These Digital SAT Algebra questions solve like linear equations with one extra rule to track — when to flip the inequality sign — plus compound ranges, word problems, and two-variable solution regions.
A linear inequality solves exactly like a linear equation, with one exception that trips up a lot of students: multiplying or dividing both sides by a negative number reverses the inequality sign. Everything else — distributing, combining like terms, moving terms across the inequality — works precisely as it does for equations.
Use Isolate, Flip, Test. First, isolate the variable using the same simplify-and-organize moves from linear equations. Second, the moment you divide or multiply by a negative number to finish isolating, flip the inequality sign — and only then. Third, test a point: plug a value from your solution set back into the original inequality to confirm it holds. This catches a missed flip before it costs you the question.
Walk through an inequality that needs a flip.
Solve −2x − 3 > 7.
That single flip rule handles most one-variable items. The next section covers the version that hides it inside parentheses — where students who know the rule still get caught.
The flip rule is simple to state and easy to forget under pressure, especially when a negative sign is hiding inside parentheses instead of sitting in front of the variable. Distribute first, exactly like you would for an equation — the flip only happens at the moment you divide or multiply by a negative, never at the moment you distribute one.
Take −(2x − 5) ≥ 9. Distribute the negative across both terms first: −2x + 5 ≥ 9. Only now do you isolate: subtract 5 to get −2x ≥ 4, then divide by −2 and flip to get x ≤ −2. Distributing the negative did not itself require a flip — it was the division by −2 that did.
A second common slip: flipping when you add or subtract a negative number. Adding or subtracting never flips an inequality, no matter the sign of the number involved — only multiplication and division by a negative do that.
Solve −(2x − 5) ≥ 9.
A compound inequality sandwiches the variable between two bounds, like −5 < 2x + 1 ≤ 9. Solve it by applying the exact same operation to all three parts at once, not just the middle and one side. Subtract 1 from every part: −6 < 2x ≤ 8. Divide every part by 2: −3 < x ≤ 4. If that division had been by a negative number, both inequality signs would flip — and swap sides.
Word problems add a translation step before any of this. Watch for "at least" (≥), "at most" or "no more than" (≤), "more than" (> , strict), and "fewer than" (<, strict). A parking garage that charges a $5 flat fee plus $2 per hour, for a customer who wants to pay no more than $17, becomes 5 + 2h ≤ 17. Solve it exactly like an equation: subtract 5 to get 2h ≤ 12, divide by 2 (no flip — the coefficient is positive) to get h ≤ 6 hours.
Once you have a solution set, sanity-check it against the story the same way you would for a one-variable equation: negative hours, negative distances, or a percent above 100% usually means a sign or direction got flipped when it shouldn't have.
A linear inequality in two variables, like y ≥ 2x − 1, doesn't have a single solution — it has infinitely many, filling an entire half of the coordinate plane. The boundary line is exactly the equation you'd get by replacing the inequality sign with an equals sign; everything on one side of that line satisfies the inequality, and everything on the other side doesn't.
To decide which side is the solution region without graphing, pick any point not on the boundary line — the origin (0, 0) is usually easiest — and substitute it into the inequality. If it makes the inequality true, that point's side is the solution region; if false, the other side is. The boundary itself is included (solid line) when the inequality has ≤ or ≥, and excluded (dashed line) when it's strict, < or >.
A system of two inequalities — tested on the Digital SAT with "which point satisfies both" questions — asks for the overlap of both regions. Rather than graphing, it's usually faster to just substitute each answer choice's point into both inequalities directly and see which point makes both true.
Speed on this skill comes from trusting the flip rule instead of re-deriving it every time. Mix a plain solve, a flip, a compound inequality, and a point-check below.
Answer each before revealing it.
Solve: 5x − 2 < 18
Solve: −x/3 ≥ 2
Solve the compound inequality: −1 ≤ 3x + 2 < 11
Which of x = −2, x = 0, x = 5 satisfy −2x + 4 > 0?
Translate: "A subscription costs $12 plus $3 per extra user, and the total must not exceed $45." Write the inequality for u extra users.
Does the point (2, 5) satisfy y ≤ 3x − 1?
Inequalities are the last piece of the linear toolkit — combined with systems of linear equations, they cover nearly every straight-line question the Digital SAT can ask.
Here are three Linear Inequalities questions from our problem sets, from easier to harder, in the same format as the Digital SAT. Answer each one to see the full explanation.
Five shorter questions to finish: a flip, a compound inequality, a strict inequality, a budget word problem, and a system of two inequalities. Test a point on each one.
Solve −4x + 9 ≤ −3.
Solve −7 < 2x − 3 ≤ 5.
Which value of x satisfies 3x + 1 > 4x − 5?
A rental car costs $35 plus $0.20 per mile. If Maria's budget is at most $75, what is the maximum number of miles she can drive?
Which point satisfies both y ≥ 2x − 1 and y < −x + 6?
Related: Linear equations in one variable · Linear equations in two variables · Linear functions · Systems of linear equations · SAT Math overview