PSAT Linear Inequalities: At Least, At Most, and Limits

For a linear-inequality word problem, define the unknown, translate the limit, and solve while keeping the same set of allowed values. Reverse the inequality when multiplying or dividing both sides by a negative number. If the unknown counts objects, finish by checking which whole numbers satisfy the original limit.

College Board’s Assessment Framework for the Digital SAT Suite includes linear inequalities in the PSAT/NMSQT Algebra domain. This guide focuses on one-variable problems. Every example below is original; the prices and situations are hypothetical, with no extra taxes or fees.

Decide whether the boundary is allowed

Use this table when a quantity $n$ is compared with $12$. Ask whether exactly $12$ would satisfy the sentence before choosing the symbol.

Write the whole quantity being limited on one side. In a budget problem, that usually means total cost, including any fixed charge. Label the variable with its units and decide whether it can be fractional.

A maximum: how many kits fit the budget?

A school club has $$94$ for a workshop. There is an $$18$ setup charge and each kit costs $$7$. What is the greatest number of complete kits the club can buy?

Let $n$ be the number of kits, a nonnegative whole number. Total cost must be no more than $94$ dollars.

$$18+7n\le94$$

Subtract $18$ from both sides, then divide by positive $7$. The inequality keeps its direction: $7n\le76$, so $n\le\frac{76}{7}$.

Since $\frac{76}{7}$ lies between $10$ and $11$, the greatest allowed whole number is $10$. Check both the proposed maximum and the next count:

• Ten kits cost $18+7(10)=88$ dollars, within the budget.
• Eleven kits cost $18+7(11)=95$ dollars, exceeding the budget.

The model permits whole-number counts from $0$ through $10$, and the question asks for the greatest. Rounding the fractional boundary to the nearest whole number would give $11$, which fails the original condition.

A minimum: how many tickets reach the target?

A club has already collected $$37$ and receives $$5$ for each additional ticket sold. It wants to collect at least $$120$ in total. What is the smallest number of additional tickets it must sell?

Let $t$ count additional tickets. The model is $37+5t\ge120$. Subtracting $37$ and dividing by $5$ gives:

$$t\ge\frac{83}{5}=16.6$$

The smallest allowed whole number is $17$. Selling $16$ tickets gives $37+5(16)=117$ dollars, below the target. Selling $17$ gives $37+5(17)=122$ dollars, which reaches it.

The kit problem and ticket problem require different choices at a fractional boundary. Choose the whole number that still satisfies the inequality. OpenStax’s applications lesson explains why context determines that choice.

Reverse the sign when dividing by a negative

Solve $17-4x>5$ for real values of $x$. Subtract $17$ from both sides to get $-4x>-12$. Dividing both sides by $-4$ reverses the inequality:

$$x<3$$

Check in the original expression: $x=2$ gives $17-4(2)=9>5$, while $x=4$ gives $1>5$, which is false. The boundary $x=3$ gives $5>5$, also false, so it is excluded.

Why does the direction reverse? Start with $2<7$. Multiplying both sides by $-1$ produces $-2>-7$, because the order of the two numbers reverses. The operation is explained in OpenStax’s linear-inequality lesson.

Simply getting negative numbers does not trigger a reversal. Subtracting $9$ from both sides of $2<7$ gives $-7<-2$, with the same direction. The trigger is multiplying or dividing both sides by a negative quantity.

If keeping operations balanced is still difficult, work through the linear-equation guide first, then apply the additional sign rule here.

Two limits must hold at the same time

A display needs at least $26$ but no more than $42$ cards. A student already has $7$ cards and can add full packs of $5$, using every card in each pack. Which numbers of packs are allowed?

Let $p$ be the nonnegative whole number of packs. Both limits apply to the total, $7+5p$:

$$26\le7+5p\le42$$

Subtract $7$ from all three parts, then divide every part by positive $5$. This gives $19\le5p\le35$, then $\frac{19}{5}\le p\le7$.

Because $\frac{19}{5}=3.8$, the allowed pack counts are $4,5,6,7$. Four packs give $27$ cards and seven packs give $42$; three give only $22$, and eight give $47$.

The answer must satisfy both limits. Finding a count that meets the minimum is only part of the check.

A short check before you submit

• Meaning: Does the inequality match the quantity and limit in the sentence?
• Equality: Does the wording allow the boundary itself?
• Operations: Did you reverse the direction exactly when multiplying or dividing by a negative?
• Context: Is the result an allowed count or measurement? For a whole-number minimum or maximum, check the adjacent count too.

Try four original questions

1. An order has a $$12$ delivery charge plus $$4$ per notebook. With a budget of $$58$, what is the maximum number of whole notebooks you can order?
2. A team has $14$ bottles and needs at least $65$. New bottles come in cases of $6$. What is the minimum number of full cases to buy?
3. Solve $25-3x\ge7$ for real values of $x$. Is the boundary included?
4. A drawer contains $8$ stickers. You add complete bundles of $4$ stickers and want fewer than $36$ stickers in the drawer. What is the maximum number of bundles you can add?

When an answer fails, record whether the problem was translating the words, handling a negative divisor, including the boundary, or choosing a whole number. The practice error-log template helps turn that observation into a specific correction.

Try a free PSAT practice question with Kibo when you are ready to continue.

Prepared with AI assistance. Original examples were checked with separate calculations; no human review is claimed. Sources checked October 26, 2026. Kibo is not affiliated with or endorsed by College Board or National Merit Scholarship Corporation.