Digital SAT Math · Algebra

Systems of linear inequalities

A system of inequalities does not ask for a single pair $(x, y)$ . It asks for a region — every point that keeps every inequality true at once. That one shift changes the fastest move on the item: before you graph anything, test the candidate. One failure eliminates the point; only the overlap survives.

On the test


Domain	Algebra (score report)
What it looks like	Two (sometimes three) linear inequalities; a shaded region in the $xy$ -plane; or a short story with two simultaneous limits
Often asked	“Which ordered pair is a solution…?”, “Which system describes the shaded region…?”, “What is the greatest integer value of $x$ for which there exists a $y$ …?”
Format	Multiple choice and student-produced response (a single number — never a whole region)
Calculator	Desmos shades the overlap if you type both inequalities; it still will not pick your symbol or decide a dashed boundary for you

Recognition cues: two inequality symbols stacked in a box, the phrase solution set or shaded region, solid vs dashed boundaries, and stems that say both inequalities or the system.

Pattern recognition

Four shapes cover almost every item:

Point in region — a candidate $(x, y)$ is given (or offered in the choices); decide whether it satisfies every inequality.
Which system — a shaded picture is given; name the pair of inequalities (boundaries + sides + solid/dashed).
Build from words — two simultaneous limits in a story (capacity and budget, total and minimum) become two symbols.
Existence / greatest value — “greatest integer $x$ for which there exists a $y$ ” means the vertical slice at that $x$ must still hit the overlap.

Method

Name the ask. Is the item offering a point, a system, a graph, or a greatest/least value? Underline that before you rearrange anything.
Test-point first (when a candidate is in play). Substitute the ordered pair into each inequality on its own. A single false statement eliminates the point — you never need the graph.
When you must graph, shade one inequality at a time and keep only the overlap. Solid boundary $\Rightarrow$ the line is included ( $\le$ or $\ge$ ); dashed $\Rightarrow$ the line is out ( $<$ or $>$ ). The solution is the intersection, never the union.
For “which system describes the region”: read both boundaries (slope and intercept), read solid vs dashed, then test one interior point from the dark overlap to lock the sides.
For existence asks: at a fixed $x$ , the lower bound on $y$ must sit at or below the upper bound. Solve that one-variable inequality for the largest (or smallest) allowed $x$ .

What you see	Fast move
Four ordered pairs as choices	Test each pair in both inequalities; stop at the first double-yes
A shaded region with solid/dashed lines	Boundaries first, style second, one interior point third
A story with two limits	Write both symbols before simplifying either one
“Greatest $x$ for which there exists a $y$ ”	Require lower bound $\le$ upper bound; solve for $x$
Ugly coefficients	Type both inequalities in Desmos and click a test point

Worked example 1 — test the point, skip the graph

Stem. Which ordered pair $(x, y)$ is a solution to the system?

y \ge x + 2 \qquad x + y \le 10

Step 1 — try $(2, 5)$ . First inequality: $5 \ge 2 + 2$ is $5 \ge 4$ , true. Second: $2 + 5 \le 10$ is $7 \le 10$ , true. Both yes.

Step 2 — kill the near-misses by the same rule. $(2, 3)$ : $3 \ge 4$ is false — eliminated by the first inequality alone. $(4, 7)$ : $7 \ge 6$ is true, but $4 + 7 = 11 \le 10$ is false — eliminated by the second alone. A point that fails either constraint is not a solution of the system.

Check. The boundary intersection of $y = x + 2$ and $x + y = 10$ is $(4, 6)$ . That point is on both boundaries and satisfies both inclusive symbols, so it is also a solution — but the item only needed any one ordered pair from the region, and $(2, 5)$ is inside. Answer: $(2, 5)$ .

Trap watch. Taking $(4, 7)$ because it “looks high enough” on the first inequality is the only-one trap. The second inequality is not optional.

Worked example 2 — read a region (boundaries, then a test point)

Stem. In the $xy$ -plane, a region is bounded by the solid line $y = x$ (shading on or above it) and the solid line $y = -\frac{1}{2}x + 3$ (shading on or below it). Which system describes the region?

Step 1 — boundaries. Solid lines mean inclusive symbols. Above $y = x$ is $y \ge x$ . Below $y = -\frac{1}{2}x + 3$ is $y \le -\frac{1}{2}x + 3$ .

