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Digital SAT Math · Algebra

Linear inequalities

Digital SAT Math · Algebra

An inequality is an equation that lost its balance: the algebra is identical, but two extra decisions ride along — which direction the symbol points, and whether the boundary value itself counts. Almost every miss on this skill is one of those two decisions, not a slip in the arithmetic. The test knows it, which is why the wrong answers on inequality items are usually your own answer with the symbol turned around or the endpoint quietly excluded.

On the test

DomainAlgebra (score report)
What it looks likeSolve for a range, translate a limit from a sentence, or read a solution set off a graph
Often asked“Which inequality represents…?”, “What is the greatest/least value…?”, “Which is a solution to…?”
FormatMultiple choice and student-produced response (a single number, never a range)
CalculatorAllowed throughout; Desmos will shade a region for you, but it will not choose your symbol

Recognition cues: at least, at most, no more than, no fewer than, minimum, maximum, exceeds, under budget, up to, must remain, plus any question that asks for the greatest or least value that works.

Pattern recognition

Four shapes cover nearly everything:

  1. Solve — isolate the variable and report a range (x ≤ −4).
  2. Translate — turn a sentence limit into symbols, then stop (build only, do not solve).
  3. Count — solve, then answer with a whole number: greatest boxes, least months, maximum kits.
  4. Read — a number line or a shaded region is given; name the inequality, or test whether a point belongs.

Method

  1. Translate the limit word first. Write the symbol and mark inclusion before touching algebra (table below). If the sentence says must remain, ask what quantity must remain and set that expression against its floor or ceiling.
  2. Solve as if it were an equation — same moves, same order.
  3. Flip only on a negative multiply or divide. Adding a negative number changes nothing. Better still, avoid the flip: move the variable to whichever side keeps its coefficient positive.
  4. Test one point. Pick any number from your range and any number outside it, and substitute into the original. Two substitutions catch a reversed symbol instantly.
  5. Match the ask. A range is not the answer to “how many”: round toward the inside of your range — down under a ceiling (x ≤ 7.5 → 7), up above a floor (x ≥ 78.6 → 79) — and never round a count in the direction that breaks the constraint.
Words in the stemSymbolBoundary included?
at least, no fewer than, minimum of, or more≥\geYes
at most, no more than, maximum of, up to, cannot exceed≤\leYes
more than, greater than, exceeds, over>>No
less than, fewer than, under, below<<No
between aa and bba<x<ba < x < bNo, unless inclusive is stated

Worked example 1 — solve without flipping

Stem. Which values of xx satisfy 12−5x≤3x−2012 - 5x \le 3x - 20?

Step 1 — pick the side that keeps xx positive. The xx terms are −5x-5x and 3x3x. Adding 5x5x to both sides leaves a positive coefficient and removes the flip risk entirely:

12≤8x−2012 \le 8x - 20

Step 2 — isolate.

32≤8x⇒4≤x⇒x≥432 \le 8x \quad\Rightarrow\quad 4 \le x \quad\Rightarrow\quad x \ge 4

Step 3 — the same problem the other way. Subtracting 3x3x instead gives −8x≤−32-8x \le -32; dividing by −8-8 reverses the symbol and returns x≥4x \ge 4. Two routes, one answer — the flip is a bookkeeping step, not a different fact.

Check. At x=4x = 4: left =12−20=−8= 12 - 20 = -8, right =12−20=−8= 12 - 20 = -8, and −8≤−8-8 \le -8 holds. At x=0x = 0: 12≤−2012 \le -20 is false, so 0 is correctly outside.

Trap watch. Dividing −8x≤−32-8x \le -32 by −8-8 without flipping gives x≤4x \le 4 — the exact complement of the truth, and always one of the four choices.

Worked example 2 — a limit in words, an answer in whole things

Stem. A summer camp charges a $65 registration fee plus $18 per activity session. Dara has $200 to spend in total. What is the greatest number of sessions she can sign up for?

Step 1 — translate. “Has $200 to spend” is a ceiling, and spending exactly $200 is allowed:

18s+65≤20018s + 65 \le 200

Step 2 — solve.

18s≤135⇒s≤7.518s \le 135 \quad\Rightarrow\quad s \le 7.5

Step 3 — round toward the inside. Sessions are whole. The range is everything at or below 7.5, so the largest whole number inside it is 7.

