A compound inequality gives two comparisons that must be true at the same time. Solve the left and right comparisons separately, then keep only the values allowed by both. Watch for a sign flip when dividing by a negative number. Desmos can graph both comparisons to check where their shaded regions overlap.
Hints
- Hint 1
A compound inequality joins two comparisons with and: the middle expression must be greater than the left expression and no greater than the right expression. Which comparison gives a lower limit on ?
- Hint 2
In the left comparison, isolating leaves it multiplied by a negative number. Dividing by that number reverses the inequality sign. Which side of the resulting cutoff is allowed?
- Hint 3
The right comparison gives an upper limit, and its sign allows equality. Once you have both limits, keep values that meet both conditions, not values that meet either one.
Step-by-step
Solve both comparisons and check their overlap
Step 1Start the left comparison
The chain requires both comparisons to hold. Start with . Subtract from both sides:
- Step 2
Collect the terms with k
Subtract from both sides, keeping the strict inequality:
- Step 3
Find the lower limit
Divide by . Dividing by a negative reverses the order, so flip the sign:
The cutoff itself is excluded because the comparison is strict.
- Step 4
Start the right comparison
Now use the other requirement, . Subtract from both sides:
- Step 5
Collect the terms with k
Add to both sides. Addition does not reverse the inequality:
- Step 6
Find the upper limit
Divide by positive , so the sign stays the same:
The upper cutoff is included because the original comparison allows equality.
- Step 7
Keep the overlap
Type both comparisons in Desmos, using in place of because Desmos graphs inequalities using . Their shaded regions overlap between the two cutoffs. Both comparisons must hold, so keep only their overlap. The lower end is excluded and the upper end is included, giving all possible values . Choice B.