 2 = 8 and 5x − 2 = −8. If you split too early with the +3 still present, you get the wrong right-hand side values (±11 instead of ±8) and both solutions will be incorrect. 'Isolate first, split second' is the non-negotiable order of operations.
Question 5 Short Answer
Explain why solving |2x − 3| = 7 requires two separate equations rather than one, connecting your explanation to the definition of absolute value as distance.
Think about your answer, then reveal below.
Model answer: Absolute value measures distance from zero on the number line. The equation |2x − 3| = 7 asks: for what values of x is the expression (2x − 3) exactly 7 units from zero? Two numbers are 7 units from zero: +7 and −7. Therefore the expression inside can equal either +7 or −7, producing two equations: 2x − 3 = 7 (giving x = 5) and 2x − 3 = −7 (giving x = −2). Considering only one case misses half the solutions by ignoring the symmetry of distance around zero.
The two-case technique is not an arbitrary algebraic rule — it follows directly from what absolute value means. Because distance has two directions (positive and negative from zero), any equation setting a distance equal to a positive number has two candidate expressions. This 'case analysis' thinking — identifying a condition with two possibilities and solving each — is a pattern that transfers to piecewise functions, inequalities, and proof techniques encountered throughout later mathematics.