Linear inequalities

This skill point covers linear inequalities in one and two variables: solving them, writing them from a described constraint, reading a solution set off a graph, and handling systems of them. The algebra is the same algebra as linear equations, with one extra rule and one extra kind of answer — a range or a region rather than a single number.

The Digital SAT asks it in four recognisable shapes. Solve-and-report: "which of the following gives all values of x." Translate: a scenario with a budget, a minimum, or a capacity, and four candidate inequalities or systems. Graph-read: an inequality or system with choices describing the boundary and the shading, or a set of candidate points. And the hard variant, which asks for the greatest or least value something can take inside a solution region.

College Board publishes the domain weight, not the skill weight: Algebra is about 35% of the scored Math questions — 13 to 15 of them — which makes it the largest domain on the section. How those split across the five Algebra skills is not published. The prep-industry figure of roughly one to three inequality questions per form is an inference from released material, not an official number; treat it as a rough prior rather than something to plan around.

One structural fact worth carrying into every item below: because the answer is a range or a region rather than a number, wrong answers on this topic are usually the right magnitude pointing the wrong way. That is a much harder error to notice than a wrong number, which is precisely why the test uses it.

If you can solve −3x + 7 < 1 without pausing over which way the symbol should end up, skip to the next block; nothing here will be new. If you can't, this is the block that matters most on the page, and it builds from nothing.

The four symbols, said out loud. x < 5 is "x is strictly less than 5" — 5 itself does not count. x ≤ 5 is "x is at most 5" — 5 does count. x > 5 is "strictly greater than." x ≥ 5 is "at least 5." The line under the symbol is the only thing that decides whether the endpoint is in or out, and on the SAT that single line is worth full marks or zero surprisingly often.

An inequality's answer is usually infinitely many numbers, not one. x > 3 names every number to the right of 3 on the number line: 3.001, 4, 900. On a number line you mark 3 with an open circle (not included) and shade to the right; for x ≥ 3 the circle is filled. Hold onto that open-versus-filled distinction — in two variables it becomes dashed versus solid, and it is the same idea both times.

Solving is the same procedure you already use on equations. Add or subtract anything from both sides. Multiply or divide both sides by any positive number. All of those steps leave the symbol exactly as it is, including when the number you add or subtract is itself negative — adding −7 to both sides never flips anything.

Exactly one step behaves differently: multiplying or dividing both sides by a negative number reverses the direction of the symbol. −2x < 6 becomes x > −3, not x < −3. Nothing else in the procedure has this property, and forgetting it is the single most common way this question type is missed.

The strong path avoids the rule instead of remembering it. Move the variable terms to whichever side keeps their coefficient positive, and you will never divide by a negative in the first place. Weak path: 4 − 3x > 10 → −3x > 6 → divide by −3 and remember to flip → x < −2. Strong path: 4 − 3x > 10 → 4 > 3x + 10 → −6 > 3x → −2 > x, which reads x < −2. Same answer, one fewer opportunity to be wrong. Use the strong path by default; know the rule for the times a question hands you the negative coefficient anyway.

The ten-second check that catches almost every direction error: take one number from inside your answer and one from outside, and substitute both into the original inequality. The inside number must satisfy it and the outside number must fail. For x < −2 above, test x = −3: 4 − 3(−3) = 13 > 10 ✓. Then test x = 0: 4 > 10 ✗. Both behaved correctly, so the direction is right.

Compound inequalities — a value trapped between two bounds, like −5 < 2x + 1 ≤ 9 — are solved by doing the same operation to all three parts at once: subtract 1 everywhere to get −6 < 2x ≤ 8, then divide by 2 everywhere to get −3 < x ≤ 4. If you ever divide all three parts by a negative, both symbols reverse and the two ends swap places.

y ≥ 2x − 1 has no single solution. It has a solution set: every ordered pair (x, y) that makes the statement true — (0, 0), (0, 5), (4, 100), infinitely many. Plotted, those points fill in a half of the plane, bounded by the line y = 2x − 1.

Drawing it is two decisions, taken separately. First, the boundary: replace the inequality symbol with = and graph that line. It is solid when the symbol is inclusive (≤ or ≥), because points on the line satisfy the statement, and dashed when the symbol is strict (< or >), because they don't. Second, the shading: which of the two sides is the solution set.

Settle the shading with a test point, never by staring at the symbol. Pick any point that is not on the boundary, substitute it into the original inequality, and see whether the statement comes out true. If it does, shade that point's side. If it doesn't, shade the other one. Use (0, 0) whenever the boundary doesn't pass through it — the arithmetic collapses to comparing 0 against a constant and there is nothing to slip on. If the line does run through the origin, pick something equally cheap, like (1, 0).