Step 2 — lock the sides with an interior point. The point $(1, 2)$ sits in the overlap: $2 \ge 1$ holds, and $2 \le -\frac{1}{2}(1) + 3 = 2.5$ holds. Any choice that flips a side will reject $(1, 2)$ .

Step 3 — reject style errors. Replacing either $\ge$ or $\le$ with a strict symbol would require a dashed boundary. Both lines are solid, so both symbols stay inclusive.

Answer: $y \ge x$ and $y \le -\frac{1}{2}x + 3$ .

Trap watch. Reading the darker overlap as the union of the two half-planes produces a system with “or” logic that the test never writes. The solution set is where the shadings stack.

Worked example 3 — greatest $x$ that still leaves a $y$

Stem. What is the greatest integer value of $x$ for which there exists a real number $y$ satisfying both

y \ge 2x - 3 \qquad y \le 11 - x\,?

Step 1 — existence condition. For a fixed $x$ , some $y$ must sit between the lower and upper bounds, so the lower bound cannot exceed the upper bound:

2x - 3 \le 11 - x

Step 2 — solve.

3x \le 14 \quad\Rightarrow\quad x \le \frac{14}{3} \approx 4.67

Step 3 — greatest integer inside the range. $x = 4$ works: then $y \ge 5$ and $y \le 7$ , so any $y$ in $[5, 7]$ is fine. $x = 5$ fails: $y \ge 7$ and $y \le 6$ is empty.

Answer: 4.

Trap watch. Reporting $14/3$ or $5$ (the ceiling) both miss. The ask is the greatest integer $x$ that still leaves a nonempty vertical slice of the region.

Practice

Answer before opening the explanation. Two items are student-produced response (type one number). Three items depend on a figure — read solid vs dashed before you read the choices. Every wrong choice below is a specific error someone actually makes on this skill.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

Which ordered pair (x, y) is a solution to the system of inequalities below? y ≥ x + 2 x + y ≤ 10

Show the answer Choice A

Why it is right

Test each inequality separately. For (2, 5): the first says 5 ≥ 2 + 2, which is 5 ≥ 4 and true; the second says 2 + 5 ≤ 10, which is 7 ≤ 10 and true. Both hold, so (2, 5) is in the solution region.

Why each other choice fails

Choice B: Fails the first inequality alone: 3 ≥ 2 + 2 is 3 ≥ 4, which is false. A point that fails either constraint is not a solution of the system.
Choice C: Fails the first inequality: 1 ≥ 8 + 2 is false. (The second happens to hold: 8 + 1 ≤ 10, which is the only-one trap in reverse.)
Choice D: Satisfies the first inequality (7 ≥ 6) but fails the second: 4 + 7 = 11, and 11 ≤ 10 is false. Accepting a point that works for only one inequality is the classic miss on this skill.

Question 2 Standard

Which ordered pair (x, y) is a solution to the system of inequalities below? y > 3x − 4 2x + y ≤ 8

Show the answer Choice B

Why it is right

Substitute (1, 2) into each inequality. First: 2 > 3(1) − 4 is 2 > −1, true. Second: 2(1) + 2 ≤ 8 is 4 ≤ 8, true. Both hold, so (1, 2) is a solution.

Why each other choice fails

Choice A: Lies on the boundary of the first inequality: −1 > 3(1) − 4 is −1 > −1, which is false because the symbol is strict. A dashed-style boundary excludes its own points.
Choice C: Fails the first inequality: 1 > 3(3) − 4 is 1 > 5, false. The second holds (7 ≤ 8), so this is the only-one trap.
Choice D: Satisfies the first (9 > −4) but fails the second: 0 + 9 ≤ 8 is false. One yes is not enough.

Question 3 Standard

A food truck sells burritos for $4 each and tacos for $2 each. The truck can prepare at most 80 items in total and needs at least $200 in sales. Which system of inequalities represents this situation, where b is the number of burritos sold and t is the number of tacos sold?

Show the answer Choice C

Why it is right

Two simultaneous limits need two inequalities. Item count is a ceiling: b + t ≤ 80 (exactly 80 items is still allowed). Sales are a floor: each burrito contributes $4 and each taco $2, so 4b + 2t ≥ 200. Both directions match the words "at most" and "at least."

