Absolute value equations and inequalities, in context
Grade 11 Math Worksheet - Challenge Me

Circle the best answer.
Bo turns a line y = mx about the origin and counts where it meets the V y = |x − 2|. For which slopes m are there exactly two solutions?
- 0 < m < 1: it cuts both arms, which have slopes ±1
- −1 < m < 1: any line flatter than the arms
- m > 1: a steep line climbs past both arms
- m > 0: each branch gives an x for a positive slope
Esme solves |x² − 2x − 3| = x + 1 by splitting it into branches and checking each answer. How many solutions are there?
- Two, x = 2 and 4: x = −1 gives 0 = 0, which does not count
- Two, x = −1 and 4: the other branch has no real roots
- Four: each branch gives two roots
- Three: x = −1, 2 and 4, with −1 from both branches
Jun counts solutions of |x − 1| − |x − 5| = k: the distance to 1 minus the distance to 5. Which counts for k = 2, 4 and 6 are right?
- One, infinitely many (every x ≥ 5) and none
- One, one and none: k = 4 is reached at x = 5 alone
- One, infinitely many and one
- Two, one and none: two branches for k = 2
Clients sit at miles 1, 4 and 10. Kira puts a depot at x to make |x − 1| + |x − 4| + |x − 10| as small as possible. Where, and what total?
- x = 4, the middle client: total 9
- x = 5.5, halfway from 1 to 10: total 10.5
- Any x from 1 to 10, since |x − 1| + |x − 10| = 9 there
- x = 5, the mean of 1, 4 and 10: total 10
Luz needs the solution set of |x − 2| + |x − 8| ≤ c to be a stretch of the number line 10 units long. Which c does it?
- c = 10: the set runs from 5 − c/2 to 5 + c/2
- c = 5: half of 10 on each side of the middle
- Any c ≥ 6, since the set is then a stretch
- c = 16: the gap 6 plus the length 10
Cable to a shed at 1 costs $1 per m and to a barn at 7 costs $2 per m. Where should Milo put x to make |x − 1| + 2|x − 7| least?
- Anywhere from 1 to 7, since the cost stays $6 on that stretch
- At x = 7: left of 7 the cost falls by $1 per meter moved
- At x = 4, the midpoint of 1 and 7
- At x = 5, two thirds of the way toward the barn
A stop x needs |x − 2| + |x − 8| ≤ 8, near two schools, and |x − 9| ≥ 2, off a crossing. Which marks does Noah allow?
- 1 ≤ x ≤ 9
- 2 ≤ x ≤ 7
- 1 ≤ x ≤ 7 or x ≥ 11
- 1 ≤ x ≤ 7
Read this first
Branch 1
2x − 1 > x + 4, so x > 5. Branch 2: 2x − 1 < −x + 4, so x < 5/3. Answer: x < 5/3 or x > 5.
Ren checks this worked solution of |2x − 1| > x + 4. Which line holds the first error?
- Branch 2: it must be 2x − 1 < −(x + 4), giving x < −1
- Branch 1: it should be 2x − 1 < x + 4
- No line: x = 6 passes, since 11 > 10
- Answer: the branches should be joined with and: 5/3 < x < 5
Tala claims y = |f(x)| and y = f(|x|) cross the x-axis at the same points, for f(x) = x² − 2x − 3 = (x − 3)(x + 1). Which verdict is right?
- False: f(|x|) has zeros at ±1 and ±3
- True: bars do not change where a graph meets zero
- False: f(|x|) = 0 needs |x| = 3, so x = ±3; −1 drops out
- False: |f(x)| has no zeros, since its outputs are positive
Step 1
For x ≥ 5, 2x − 6 = 6, so x = 6. Step 2: for x ≤ 1, 6 − 2x = 6, so x = 0. Step 3: for 1 < x < 5, the total is 4, so none. Answer: x = 0 or x = 6.
Zee checks a worked solution of |x − 1| + |x − 5| = 6. What is the verdict?
- No error: the spare 2 is shared, 1 past each end, giving 0 and 6
- Steps 1 and 2: the answers are 3 ± 6, so −3 and 9
- Step 3: x = 3, the midpoint, also works
- Steps 1 and 2: the spare 2 goes past each end, so −1 and 7