Mechanism: one test point is enough because a linear expression can only get from "less than" to "greater than" by passing through "equal to," and the set of points where it is equal to the constant is exactly the boundary line. Cross that line and the comparison flips; stay on one side and it never can. That is why the plane splits into precisely two regions, and why a single substitution settles an infinite set.

The reason the test point matters more than the symbol: the SAT frequently gives you the inequality in a form that is not solved for y, such as 4x − 2y > 10. Isolating y there means dividing by −2, which reverses the symbol to y < 2x − 5 — so a question with a > in it shades below. The symbol on the page is not the symbol that governs the picture until y is alone with a positive coefficient. A test point is immune to that entirely.

In Bluebook, the built-in Desmos calculator will shade a two-variable inequality, or an entire system, if you type it in. On "which point is in the solution set" and "which system is graphed" items that is a legitimate and often decisive move. It costs twenty to forty seconds of typing, so treat it as the tool for items you are unsure about rather than a default — and note that it offers no protection at all against answering a question about the wrong quantity, which is where most points on this topic are actually lost.

A system of inequalities is a set of conditions that must all hold at once. Its solution set is the intersection of the individual regions — the overlap, and only the overlap. A point in the shaded area of one inequality and outside another is not a solution to anything.

The translation table earns more points on this topic than the algebra does. "At least" and "no less than" and "a minimum of" are ≥. "At most," "no more than," "cannot exceed," and "a maximum of" are ≤. "More than" and "fewer than" alone are strict, > and <. The two that get misread under time pressure are the double negatives: "no fewer than 12" means ≥ 12, and "not more than 12" means ≤ 12.

Building a system from a scenario is mechanical once you name the units. Each constraint is one sentence about one quantity: a total cost in dollars compared against a budget in dollars, a total number of people compared against a required headcount. Write each constraint separately, check that every term in it carries the same unit as the bound it is being compared to, and only then look at the choices.

On "which system is graphed here" items, run three independent checks per boundary: the line itself (slope and intercept), its style (solid or dashed), and which side is shaded. Answer choices in this family almost always agree on two of the three and differ on one. Find the axis they differ on before you start verifying anything, and one check eliminates two choices at once.

In real-world systems the quantities are often counts, which cannot be negative. Some items write x ≥ 0 and y ≥ 0 into the system explicitly and some leave it to context. If two answer choices differ only by including those constraints, the question is testing whether you noticed the quantities were counts.

At 1500 the flip rule is not the problem; the edge cases and the execution are. The first edge case is the one the mechanism block named: multiplying or dividing both sides by an expression containing a variable. If the sign of that expression is unknown, the step has no defined effect on the symbol and cannot be taken. When a question appears to require it, the intended path is almost always different — clear the fraction by moving terms, or test the answer choices — because the SAT does not write items whose only route runs through an undefined step.

The second is answering the right question about the wrong quantity, which costs more points at the top of the scale than any misconception does. The item asks for the greatest number of hours and you report the cost. It asks for the maximum value of y and you report the x you solved for on the way. It asks how many more of one thing than another and you report a total. Before selecting, re-read the final clause of the prompt and say what the answer is a count of, and in what unit.

Third: boundary inclusion and rounding are two separate decisions that only interact in one place. Inclusion is settled by the symbol — ≤ includes the endpoint, < does not. Rounding is settled by the context — a "greatest number that fits inside a budget" rounds down, a "least number needed to reach a target" rounds up, regardless of what the decimal looks like. The one case where they collide is when the division lands on an exact integer: at 12.0 with ≤ the answer is 12, and with < it is 11. If your arithmetic comes out whole, the strict-versus-inclusive distinction has just become load-bearing.

Fourth, the maximum-and-minimum family. When a system's solution region is bounded and a question asks for the greatest or least value of a variable inside it, the answer sits at a corner — the intersection of two boundary lines — so drop the inequality symbols, solve those two boundaries as a system of equations, and you have the point. Then confirm the corner is actually included: if either boundary is strict, that exact point is outside the region and the question is either asking about a different corner or has been engineered so the extremum is approached but never reached.

Fifth, the slips that are not misconceptions at all. Distributing a negative across parentheses and missing the second term: −3(2x − 5) is −6x + 15, and the sign on the 15 is the one that gets dropped. A rate given per week when the variable counts days. A constraint stated in cents against answer choices in dollars. A system where the two variables are defined in the opposite order to the one the choices use. None of these are knowledge failures, which is why they survive to test day — they are caught only by a deliberate re-read of the setup, not by knowing more algebra.

Finally, on hard graph-reading items, remember that a strict boundary drawn as a dashed line changes which answer is correct without changing the picture much. Two answer choices differing only in ≤ versus < are not a formatting detail; they are the entire question, and the item was written by someone who knew exactly which one you would notice under time pressure.