Why each other choice fails

Choice A: Treats the $200 sales target as a ceiling (≤) instead of a floor. "At least $200" requires ≥, not ≤.
Choice B: Reverses the capacity constraint. "At most 80 items" is b + t ≤ 80, not ≥ 80.
Choice D: Swaps the prices: it charges $2 per burrito and $4 per taco. The $4 coefficient must multiply b.

Question 4 Standard

Which system is equivalent to the system below? −2y ≥ 4x − 12 x − y < 1

Show the answer Choice D

Why it is right

Rearrange each inequality for y. First: divide −2y ≥ 4x − 12 by −2 and flip the symbol, getting y ≤ −2x + 6. Second: x − y < 1 is −y < 1 − x; multiplying by −1 flips again, so y > x − 1. The equivalent system is y ≤ −2x + 6 and y > x − 1.

Why each other choice fails

Choice A: Divides the first inequality by −2 without flipping, producing y ≥ −2x + 6 — the complement of the true first constraint.
Choice B: Gets the first inequality right but reverses the second: x − y < 1 rearranges to y > x − 1, not y < x − 1.
Choice C: Drops a sign when dividing: −2y ≥ 4x − 12 becomes y ≤ 2x − 6 instead of y ≤ −2x + 6, which is a different half-plane.

Question 5 Standard

In the xy-plane, the darker shaded region shown is the solution set of a system of linear inequalities. Which system describes the shaded region?

The darker overlap is the solution set. Both boundary lines are solid.

Show the answer Choice C

Why it is right

Read the boundaries first. One solid line is y = x; the other solid line is y = −0.5x + 3. Solid lines mean inclusive symbols. The darker overlap lies on or above y = x and on or below y = −0.5x + 3. The marked point (1, 2) confirms both sides: 2 ≥ 1 and 2 ≤ −0.5(1) + 3 = 2.5. The system is y ≥ x and y ≤ −0.5x + 3.

Why each other choice fails

Choice A: Shades the wrong side of y = x. The point (1, 2) is above that line, so the first inequality must be y ≥ x, not y ≤ x.
Choice B: Shades the wrong side of y = −0.5x + 3. The overlap sits below that line, so the second inequality is y ≤ −0.5x + 3, not ≥.
Choice D: Uses strict symbols, which would require dashed boundaries. Both drawn lines are solid, so the endpoints on each line belong to the solution set.

Question 6 Harder

Which ordered pair (x, y) is a solution to both inequalities in the system below? y ≤ −x + 8 y ≥ 2x − 2

Show the answer Choice B

Why it is right

Test (2, 4) in each inequality. First: 4 ≤ −2 + 8 is 4 ≤ 6, true. Second: 4 ≥ 2(2) − 2 is 4 ≥ 2, true. Both hold, so (2, 4) is in the overlap.

Why each other choice fails

Choice A: Satisfies only the first inequality: 1 ≤ −2 + 8 is true, but 1 ≥ 2(2) − 2 is 1 ≥ 2, false. One yes does not make a solution of the system.
Choice C: Satisfies only the second inequality: 8 ≥ 2(1) − 2 is true, but 8 ≤ −1 + 8 is 8 ≤ 7, false. This point sits above the upper boundary.
Choice D: Fails both: 5 ≤ −4 + 8 is 5 ≤ 4, false, and 5 ≥ 2(4) − 2 is 5 ≥ 6, also false. The point is outside the wedge on both sides.

Question 7 Harder

In the xy-plane, the darker shaded region shown is the solution set of a system of linear inequalities. Which system describes the shaded region?

The darker overlap is the solution set. One boundary is dashed; the other is solid.

Show the answer Choice A

Why it is right

The rising boundary has slope 2 and y-intercept −1, and it is dashed, so the first inequality is strict: points on y = 2x − 1 are out. Shading is above that line, so y > 2x − 1. The falling boundary has slope −1 and y-intercept 4, and it is solid, so the second inequality is inclusive: y ≤ −x + 4. The marked point (1, 2) checks both: 2 > 2(1) − 1 is 2 > 1, and 2 ≤ −1 + 4 is 2 ≤ 3.

