11th grade Absolute value equations and inequalities, in context worksheet

Absolute value equations and inequalities, in context

Grade 11 Math Worksheet - Challenge Me

Circle the best answer.

1. 1.

   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?

   • m > 0: each branch gives an x for a positive slope
   • m > 1: a steep line climbs past both arms
   • 0 < m < 1: it cuts both arms, which have slopes ±1
   • −1 < m < 1: any line flatter than the arms
2. 2.

   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?

   • Two, one and none: two branches for k = 2
   • One, infinitely many and one
   • One, one and none: k = 4 is reached at x = 5 alone
   • One, infinitely many (every x ≥ 5) and none
3. 3.

   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?

   • Any x from 1 to 10, since |x − 1| + |x − 10| = 9 there
   • x = 5, the mean of 1, 4 and 10: total 10
   • x = 5.5, halfway from 1 to 10: total 10.5
   • x = 4, the middle client: total 9
4. 4.

   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?

   • Any c ≥ 6, since the set is then a stretch
   • c = 5: half of 10 on each side of the middle
   • c = 16: the gap 6 plus the length 10
   • c = 10: the set runs from 5 − c/2 to 5 + c/2
5. 5.

   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?

   • At x = 7: left of 7 the cost falls by $1 per meter moved
   • Anywhere from 1 to 7, since the cost stays $6 on that stretch
   • At x = 5, two thirds of the way toward the barn
   • At x = 4, the midpoint of 1 and 7
6. 6.

   A stop x needs |x − 2| + |x − 8| ≤ 8, near two schools, and |x − 9| ≥ 2, off a crossing. Which marks does Noah allow?

   • 2 ≤ x ≤ 7
   • 1 ≤ x ≤ 9
   • 1 ≤ x ≤ 7
   • 1 ≤ x ≤ 7 or x ≥ 11

7. Read this first

   Line 1

   Square both sides: (x − 3)² = (x − 5)². Line 2: x² − 6x + 9 = x² − 10x + 25. Line 3: 4x = 16, so x = 4. Line 4: the solution is x = 4.

8. 7.

   Pia checks this worked solution of |x − 3| = x − 5. Which line holds the first error?

   • Line 1: squaring both sides of a bars equation is not allowed
   • No line: squaring keeps the solutions, so x = 4 stands
   • Line 3: 4x = 16 should give x = −4
   • Line 4: x = 4 gives 1 on the left and −1 on the right: no solution

9. 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.

10. 8.

    Ren checks this worked solution of |2x − 1| > x + 4. Which line holds the first error?

    • No line: x = 6 passes, since 11 > 10
    • Branch 2: it must be 2x − 1 < −(x + 4), giving x < −1
    • Branch 1: it should be 2x − 1 < x + 4
    • Answer: the branches should be joined with and: 5/3 < x < 5
11. 9.

    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|) = 0 needs |x| = 3, so x = ±3; −1 drops out
    • True: bars do not change where a graph meets zero
    • False: f(|x|) has zeros at ±1 and ±3
    • False: |f(x)| has no zeros, since its outputs are positive

12. 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.

13. 10.

    Zee checks a worked solution of |x − 1| + |x − 5| = 6. What is the verdict?

    • Step 3: x = 3, the midpoint, also works
    • Steps 1 and 2: the spare 2 goes past each end, so −1 and 7
    • Steps 1 and 2: the answers are 3 ± 6, so −3 and 9
    • No error: the spare 2 is shared, 1 past each end, giving 0 and 6