Check. 7 sessions cost 18(7)+65=191≤20018(7) + 65 = 191 \le 200. 8 sessions cost 18(8)+65=20918(8) + 65 = 209, over budget. Answer: 7 sessions.

Trap watch. Rounding 7.5 up to 8 is the standard miss — it is the only rounding rule where “nearest” is wrong. Computing 200/18≈11.1200/18 \approx 11.1 forgets that the registration fee eats $65 of the budget before any session is bought.

Worked example 3 — a compound inequality

Stem. Which values of xx satisfy −7≤4−x2<5-7 \le 4 - \dfrac{x}{2} < 5?

Step 1 — operate on all three parts. Subtract 4 everywhere:

−11≤−x2<1-11 \le -\frac{x}{2} < 1

Step 2 — multiply by −2-2 and reverse both symbols.

22≥x>−222 \ge x > -2

Step 3 — rewrite smallest-first. Reading the chain from the left, −2<x≤22-2 < x \le 22. Notice what travelled: the ≤\le started attached to the −11-11, and −11⋅(−2)=22-11 \cdot (-2) = 22, so 22 is the included end. Strictness belongs to the number, not to the side of the page it was printed on.

Check. At x=22x = 22: 4−11=−74 - 11 = -7, and −7≤−7-7 \le -7 holds — 22 is in. At x=−2x = -2: 4+1=54 + 1 = 5, and 5<55 < 5 is false — −2-2 is out, exactly as written.

Trap watch. Reversing the symbols but leaving the inclusive end on the left produces −2≤x<22-2 \le x < 22: right endpoints, both boundaries wrong.

Practice

Answer first, then open the explanation. Two items are student-produced response (type the number — note that an SPR answer is always a single value, never a range). Two items depend on a figure; read the circle and line styles before you read the choices. Every wrong choice below is a specific error, not filler.

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

Question 1 Warm-up

A parking garage charges a $4 entry fee plus $3 for each hour parked. Elena wants her total charge to be no more than $25. Which inequality represents the possible numbers of hours h that Elena can park?

Show the answer Choice A

Why it is right

Each hour adds the same $3, so 3 multiplies h; the $4 entry fee is paid once no matter how long Elena stays, so it is a constant added on. Her total charge is 3h + 4. The phrase "no more than $25" allows a total of exactly $25 but nothing above it, which is the inclusive symbol ≤. The inequality is 3h + 4 ≤ 25.

Why each other choice fails

Choice B
Uses the strict symbol <, which would rule out a total of exactly $25. "No more than" includes the boundary — a $25 charge is still acceptable.
Choice C
Reverses the direction. 3h + 4 ≥ 25 describes charges of at least $25, the opposite of the ceiling Elena wants.
Choice D
Swaps the roles of the two numbers: it charges $4 for every hour and adds the $3 entry fee once.

Question 2 Standard

Which of the following gives all values of x that satisfy -4x + 9 ≥ 25?

Show the answer Choice C

Why it is right

Subtract 9 from both sides: -4x ≥ 16. Dividing both sides by -4 is a division by a negative number, so the direction of the inequality reverses: x ≤ 16 / (-4) = -4. Check a value on each side: x = -5 gives -4(-5) + 9 = 29, and 29 ≥ 25 is true; x = 0 gives 9 ≥ 25, which is false. The solutions lie at and to the left of -4.

Why each other choice fails

Choice A
Divides by -4 without reversing the symbol. This choice allows x = 0, but -4(0) + 9 = 9 and 9 ≥ 25 is false.
Choice B
Reverses the symbol correctly but divides 16 by 4 instead of -4, losing the sign of the boundary. This choice also allows x = 0, which fails the original inequality.
Choice D
Makes both errors: keeps the original direction and drops the negative sign. A value it allows, x = 10, gives -40 + 9 = -31, far below 25.

Question 3 Standard

Which of the following gives all values of x that satisfy 5(x - 3) < 2x + 6?

Show the answer Choice B

Why it is right

Distribute the 5 across both terms in the parentheses: 5x - 15 < 2x + 6. Subtract 2x from both sides: 3x - 15 < 6. Add 15 to both sides: 3x < 21. The final division is by the positive number 3, so the direction stays the same and x < 7. Check x = 0: 5(-3) = -15 and -15 < 6 is true. Check x = 8: 5(5) = 25 and 25 < 22 is false.

