Solve one-variable inequalities

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Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 28 minutes
Domains: Algebra
Techniques: Inequality-reversalCompound-inequalitiesInterval-endpointsInteger-bounds

What you’ll learn

Spot when one quantity can take a whole range of values.
Solve a linear inequality and explain why multiplying or dividing by a negative flips the sign.
Read strict, inclusive and compound intervals.
Let Desmos work out a messy cutoff, then pick the right whole-number minimum or maximum yourself.

Prerequisites

Solve linear equations0%
Use equivalent algebraic steps to isolate one unknown before tracking how those steps affect an inequality.

Why this matters on the SAT

Find the last value that still works

Lots of SAT questions ask how long, how many or how much something can go before it hits a limit. An inequality answers three questions about that limit. Where is the boundary? Which side of it is allowed? Does the boundary itself count?

When only whole numbers make sense, those three answers tell you which whole number to pick. Here's a typical one.

SAT example

A tank contains $130$ liters of water. A pump adds water at a constant rate of $18$ liters per minute. The tank can hold no more than $425$ liters.

What is the greatest whole number of minutes the pump can operate without the tank exceeding its capacity?

A
$15$
B
$16$
C
$17$
D
$18$

Solution to the example

Let $m$ be the number of minutes. The tank starts at $130$ liters and gains $18$ liters each minute, and it can't go over $425$ , so

130+18m\le425.

Subtract $130$ from both sides, then divide by $18$ :

\begin{aligned} 18m&\le295\\[1.4em] m&\le\frac{295}{18}. \end{aligned}

That boundary is about $16.39$ minutes, and the $\le$ allows every time at or below it. The greatest whole number in that range is $16$ , so the answer is B. Check it: after $16$ minutes the tank holds $418$ liters, but after $17$ minutes it would hold $436$ , which is over capacity.

In the graph, $x$ is minutes and $y$ is liters. The line $y=130+18x$ is the water in the tank, and $y=425$ is the capacity. They meet at the boundary. The third expression works out that boundary's awkward $x$ -coordinate, and the labeled point shows where the decimal sits on the graph.

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The lines cross at the boundary, about

16.39

minutes. Any time up to that point keeps the water at or below

425

liters, and

16

is the last whole minute that does.

Recognize a one-quantity limit

Use a one-variable inequality when one unknown quantity has a lower limit, an upper limit or both. The answer is a range of values, not usually one exact number.

Before you solve, turn the limit words into a symbol:

The question says	Symbol	Does the boundary count?
at least, no less than	$\ge$	Yes, it's included.
more than, greater than	$>$	No, it's left out.
at most, no more than	$\le$	Yes, it's included.
less than, fewer than	$<$	No, it's left out.

An interval is an unbroken stretch of the number line. It helps to picture each one:

$x\le4$ is everything to the left of $4$ , including $4$ itself.
$x>4$ is everything to the right of $4$ , but not $4$ .
$-2<x\le5$ is everything between $-2$ and $5$ , leaving out $-2$ but including $5$ .

On a number line, a closed endpoint means the value is included and an open endpoint means it's left out. So the symbol tells you two things: which way the values run, and whether the endpoint counts.

Check your understanding:

A venue can hold no more than $480$ people. If $p$ is the number of people inside, what’s the inequality, and is $480$ included?

One quantity or two?

Everything here is about one quantity and the range it can take. If two quantities change together and the question asks about a boundary line, a shaded region, limits that overlap or an ordered pair that works, turn to Graph and model two-variable inequalities.

Keep track of the order as you solve

You solve an inequality with the same moves you used in Solve linear equations to get the variable alone. The only new question is what each move does to the order of the two sides. Does it keep the order, or flip it?

Adding or subtracting the same number on both sides keeps the order.
Multiplying or dividing both sides by a positive number keeps the order.
Multiplying or dividing both sides by a negative number flips the order, so the sign flips too.

Why a negative flips the sign

Start with something true:

2<5.

Multiply both numbers by $-1$ and you get $-2$ and $-5$ . Which one is bigger now? On the number line, $-2$ sits to the right of $-5$ , so the true statement is

-2>-5.

Multiplying by a negative mirrors every number across $0$ . Whatever was farther right lands farther left, and whatever was farther left lands farther right. The order swaps, so the sign has to flip. Dividing by a negative undoes that same mirror flip, so it swaps the order too. A short way to remember it: only a negative multiply or divide flips the sign.

Here's how that plays out in

7-3x>19.

Subtract $7$ from both sides, which leaves the sign alone. Then divide by $-3$ , which flips it:

\begin{aligned} -3x&>12\\[1.4em] x&<-4. \end{aligned}

The sign changes on that one line, where you divide by $-3$ , and nowhere else. To be sure, test one value on each side of $-4$ . At $x=-5$ , the left side is $22$ , and $22>19$ is true. At $x=-3$ , it's $16$ , and $16>19$ is false. The allowed value works and the rejected one doesn't, so $x<-4$ is right.

Check your understanding:

Solve $-5x+4\le19$ . At which step does the sign flip?

Common mistake:

A common slip is flipping the sign because a negative number shows up somewhere, like the $-5$ in $-5x+4\le19$ while you’re still subtracting the $4$ . The sign flips only when you multiply or divide both whole sides by a negative. Mark that step, flip the sign on that line, and then test a value from your answer in the original inequality.

Solve a compound inequality as one chain

A compound inequality puts more than one condition together. Take the two-sided chain

-10<2-4x\le14.

It says the middle is greater than $-10$ and at most $14$ , both at once. So keep all three parts together, and do every step to the left side, the middle and the right side.

Subtract $2$ from all three parts:

-12<-4x\le12.

Now divide all three parts by $-4$ . Both signs flip:

3>x\ge-3.

