How do I know which side of the line to shade?

Solve for y if possible: y > or y ≥ shades above the line, y < or y ≤ shades below. Otherwise, pick a test point like (0,0), substitute it, and shade the side containing the point if the inequality is true.

What is the difference between a solid and dashed boundary line?

A solid line is used for ≤ and ≥, meaning points on the line are included in the solution. A dashed line is used for , meaning points on the line are excluded. The exam uses boundary points as distractors to test this.

How do I check if a point solves a system of inequalities?

Substitute the point's coordinates into every inequality. The point is a solution only if all inequalities are true at once. Stop as soon as one fails — that point cannot be a solution regardless of the others.

DSAT-1.5

Linear Inequalities in One or Two Variables

Q: What is the difference between a solid and dashed boundary line?

A solid line is used for ≤ and ≥, meaning points on the line are included in the solution. A dashed line is used for , meaning points on the line are excluded. The exam uses boundary points as distractors to test this.

Master linear inequalities on the Digital SAT: solve one-variable inequalities, graph two-variable regions, test solutions, and handle real-world constraint problems.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Linear Inequalities in One or Two Variables, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Linear Inequalities in One or Two Variables free

What this lesson covers

Linear inequalities look almost identical to linear equations, but one small rule — flipping the sign when you multiply or divide by a negative — trips up more test-takers than almost anything else in the Heart of Algebra content. On the Digital SAT you'll be asked to solve inequalities, test whether a point satisfies a system, and translate word problems into constraints like "at least," "no more than," and "between."

This lesson gives you the mechanics for one-variable inequalities, the geometry of two-variable solution regions, and a reliable method for constraint problems. By the end you'll know exactly what the exam is checking and how to avoid the classic sign-flip and boundary-line mistakes.

Solving Inequalities in One Variable

Solving a linear inequality uses the same algebra as solving an equation: isolate the variable using inverse operations. The one crucial difference is the flip rule.

When you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality symbol. Adding or subtracting any number, and multiplying or dividing by a positive number, never changes the direction.

For example, solve

-3x + 5 > 14

. Subtract 5:

-3x > 9

. Now divide by

-3

and flip:

x < -3

Operation	Effect on symbol
Add or subtract any value	no change
Multiply/divide by positive	no change
Multiply/divide by negative	reverse it

The Digital SAT often hides this in a multi-step problem or asks for the "greatest possible value" or "least possible value" of

x

. If the answer is

x < -3

, the greatest integer value is

-4

, not

-3

, because the inequality is strict. Watch the difference between

<

(open, endpoint excluded) and

\leq

(closed, endpoint included). A common misconception is treating the boundary as always included — always read the symbol.

Graphing Two-Variable Inequalities

A two-variable linear inequality such as

y \leq 2x - 1

describes a region of the coordinate plane, not just a line. To graph it, first graph the boundary line

y = 2x - 1

, then shade one side.

The boundary is solid for

\leq

\geq

(points on the line count as solutions) and dashed for

<

>

(points on the line are excluded). To decide which side to shade, pick a simple test point not on the line — usually the origin

(0,0)

— and check whether it makes the inequality true. If it does, shade the side containing that point; if not, shade the other side.

For

y \leq 2x - 1

, test

(0,0)

: is

0 \leq 2(0) - 1 = -1

? No. So shade the side away from the origin (below-right).

A quick shortcut once the inequality is solved for

y

y >

y \geq

means shade above the line;

y <

y \leq

means shade below. This only works reliably after isolating

y

. The Digital SAT rarely asks you to physically draw a graph, but it frequently asks which graph matches an inequality or which point lies in a shaded region, so knowing boundary style and shading direction is essential.

Testing Points and Systems of Inequalities

A very common Digital SAT task gives you a system of two or more inequalities and asks which ordered pair is a solution. A point is a solution only if it satisfies every inequality in the system simultaneously. The graphical solution is the overlap (intersection) of all the shaded regions.

The efficient strategy: substitute the point's coordinates into each inequality and check. As soon as one inequality fails, reject that point — you don't need to check the rest.

Consider the system

y > x + 1

and

y \leq -2x + 6

. Test

(1, 3)

: first,

3 > 1 + 1 = 2

true; second,

3 \leq -2(1) + 6 = 4

true. Both hold, so

(1,3)

is a solution.

Watch strict versus non-strict symbols at boundaries. A point lying exactly on a dashed boundary line fails, but on a solid line it passes. The exam loves to include a distractor point that sits on a boundary to test whether you know the difference between

<

and

\leq

. Also verify you substitute

x

and

y

into the correct positions — swapping them is a frequent careless error.

Real-World Constraint Problems

Word problems translate everyday phrases into inequality symbols. Learning this dictionary is half the battle.

Phrase	Symbol
at least, no less than, minimum	$\geq$
at most, no more than, maximum	$\leq$
more than, greater than, exceeds	$>$
fewer than, less than, under	$<$

A typical scenario: a food truck sells tacos for 3 dollars and burritos for 5 dollars and must earn at least 60 dollars. With

t

tacos and

b

burritos, the constraint is

3t + 5b \geq 60

. If there's also a limit of 25 total items, add

t + b \leq 25

. Both variables are often implicitly non-negative because you can't sell negative food, giving

t \geq 0

and

b \geq 0

.