Why each other choice fails

Choice A
Reverses the symbol even though every operation used was subtraction, addition, and division by the positive number 3. The direction changes only when both sides are multiplied or divided by a negative.
Choice C
Multiplies only the x by 5, writing 5x - 3 < 2x + 6 and reaching 3x < 9. The 5 must also multiply the -3 inside the parentheses.
Choice D
Subtracts 15 from both sides instead of adding it: 3x - 15 < 6 becomes 3x < -9. A -15 is removed by adding 15.

Question 4 Standard

Which ordered pair (x, y) is a solution to the inequality y ≤ 2x - 5?

Show the answer Choice D

Why it is right

There is nothing to solve here: substitute each pair and read whether the resulting numerical statement is true. For (4, 1), the right side is 2(4) - 5 = 3, and 1 ≤ 3 is true, so (4, 1) belongs to the solution set. Every other pair turns the inequality into a false statement, so testing is both the fastest and the safest method.

Why each other choice fails

Choice A
Puts the numbers in the wrong slots. With x = 1 and y = 4, the right side is 2(1) - 5 = -3, and 4 ≤ -3 is false.
Choice B
Works only if the -5 is dropped: 1 ≤ 2(2) = 4 looks true, but the actual right side is 2(2) - 5 = -1 and 1 ≤ -1 is false.
Choice C
Satisfies the reversed inequality. The right side is 2(6) - 5 = 7, and 8 ≥ 7, not 8 ≤ 7.

Question 5 Harder

The number line shown graphs the solution set of one of the inequalities below. Which inequality has this solution set?

-5 -4 -3 -2 -1 0 1 2 3 4 5 x
The solution set graphed on a number line. The circle at 2 is filled.
Show the answer Choice B

Why it is right

Read the graph before solving anything: the circle sits at 2, it is filled, and the shading runs to the left. That is the solution set x ≤ 2, boundary included. Now solve the candidates: -2x + 1 ≥ -3 gives -2x ≥ -4, and dividing by -2 reverses the symbol, so x ≤ 2. The endpoint checks exactly: -2(2) + 1 = -3 and -3 ≥ -3 is true, which is why the circle is filled rather than open.

Why each other choice fails

Choice A
Solves to x ≥ 2: the same endpoint but the opposite ray. Its graph would shade to the right of 2.
Choice C
Solves to x < 2. The strict symbol excludes the endpoint, which is drawn as an open (unfilled) circle, not the filled one shown.
Choice D
Solves to -2x ≥ 2, that is x ≤ -1. The direction and the closed endpoint match the graph, but the boundary is at -1, not at 2.

Question 6 Standard

An empty delivery van weighs 3,200 pounds. The van is loaded with crates that weigh 95 pounds each, and the combined weight of the van and its crates must be no more than 8,000 pounds. What is the greatest number of crates that can be loaded?

Show the answer Choice C

Why it is right

Let c be the number of crates. The combined weight is 95c + 3200, so 95c + 3200 ≤ 8000, which gives 95c ≤ 4800 and c ≤ 4800 / 95 = 50.52... Crates are whole objects and the inequality caps c from above, so the answer rounds down to 50: fifty crates weigh 4,750 pounds for a total of 7,950 pounds, while fifty-one weigh 4,845 pounds for a total of 8,045 pounds.

Why each other choice fails

Choice A
Rounds 50.52 up. Rounding up crosses the ceiling the inequality sets: 51 crates put the loaded van 45 pounds over the 8,000-pound limit.
Choice B
Divides 8,000 by 95 and ignores the van's own 3,200 pounds. Only the remaining 4,800 pounds of capacity is available for crates.
Choice D
Divides the van's empty weight, 3,200, by 95. The empty weight is not capacity for crates; it is the part of the limit already used up.

Question 7 Harder

Which of the following gives all values of x that satisfy -5 < 3 - 2x ≤ 11?

Show the answer Choice D

Why it is right

Subtract 3 from all three parts: -8 < -2x ≤ 8. Divide all three parts by -2; because -2 is negative, both symbols reverse: 4 > x ≥ -4. Rewriting from smallest to largest gives -4 ≤ x < 4. Each strictness travels with the number it came from: the ≤ was attached to the 8, and 8 / (-2) = -4, so -4 is the included end. Check x = -4: 3 + 8 = 11 and 11 ≤ 11 is true. Check x = 4: 3 - 8 = -5, and -5 < -5 is false.