That's correct, but it reads backward. Rewrite it from least to greatest:

-3\le x<3.

Now it's easy to picture on a number line: every real number from $-3$ up to $3$ , including $-3$ but not $3$ .

Common mistake:

Flipping only one sign is the classic slip here. Dividing a chain by a negative, like the $-4$ here, changes both comparisons, so both signs flip. Then write the chain from least to greatest and check each endpoint on its own.

Calculate the boundary, then interpret it

Method	When it works best	What it can't do
Hand	Use it for direct and compound inequalities, where you want to see every sign flip and keep the endpoints exact.	The arithmetic gets slow when the boundary is a messy fraction or decimal.
Desmos	Type a messy cutoff like `295/18` straight into Desmos to get its decimal.	It can't tell you which way the sign points or which whole number the story allows.
Hybrid	Solve by hand down to the exact boundary, get its decimal from Desmos, then pick and check the whole number yourself.	Your reasoning, not the calculator, decides the answer you submit.

Save the hybrid for when the boundary arithmetic is messy. You also usually don't need to rebuild a graph like the one in the opening. Once the variable is alone, the inequality already shows which values are allowed.

Here's the hybrid on the tank problem:

Write the inequality and get $m$ alone by hand: $m\le\frac{295}{18}$ .
Type the whole cutoff, 295/18, into Desmos. It shows about $16.39$ .
Go back to the sign and the story. The $\le$ allows times at or below $16.39$ , and the question wants whole minutes, so pick $16$ .
Test your pick in the original condition. Plugging $16$ and $17$ into 130+18x shows that $16$ minutes stays within capacity and $17$ doesn't.

Choose the first or last allowed integer

Step 3 is where it's easy to slip. Don't round the boundary the usual way. Start at the boundary, move in the direction the inequality allows, and take the first allowed whole number you reach. Don't round; walk the way the sign points.

What you solved	The question asks for	The whole number
$n\ge6.2$	the minimum	$7$
$n>6$	the minimum	$7$
$n\le6.8$	the maximum	$6$
$n<6$	the maximum	$5$

The last row is the sneaky one. The strict $n<6$ leaves out $6$ itself, so the largest whole number allowed is $5$ , even though the boundary is already a whole number.

Check your understanding:

You solve a package problem and get $p\ge\frac{317}{24}$ . Here $p$ counts whole packages, and the question asks for the minimum. What does Desmos do for you, and what do you decide yourself?

Example: Find the first whole-number solution

Worked example

A warehouse begins the day with $143$ unfilled orders. A worker fills $12$ orders per hour. The warehouse wants to end with no more than $50$ orders still unfilled.

What is the minimum whole number of hours the worker must work?

A
$7$
B
$7.75$
C
$8$
D
$9$

Step 1

Turn the limit into an inequality

Let $h$ be the number of hours worked. The worker fills $12$ orders an hour, so after $h$ hours the number of unfilled orders is

143-12h.

“No more than $50$ ” means that amount can be $50$ or less:

143-12h\le50.

Step 2

Get h alone

Subtract $143$ from both sides:

-12h\le-93.

Now divide both sides by $-12$ . You're dividing by a negative, so the sign flips:

h\ge\frac{93}{12}=7.75.

That direction makes sense: the more hours the worker puts in, the fewer orders are left, so the question is how many hours are enough.

Step 3

Walk up to the first whole number

The allowed values start at $7.75$ and go up. The question wants a whole number of hours, so walk up from $7.75$ to the first whole number:

h=8.

The boundary $7.75$ isn't a whole number, and $7$ falls below it, outside the allowed range.

Step 4

Check the whole numbers on each side

At $7$ hours, $143-12(7)=59$ orders are left, which is too many. At $8$ hours,

143-12(8)=47\le50.

So $8$ is the first whole number that works, and the answer is C.

Common mistake:

Choosing $7.75$ gives the boundary, but the question asks for a whole number. Choosing $7$ rounds the wrong way and leaves $59$ orders. The solved sign, $h\ge7.75$ , points up, so walk up to $8$ and check it.

Practice problems

Your turn. In each one, watch which way the sign points and whether each endpoint counts.

Flip the sign at the right step

Practice problem

Which choice gives all values of $x$ that satisfy

11-4(2x+1)>x+16?

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Solve by hand so you can see the sign flip. If you like, work out both sides at

x=-1

here to check the strict endpoint, then test

x=-2

as an allowed value.

Count the integers in a chain

Practice problem

How many integer values of $x$ satisfy

-4<5-3x\le17?

Your answer

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Work the chain by hand. If it helps, plug a possible endpoint into the original chain here before you count.

Model a score and find the minimum

Practice problem

On a $64$ -question competition, a student answers every question. The student earns $7$ points for each correct answer and loses $2$ points for each incorrect answer.

What is the minimum number of questions the student must answer correctly to earn at least $300$ points?

Your answer

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Build and solve the inequality yourself. Then type 428/9 here and let the sign tell you which whole number comes first.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

An inequality describes a range of values, not usually one exact answer.
Adding, subtracting, or multiplying or dividing by a positive keeps the order. Multiplying or dividing by a negative flips it, because it mirrors every number across $0$ .
In a compound inequality, do each step to all three parts, and flip both signs when you multiply or divide by a negative.
Strict symbols leave the boundary out; inclusive symbols include it.
For a whole-number minimum or maximum, don't round. Move the way the sign allows and check the whole numbers on each side.
Decide the sign, the direction and the whole number yourself. Let Desmos work out a messy cutoff when that saves you arithmetic.

Next lesson

Graph and model two-variable inequalities

Turn two-variable constraints into boundaries and shaded regions, then identify feasible points and extrema.

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