Why each other choice fails

Choice B: Reads the dashed boundary as solid. A dashed line excludes its own points, so the first symbol must be > rather than ≥. The point (1, 1) on that line is not in the solution set.
Choice C: Reads the solid boundary as dashed. The upper line is solid, so points on y = −x + 4 are included and the symbol is ≤, not <.
Choice D: Shades the wrong side of the dashed line. Testing (1, 2) in y < 2x − 1 gives 2 < 1, which is false, so that choice cannot describe the region containing (1, 2).

Question 8 Harder

Which system is equivalent to the system below? −4y < 8x − 12 2x − y ≥ 1

Show the answer Choice D

Why it is right

Solve each inequality for y. First: divide −4y < 8x − 12 by −4 and flip: y > −2x + 3. Second: 2x − y ≥ 1 is −y ≥ 1 − 2x; multiply by −1 and flip: y ≤ 2x − 1. The equivalent system is y > −2x + 3 and y ≤ 2x − 1.

Why each other choice fails

Choice A: Divides the first inequality by −4 without flipping the symbol, producing y < −2x + 3 instead of y > −2x + 3.
Choice B: Rearranges the second inequality without flipping when multiplying by −1, producing y ≥ 2x − 1 instead of y ≤ 2x − 1.
Choice C: Misses the flip on both inequalities — the double error that recreates the complement of the true solution region.

Question 9 Harder Student-produced response

What is the greatest integer value of x for which there exists a real number y satisfying both inequalities below? y ≥ 2x − 3 y ≤ 11 − x

Your answer

Show the answer 4

Why it is right

For a fixed x, some y must satisfy both bounds at once, so the lower bound cannot exceed the upper bound: 2x − 3 ≤ 11 − x. Collecting terms gives 3x ≤ 14, so x ≤ 14/3 ≈ 4.67. The greatest integer at or below 4.67 is 4. At x = 4 the slice is 5 ≤ y ≤ 7, which is nonempty. At x = 5 the slice is 7 ≤ y ≤ 6, which is empty.

Answers students type instead

3: Stops one integer early, perhaps testing only x = 3 or mis-solving 3x ≤ 14 as x ≤ 3.
5: Rounds 14/3 up to the next integer. The existence condition is x ≤ 14/3, so 5 is outside the range and leaves no y.
14/3: Reports the exact boundary of the existence inequality instead of the greatest integer value the stem asked for.

Question 10 Hardest

In the xy-plane, the darker shaded region shown is the solution set of a system of linear inequalities. Which of the labeled points lies in the solution set?

The darker overlap is the solution set. Points P, Q, and R are labeled.

Show the answer Choice B

Why it is right

The system drawn is y ≥ 0.5x + 1 and y ≤ −x + 6 (both boundaries solid). Point P is (2, 3): 3 ≥ 0.5(2) + 1 = 2, true, and 3 ≤ −2 + 6 = 4, true. P is in the darker overlap. Q is (2, 1): 1 ≥ 2 is false, so Q fails the lower boundary. R is (4, 3): 3 ≥ 0.5(4) + 1 = 3 is true, but 3 ≤ −4 + 6 = 2 is false, so R fails the upper boundary.

Why each other choice fails

Choice A: Q sits below the lower boundary. At x = 2 the lower line is at y = 2, and Q has y = 1, so it fails y ≥ 0.5x + 1.
Choice C: R sits above the upper boundary. At x = 4 the upper line is at y = 2, and R has y = 3, so it fails y ≤ −x + 6.
Choice D: P is inside the overlap, so it is not true that none of the labeled points works.

Question 11 Hardest

In the xy-plane, the solution region of a system is the set of points on or above the line y = x − 1 and strictly below the line y = −2x + 8. Which system describes the region?

Show the answer Choice A

Why it is right

"On or above" y = x − 1 is the inclusive half-plane y ≥ x − 1. "Strictly below" y = −2x + 8 is the open half-plane y < −2x + 8. Together: y ≥ x − 1 and y < −2x + 8. An interior check at (2, 2): 2 ≥ 2 − 1 holds, and 2 < −4 + 8 = 4 holds.