Why each other choice fails

Choice A
Divides by 2 instead of -2, so nothing reverses and the strict and inclusive ends stay where they were printed. Test x = 4: 3 - 2(4) = -5, and -5 < -5 is false, so 4 cannot be included.
Choice B
Adds 3 to the outer parts instead of subtracting it, producing -2 < -2x ≤ 14 and then -7 ≤ x < 1. Test x = -7: 3 + 14 = 17, which is not ≤ 11.
Choice C
Has both endpoints right but marks both as included. The original chain opens with a strict <, so x = 4 must be excluded: 3 - 2(4) = -5 and -5 < -5 is false.

Question 8 Harder

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

-3 -2 -1 0 1 2 3 4 5 -6 -4 -2 0 2 4 6 8 10 (2, 0) x y
The shaded region is the solution set. The boundary line is dashed.
Show the answer Choice A

Why it is right

Read the boundary first. It crosses the y-axis at (0, 1) and rises 2 units for every 1 unit to the right, so the boundary line is y = 2x + 1. The line is dashed, which means points on the line are not solutions, so the symbol must be strict. The shading lies below the line, and the marked point (2, 0) settles the direction: 2(2) + 1 = 5, and 0 < 5 is true. The inequality is y < 2x + 1.

Why each other choice fails

Choice B
Reads the dashed boundary as a solid one. A solid line includes its own points in the solution set; a dashed line excludes them.
Choice C
Shades the wrong side. Testing the marked point (2, 0) gives 0 > 5, which is false, so points below the line do not satisfy y > 2x + 1.
Choice D
Inverts the slope, reading a rise of 2 over a run of 1 as 1/2. The line y = 0.5x + 1 passes through (2, 2), not through the (2, 5) that the drawn boundary passes through.

Question 9 Hardest

In the inequality -4x + k ≥ 6, k is a constant. The solution to the inequality is x ≤ 3. What is the value of k?

Show the answer Choice B

Why it is right

Solve with k carried along. Subtract k from both sides: -4x ≥ 6 - k. Divide by -4, which reverses the symbol: x ≤ (6 - k) / (-4) = (k - 6) / 4. The published solution is x ≤ 3, so the two boundaries must match: (k - 6) / 4 = 3, giving k - 6 = 12 and k = 18. Check by substituting: -4x + 18 ≥ 6 becomes -4x ≥ -12, and dividing by -4 with a flip gives x ≤ 3.

Why each other choice fails

Choice A
Divides by 4 instead of -4, so the symbol never reverses; setting (6 - k) / 4 = 3 gives k = -6. Substituting -6 gives -4x ≥ 12, whose solution is x ≤ -3, not x ≤ 3.
Choice C
Stops at an intermediate quantity. The last step must read -4x ≥ -12, so 12 is the size of the constant that has to move, but k itself is 6 + 12 = 18.
Choice D
Adds k to the right side instead of subtracting it: -4x ≥ 6 + k, then 6 + k = -12 and k = -18. Substituting -18 gives -4x ≥ 24, whose solution is x ≤ -6.

Question 10 Hardest

A robotics club has $560 to spend on competition kits that cost $24 each. After buying the kits, the club must have at least $80 of that money left for travel. What is the greatest number of kits the club can buy?

Show the answer Choice D

Why it is right

If the club buys n kits it spends 24n dollars and has 560 - 24n dollars left. The travel rule is "at least $80 left," so 560 - 24n ≥ 80. Subtracting 560 gives -24n ≥ -480, and dividing by -24 reverses the symbol: n ≤ 20. Twenty kits cost $480 and leave exactly $80, which "at least $80" permits, so 20 is legal; twenty-one kits cost $504 and leave only $56.

Why each other choice fails

Choice A
Treats the boundary as forbidden. "At least $80" includes exactly $80, and 20 kits leave exactly $80, so the club can still afford the twentieth kit.
Choice B
Spends the entire $560: 560 / 24 = 23.3, so 23 kits. That leaves only $8 for travel and breaks the reserve requirement.
Choice C
Adds the $80 reserve to the budget instead of subtracting it: 640 / 24 = 26.7. Twenty-six kits cost $624, which is more money than the club has.

