Free video lesson

How to Solve Compound Inequalities on the SAT

"At least 10 less than 40" is 30, and getting that number right is only half the job. The sentence carried a second phrase, and the answer that leaves it out agrees with the correct one everywhere except the place it matters.

Math · Linear inequalities1:59Published September 11, 2026

On YouTube: One Bound Is Not an Answer | SAT Compound Inequalities

What this lesson covers

  • "A less than B" is B minus A, so the phrase order reverses the subtraction
  • Turning two phrases into two bounds, then chaining them into one statement
  • Testing a value that passes one bound and fails the other
  • Operating on a chain: all three parts, and what a negative does to both ends

Questions worked in the video

  1. 0:14Student tickets s are at least 10 less than adult tickets a, and no more than a. If 40 adult tickets are sold, which describes all possible values of s?

Worked examples

The questions the video works, written out: the setup, each step, the answer and the trap.

0:14Example 1

A school play sells adult tickets and student tickets. The number of student tickets s is at least 10 less than the number of adult tickets a, and s is no more than a. If 40 adult tickets are sold, which describes all possible values of s? A) s≤30s \le 30 B) 30≤s≤4030 \le s \le 40 C) −30≤s≤40-30 \le s \le 40 D) s≥30s \ge 30

  1. Two phrases in the sentence means two bounds, and the answer needs both of them. Substitute a = 40 first.
  2. First phrase: s is at least 10 less than a. Inside a phrase, 10 less than 40 is 40 - 10, not 10 - 40; the order on the page is the reverse of the subtraction. So 10 less than 40 is 30, and at least 30 gives s≥30s \ge 30. This is the floor.
  3. Second phrase: s is no more than a. No more than 40 gives s≤40s \le 40. This is the ceiling.
  4. Chain them into one statement: 30≤s≤4030 \le s \le 40. That is 11 whole-number values, 30 through 40.
  5. Check by running values against both ends. s = 45 clears the floor (45 is at least 30) but breaks the ceiling (45 is more than 40), so any answer carrying only the floor is wrong. s = 25 breaks the floor. s = -5 would be a negative ticket count, and only choice C admits it.

Answer: B) 30≤s≤4030 \le s \le 40

Since 10 less than 40 is 30 and s cannot exceed the 40 adult tickets, s is trapped between 30 and 40 inclusive. Choice D is the star distractor: it is right about everything it says and agrees with B at every value up to 40, but it admits s = 45, which is impossible. Choice C comes from computing 10 - 40 = -30, and choice A comes from flipping at least into a ceiling; s = 25 exposes it.

1:17Example 2

Solve the chained inequality 2≤x+5≤92 \le x + 5 \le 9, then write the resulting range for −2x-2x.

  1. A chain has three parts, and whatever you do to one part you do to all three. Subtract 5 everywhere: 2−5≤x≤9−52 - 5 \le x \le 9 - 5, so −3≤x≤4-3 \le x \le 4.
  2. Multiply all three parts by -2. Multiplying by a negative flips both inequality signs: 6≥−2x≥−86 \ge -2x \ge -8.
  3. Rewrite with the smaller number on the left so the chain reads left to right: −8≤−2x≤6-8 \le -2x \le 6. Both signs flipped and the two ends swapped places.
  4. Check with x = 0, which sits inside -3 to 4: -2(0) = 0, and -8 is at most 0, which is at most 6, true. The unswapped form 6≤−2x≤−86 \le -2x \le -8 would need 6 to be at most -8, which no number satisfies.

Answer: −3≤x≤4-3 \le x \le 4, and −8≤−2x≤6-8 \le -2x \le 6

Subtracting from all three parts keeps the chain balanced, and multiplying by a negative reverses both signs, which is why the ends trade places. The trap is flipping the signs but leaving the ends in their old order, producing a chain that no number satisfies; a second trap is subtracting 5 from only the middle part.

Chapters

Lesson transcript

The narration of the video, word for word, under its chapter headings.

0:14The setup

A school play sells adult tickets and student tickets. The student count is at least ten less than the adult count, and no more than the adult count. Forty adult tickets sold. Which range describes the student tickets?

0:30Two phrases, two bounds

Two phrases, so two bounds, and you need both of them. Ten less than forty is thirty, not negative thirty. Less than inside a phrase is subtraction, and the order reverses. At least thirty gives s is greater than or equal to thirty. No more than forty gives s is less than or equal to forty. Chain them together. Now run the choices against both ends. Forty five clears the floor and blows straight through the ceiling, so any answer carrying only one bound is out. Twenty five sits under the floor, so the flipped version is out. And negative thirty would be a negative number of tickets. Only the closed range from thirty to forty survives, and that is eleven whole numbers.

1:17Three ways this goes wrong

Three ways this goes wrong. One, solving one bound, circling it, and never writing the other. Thirty or more on its own would allow a hundred students out of forty adults. Two, reading ten less than forty as ten minus forty. In a phrase, the number after less than goes second. Three, multiplying a chain by a negative and keeping the order. Both signs flip and the two ends swap.

1:42Write both

Two phrases means two bounds, and both of them have to get written down.

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