Two boundaries, one x — whatever you do, do it to all three parts.
Solve for x
−2 < 3x + 1 ≤ 10
Do the same thing to all three parts. Subtract 1: −3 < 3x ≤ 9. Divide by 3: −1 < x ≤ 3. The strict end stays on the left where it started — B swaps which end is strict, and C forgets to divide.
3x + 1 is trapped between two numbers at once.
−2 < 3x + 1 ≤ 10
lower wall · the middle · upper wall
greater than −2
−2 itself is NOT allowed
at most 10
10 itself IS allowed
The line under ≤ is the words “or equal to”. Keep each sign exactly as given — they decide the dots at the very end.
The middle has to end up as just x.
−2 < 3x + 1 ≤ 10
Missing one part is the classic slip. −2 − 1 = −3 and 10 − 1 = 9, so both walls drop by 1.
Dividing undoes the multiplying.
−3 < 3x ≤ 9
We divided by +3, so the signs stay put. Divide or multiply by a negative and both signs must flip round.
The dots do the talking: open means excluded.
for < and >
for ≤ and ≥
The gold band holds every number between the walls — fractions and decimals too, not only whole numbers.
Same answer, shorter notation.
−1 < x ≤ 3
(−1, 3]
round = excluded · square = included
| Sign | Dot | Bracket |
|---|---|---|
| < or > | open | ( or ) |
| ≤ or ≥ | closed | [ or ] |
Infinity always gets a round bracket.
How many integers satisfy −4 < 2x − 2 ≤ 7? Solve, then list — do not subtract the ends.
−2 < 2x ≤ 9
+ 2 to all three parts
−1 < x ≤ 4.5
÷ 2
0, 1, 2, 3, 4
−1 is excluded; 4.5 is not an integer
When x appears in all three parts you cannot isolate it in the middle. Split into two inequalities and keep the overlap. Solve x − 2 < 2x + 1 ≤ x + 6.
x − 2 < 2x + 1 → −3 < x
left pair
2x + 1 ≤ x + 6 → x ≤ 5
right pair
−3 < x ≤ 5
both must hold
4.5 − (−1) = 5.5 tells you nothing about how many integers fit. List them from the first allowed integer to the last, and check each end against its sign.
Solve 2x + 1 < −3 or 3x − 2 ≥ 7.
2x < −4 → x < −2
first part alone
3x ≥ 9 → x ≥ 3
second part alone
x < −2 or x ≥ 3 = (−∞, −2) ∪ [3, ∞)
“or”: keep everything either allows
Different: an “or” is two separate solves, and the graph is two rays pointing away from each other — never squeeze it into one sandwich.
Writing the two pieces as one chain claims x is below −2 and at least 3 at the same time. No number does both; an “or” answer must keep its “or”.
Solve x + 4 < 1 and 2x > 6.
x < −3 and x > 3
each part solved
no number is in both
no solution (∅)
Different: “and” keeps only the overlap, and two rays pointing apart have none. (With “or”, x < 5 or x > 1 would cover every real number.)
Solve −3 ≤ 1 − 2x3 < 1.
−9 ≤ 1 − 2x < 3
× 3, all three parts
−10 ≤ −2x < 2
− 1
5 ≥ x > −1, so −1 < x ≤ 5
÷ (−2): flip both, then reorder
Different: clear the fraction first (a positive 3, no flip), then the negative x term does the flipping. Check: x = 5 gives −3, allowed.
Start from the range of x and apply the same steps the expression does. If −1 ≤ x ≤ 4, what is the range of 3 − 2x?
−8 ≤ −2x ≤ 2
× (−2): flip, then reorder low to high
−5 ≤ 3 − 2x ≤ 5
+ 3; ends: x = 4 gives −5, x = −1 gives 5 ✓
Two variables: if 1 < x < 6 and −2 < y < 3, find the range of x − y. Turn the subtraction into adding −y.
−3 < −y < 2
× (−1) flips and swaps the ends
1 + (−3) < x − y < 6 + 2
add lows to lows, highs to highs
−2 < x − y < 8
smallest: small x, big y
1 − (−2) = 3 and 6 − 3 = 3 would give “3 < x − y < 3”, which is impossible. Inequalities may be added, not subtracted: negate first, then add.
Compound inequalities show up as a solve, a count of integers, a number line to match, or the range of an expression like 5 − 2x or x − y. Do the same thing to every part, flip both signs for a negative, and check the two ends by putting the extreme values back in.
The open dot at −1 is exactly right.
Solve for x
1 ≤ 4 − 2x < 9
Subtract 4 to get −3 ≤ −2x < 5, then divide by −2 — and dividing by a negative flips both signs: 32 ≥ x > −52, which reads as −52 < x ≤ 32. C is what you get without flipping.
Hint: the x term is negative here. When you divide by −2, both signs flip and the two ends swap places.