Question 11 Harder Student-produced response

What is the greatest integer value of x that satisfies 9 - 2x ≥ 3x - 16?

Show the answer 5

Why it is right

Collect the x terms on the side that keeps them positive so no flip is needed: add 2x to both sides to get 9 ≥ 5x - 16, then add 16 to get 25 ≥ 5x. Dividing by the positive 5 keeps the direction, so 5 ≥ x, that is x ≤ 5. The symbol allows equality, and x = 5 checks exactly: 9 - 10 = -1 on the left and 15 - 16 = -1 on the right, so -1 ≥ -1 is true. The greatest integer solution is 5.

Answers students type instead

1
Combines the constants as 16 - 9 = 7 instead of 9 + 16 = 25, producing 5x ≤ 7 and x ≤ 1.4.
4
Assumes the boundary is excluded. At x = 5 both sides equal -1, and ≥ allows equality, so 5 itself is a solution.
25
Stops at the intermediate step 25 ≥ 5x and reports 25 instead of dividing both sides by 5.

Question 12 Hardest Student-produced response

A food pantry has already collected 1,250 pounds of food toward a goal of at least 4,000 pounds. The rest will be collected in donation boxes that hold 35 pounds each. What is the minimum number of boxes needed to reach the goal?

Show the answer 79

Why it is right

Let b be the number of boxes. The pantry will have 1250 + 35b pounds, and the goal is "at least 4,000", so 1250 + 35b ≥ 4000. Subtracting gives 35b ≥ 2750, so b ≥ 2750 / 35 = 78.57... Here the inequality bounds b from below, so the whole-number answer rounds up, not down: 79 boxes add 2,765 pounds for a total of 4,015 pounds, while 78 boxes add only 2,730 pounds for a total of 3,980.

Answers students type instead

36
Divides the 1,250 pounds already collected by 35 instead of the 2,750 pounds still needed.
78
Rounds 78.57 down out of habit. Seventy-eight boxes reach only 3,980 pounds — 20 pounds short of the goal.
115
Ignores the 1,250 pounds already collected and divides the whole goal by 35: 4000 / 35 = 114.3, rounded up to 115.

Common mistakes

  1. Forgetting the flip — dividing by a negative and keeping the symbol, which returns the exact complement of the right answer.
  2. Flipping when nothing negative was divided — subtracting a bigger number, or seeing a minus sign anywhere in the problem, does not reverse anything.
  3. Missing the boundary word — reading “no more than 25” as <25<25, or “at least 80” as >80>80, and losing the one value that decides the item.
  4. Rounding the wrong way — 7.5 rounded up under a ceiling, or 78.6 rounded down above a floor. Round toward the inside of the range, never to the nearest.
  5. Reversing the constraint in a story — writing “budget ≥\ge cost” as “cost ≥\ge budget”, or capping the quantity that was supposed to be the minimum.
  6. Answering the boundary instead of the count — reporting 7.5 sessions, or the intermediate 25 from 25≥5x25 \ge 5x instead of 5.
  7. Losing the strict/inclusive ends in a compound inequality — reversing the chain but leaving each symbol on the side it was printed on.
  8. Reading the figure’s style wrong — a hollow circle taken as filled, or a dashed boundary taken as solid, which changes << into ≤\le and vice versa.
  9. Forgetting the whole other side — checking one number inside your range but never one outside it, so a reversed symbol survives the check.

FAQ

Do I flip when I subtract a negative number? No. The direction changes only when both sides are multiplied or divided by a negative. x−(−3)>5x - (-3) > 5 is just x+3>5x + 3 > 5, symbol untouched.

Is “at least 5” the same as “more than 5”? No, and items are built on the difference. “At least 5” is ≥5\ge 5 and 5 itself qualifies; “more than 5” is >5>5 and 5 does not. On a graph, that is the difference between a filled and a hollow circle.

Is this the same skill as systems of inequalities? No. Here a single constraint produces one range or one half-plane. When two constraints must hold at once and you need their overlap, that is systems of linear inequalities — the search results mix the two, the score report does not.

Can an SPR answer be a range? Never. If an item is student-produced response, it is asking for one number — usually the greatest or least value that satisfies the inequality, which is why the boundary decision matters twice as much there.

Related skills

Coming next in this domain: systems of linear inequalities, graphs of linear equations.

Editorial team · last reviewed 2026-08-06