Linear inequalities on the SAT

Example 1: dividing by a negative

Problem Solve $5 - 3x \le 17$.

1. Subtract 5 from both sides.
   $$-3x \le 12$$
2. Divide by $-3$. Dividing by a negative flips the sign.
   $$x \ge -4$$
3. Check a value that should work, $x = 0$: $5 \le 17$ is true. Check one that should not, $x = -5$: $20 \le 17$ is false. Good.

Answer $x \ge -4$

Example 2: a word problem where you round down

Problem A delivery van can carry at most 2,000 pounds. The driver weighs 180 pounds, and each box weighs 40 pounds. What is the greatest number of boxes the van can carry with the driver?

1. Let b be the number of boxes. Total weight is at most 2,000.
   $$180 + 40b \le 2000$$
2. Subtract 180.
   $$40b \le 1820$$
3. Divide by 40.
   $$b \le 45.5$$
4. Boxes come in whole numbers, and 46 would be too heavy. Round down to 45. Check: $180 + 40(45) = 1980$, which is under 2,000.

Answer 45 boxes

Example 3: a point in a system of inequalities

Problem The point $(3, b)$ satisfies both $y \ge x + 2$ and $y < -2x + 12$. If b is an integer, what is b?

1. Put $x = 3$, $y = b$ into the first inequality.
   $$b \ge 3 + 2$$
2. Put the same point into the second inequality.
   $$b < -2(3) + 12$$
3. So $b \ge 5$ and $b < 6$. The only integer in that range is 5.

Answer $b = 5$

Example 4 (SAT-hard): two conditions in one context

Problem Jada has $60 to spend on notebooks that cost $4 each and pens that cost $1.50 each. She needs at least 8 notebooks. What is the greatest number of pens she can buy?

1. Write both conditions. Let n be notebooks and p be pens.
   $$4n + 1.5p \le 60$$
2. Pens are greatest when she buys the fewest notebooks she is allowed, so $n = 8$.
   $$32 + 1.5p \le 60$$
3. Subtract 32.
   $$1.5p \le 28$$
4. Divide by 1.5. That is about 18.67.
   $$p \le \frac{56}{3}$$
5. Round down: 19 pens would cost $32 + 28.50 = 60.50$, which is too much. 18 pens cost $32 + 27 = 59$.

Answer 18 pens