Why each other choice fails

Choice B: Swaps the inclusion styles: it makes the first boundary strict and the second inclusive. The stem says the first line is included ("on or above") and the second is excluded ("strictly below").
Choice C: Shades the wrong side of y = x − 1. "On or above" is y ≥ x − 1, not y ≤ x − 1. The test point (2, 2) fails y ≤ x − 1.
Choice D: Shades the wrong side of the second line. "Strictly below" is y < −2x + 8, not y > −2x + 8. That choice is the complementary half-plane above the second boundary.

Question 12 Hardest Student-produced response

A summer camp offers kayaking sessions k and hiking sessions h, where k and h are nonnegative integers. The camp can run at most 25 sessions in total and needs at least 70 activity credits, where each kayaking session is worth 2 credits and each hiking session is worth 5 credits. What is the least possible number of hiking sessions the camp can run?

Your answer

Show the answer 7

Why it is right

The constraints are k + h ≤ 25 and 2k + 5h ≥ 70 with k ≥ 0, h ≥ 0. For a fixed h, some integer k must satisfy k ≤ 25 − h and 2k ≥ 70 − 5h. That is possible only when the largest available k is still large enough: 2(25 − h) ≥ 70 − 5h, which simplifies to 50 − 2h ≥ 70 − 5h, then 3h ≥ 20, so h ≥ 20/3 ≈ 6.67. The least integer h is 7. Check: at h = 7, need 2k ≥ 35 so k ≥ 18, and k ≤ 18, so k = 18 works. At h = 6, need 2k ≥ 40 so k ≥ 20, but k ≤ 19 — empty.

Answers students type instead

6: Rounds 20/3 down instead of up. The existence bound is h ≥ 20/3, so 6 leaves no feasible k.
14: Solves 5h ≥ 70 as if kayaking contributed no credits (h ≥ 14), ignoring that k can still supply 2 credits each.
25: Reports the session-capacity ceiling instead of the least hiking count that still meets the credit floor.
20/3: Leaves the bound as a fraction instead of the least integer the stem asked for.

Common mistakes

Accepting a point that satisfies only one inequality — the single most expensive habit here; the other constraint is not optional.
Treating the solution as the intersection point of the two boundary lines — that point is one candidate, not the whole answer, and it may even be excluded if a boundary is dashed.
Reading the union instead of the overlap — shading either half-plane rather than where both are shaded.
Counting a dashed-boundary point as a solution — open symbols exclude the line; solid symbols include it.
Forgetting to flip when multiplying or dividing an inequality by a negative while rearranging one constraint of the system.
Reversing a side when converting “above/below” into a symbol — always lock the side with a test point from the shaded region.
Building only one inequality from a two-limit story — capacity without budget (or minimum without total) leaves the system incomplete.
Rounding an existence bound the wrong way — greatest integer $x$ with $x \le 4.67$ is 4, not 5.

FAQ

Is the intersection of the two boundary lines always a solution? Only if both symbols are inclusive ( $\le$ or $\ge$ ). If either boundary is dashed, the intersection point fails that inequality and is out.

Can I just graph every system in Desmos? For “which point is a solution” and messy coefficients, yes — and you should. For “which system matches this printed figure,” Desmos cannot see the paper; you still have to read solid/dashed and test an interior point by hand.

How is this different from linear inequalities? One inequality is one half-plane. A system requires the overlap of two or more. The score report splits them; search results often do not.

What do I enter on an SPR? One number — usually a greatest or least integer value that still leaves the region nonempty. Never try to enter a whole region or a pair.

Editorial team · last reviewed 2026-08-12

Systems of linear inequalities

On the test

Pattern recognition

Method

Worked example 1 — test the point, skip the graph

Worked example 2 — read a region (boundaries, then a test point)

Worked example 3 — greatest xxx that still leaves a yyy

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard

Why it is right

Why each other choice fails

Question 5 Standard

Why it is right

Why each other choice fails

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder

Why it is right

Why each other choice fails

Question 8 Harder

Why it is right

Why each other choice fails

Question 9 Harder Student-produced response

Why it is right

Answers students type instead

Question 10 Hardest

Why it is right

Why each other choice fails

Question 11 Hardest

Why it is right

Why each other choice fails

Question 12 Hardest Student-produced response

Why it is right

Answers students type instead

Common mistakes

FAQ

Related skills

Worked example 3 — greatest $x$ that still leaves a $y$