The Digital SAT may ask you to build the inequality, identify a viable combination, or find a maximum/minimum value within the constraints. Read carefully: "at least" versus "more than" changes whether the boundary counts. Also match units — if quantities must be whole numbers of items, the answer must be an integer even though the inequality allows any real value. Always define what each variable represents before writing the model.

Key terms

Linear inequality.: A statement comparing two linear expressions using $<$ , $>$ , $\leq$ , or $\geq$ , whose solution is a range of values rather than a single value.
Flip rule.: The requirement to reverse the inequality symbol whenever both sides are multiplied or divided by a negative number.
Boundary line.: The line obtained by replacing the inequality symbol with $=$ ; solid if the symbol includes equality, dashed if strict.
Solution region.: The set of all points in the plane that satisfy a two-variable inequality, shown as a shaded half-plane.
System of inequalities.: Two or more inequalities considered together; solutions must satisfy all of them, corresponding to the overlap of shaded regions.
Test point.: A point substituted into an inequality to determine which side of the boundary to shade; $(0,0)$ is preferred when the line does not pass through it.
Constraint.: A real-world limitation expressed as an inequality, such as a budget, capacity, or minimum requirement.
Strict inequality.: An inequality using $<$ or $>$ that excludes the boundary value, drawn with a dashed line or open endpoint.

Worked example

A student has at most 40 dollars to spend on notebooks and pens. Notebooks cost 4 dollars each and pens cost 2 dollars each. The student needs at least 3 notebooks. If

n

is the number of notebooks and

p

is the number of pens, write the system of inequalities and determine whether buying 5 notebooks and 8 pens is possible.

Start by translating each condition into an inequality.

The total cost cannot exceed 40 dollars, so

4n + 2p \leq 40

("at most" gives

\leq

).

The student needs at least 3 notebooks, so

n \geq 3

("at least" gives

\geq

).

Since you cannot buy negative pens,

p \geq 0

.

Now test the combination

n = 5

p = 8

. Check the budget:

4(5) + 2(8) = 20 + 16 = 36

. Is

36 \leq 40

? Yes.

Check the notebook requirement: is

5 \geq 3

? Yes.

Check non-negativity:

8 \geq 0

? Yes.

All three inequalities are satisfied, so buying 5 notebooks and 8 pens is possible, spending 36 dollars and leaving 4 dollars unspent.

Practice questions

What is the greatest integer value of

x

that satisfies

-2x + 7 \geq 1

2
3
4
-3

Answer: 3

Subtract 7 from both sides:

-2x \geq -6

. Divide by

-2

and flip the symbol:

x \leq 3

. The inequality includes 3 because of

\leq

, so the greatest integer value is 3. A common mistake is forgetting to flip the sign, which would wrongly give

x \geq 3

Which ordered pair is a solution to the system

y < 3x - 2

and

y \geq -x + 1

$(0, 0)$
$(2, 3)$
$(1, -1)$
$(0, 4)$

Answer: $(2, 3)$

Test

(2,3)

in the first inequality:

3 < 3(2) - 2 = 4

is true. In the second:

3 \geq -2 + 1 = -1

is true. Both hold. Check why others fail:

(0,0)

gives

0 < -2

false;

(1,-1)

gives

-1 \geq 0

false;

(0,4)

gives

4 < -2

false.

A charity event sells adult tickets for 12 dollars and child tickets for 7 dollars, and organizers want to raise more than 500 dollars. Let

a

be adult tickets and

c

be child tickets sold. Write the inequality that models the fundraising goal, and explain why the boundary is not included.

Answer: $12a + 7c > 500$

Each adult ticket contributes

12a

dollars and each child ticket

7c

dollars, so total revenue is

12a + 7c

. The phrase "more than 500 dollars" means the total must exceed 500 but not equal it, which requires the strict symbol

>

. The boundary value of exactly 500 dollars is excluded because raising exactly 500 does not count as more than 500.

FAQ

When exactly do I flip the inequality sign?: Only when you multiply or divide both sides by a negative number. Adding, subtracting, or multiplying/dividing by a positive number never changes the direction. Flipping at the wrong time — or forgetting to flip — is the single most common error on inequality questions.
How do I know which side of the line to shade?: Solve for $y$ if possible: $y >$ or $y \geq$ shades above the line, $y <$ or $y \leq$ shades below. Otherwise, pick a test point like $(0,0)$ , substitute it, and shade the side containing the point if the inequality is true.
What is the difference between a solid and dashed boundary line?: A solid line is used for $\leq$ and $\geq$ , meaning points on the line are included in the solution. A dashed line is used for $<$ and $>$ , meaning points on the line are excluded. The exam uses boundary points as distractors to test this.
How do I check if a point solves a system of inequalities?: Substitute the point's coordinates into every inequality. The point is a solution only if all inequalities are true at once. Stop as soon as one fails — that point cannot be a solution regardless of the others.

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