7.1 Linear Inequalities in One and Two Variables
Key Takeaways
Solving a one-variable linear inequality uses the same inverse operations as solving an equation, but multiplying or dividing both sides by a negative number reverses the inequality symbol.
On a number line, a strict inequality (< or >) uses an open circle and a non-strict inequality (≤ or ≥) uses a closed circle; shade in the direction that satisfies a test point.
The graph of a two-variable linear inequality has a boundary line that is dashed when the inequality is strict and solid when it is non-strict; shade the half-plane that contains a test point satisfying the inequality.
A compound inequality joined by 'and' is the intersection of two solution sets (a single overlapping interval), while one joined by 'or' is the union (either part alone is enough).
Florida's B.E.S.T. benchmark MA.912.AR.2.6 covers one-variable linear inequalities (including compound inequalities), while MA.912.AR.2.7 and MA.912.AR.2.8 cover writing and graphing two-variable linear inequalities; all three require interpreting solutions in context.
Quick Answer: A linear inequality is solved like a linear equation, except multiplying or dividing both sides by a negative number flips the inequality symbol. On the coordinate plane, graph the boundary line first—solid for ≤/≥ and dashed for </>—then test a point (the origin is fastest) to decide which half-plane to shade. Compound inequalities combine two inequalities with 'and' (intersection, one overlapping interval) or 'or' (union, either part works).
One-Variable Linear Inequalities (MA.912.AR.2.6)
A linear inequality in one variable is a statement such as 3x − 7 < 11 or −2x + 5 ≥ 13. Solving it is identical to solving the corresponding equation—combine like terms, isolate the variable term, then divide—with one critical exception: multiplying or dividing both sides by a negative number reverses the inequality symbol.
Worked example: solve −4x + 12 > 28.
- Subtract 12 from both sides:
−4x > 16. - Divide both sides by −4 and flip the symbol:
x < −4.
The solution set is all real numbers less than −4. On the number line, draw an open circle at −4 (because the inequality is strict, <) and shade to the left.
Number-Line Conventions
| Symbol | Circle Type | Meaning |
|---|---|---|
< | Open | Endpoint not included |
> | Open | Endpoint not included |
≤ | Closed (filled) | Endpoint included |
≥ | Closed (filled) | Endpoint included |
A common Florida EOC trap shows a correctly solved inequality paired with a reversed circle type. If the original inequality uses ≥, the endpoint is included, so the circle must be closed—no exceptions.
Multi-Step Inequalities with Variables on Both Sides
For 5(x − 2) + 3 ≤ 2x + 4, distribute first: 5x − 10 + 3 ≤ 2x + 4, combine like terms: 5x − 7 ≤ 2x + 4, subtract 2x from both sides: 3x − 7 ≤ 4, add 7: 3x ≤ 11, divide by 3: x ≤ 11/3. No sign flip was needed because the divisor was positive.
Two-Variable Linear Inequalities (MA.912.AR.2.7 / MA.912.AR.2.8)
A two-variable linear inequality looks like y > 2x − 3 or 2x + 3y ≤ 12. Its graph is a half-plane: the set of all points (x, y) satisfying the inequality.
Step-by-Step Graphing Procedure
- Graph the boundary line. Replace the inequality symbol with
=and graph the resulting line. Use a solid line for≤or≥(points on the line are solutions) and a dashed line for<or>(points on the line are NOT solutions). - Pick a test point not on the line—
(0, 0)is easiest when the line does not pass through the origin. - Substitute the test point into the original inequality. If it produces a true statement, shade the half-plane containing the test point. If it produces a false statement, shade the opposite half-plane.
Worked example: graph y ≤ −x + 4.
- Boundary:
y = −x + 4, slope −1, y-intercept 4. The inequality is≤, so the line is solid. - Test
(0, 0):0 ≤ −0 + 4→0 ≤ 4, which is true. Shade the half-plane containing the origin (the region below and to the left of the line).
Boundary-Line Trap
The single most frequently missed point on Florida EOC inequality items is the boundary-line type. A strict inequality (< or >) demands a dashed boundary because points on the line do not satisfy the inequality. A non-strict inequality (≤ or ≥) demands a solid boundary. If an answer choice shows y > 3x − 2 with a solid line, eliminate it immediately.
Compound Inequalities
A compound inequality joins two inequalities with the words and or or.
'And' (Intersection)
−3 ≤ 2x − 1 < 7 is shorthand for −3 ≤ 2x − 1 and 2x − 1 < 7. Solve both parts by performing the same operation on all three segments: add 1 → −2 ≤ 2x < 8, divide by 2 → −1 ≤ x < 4. The graph is a segment from −1 (closed) to 4 (open). The solution set is the overlap of the two individual solution sets.
'Or' (Union)
x < −2 or x ≥ 5 is satisfied by any number less than −2 OR any number 5 or greater. The graph has two shaded rays pointing in opposite directions. Only one part needs to be true for the full statement to hold.
Florida EOC Compound-Inequality Traps
- A three-part inequality written as
a < bx + c < dis always an 'and' statement; it never represents 'or'. - If you split
a < bx + c < dinto two separate inequalities and solve each independently, you must keep the connector as 'and'—changing it to 'or' changes the meaning. - When the coefficient of x is negative and you divide all three parts by it, every inequality symbol flips:
5 > −x > −1becomes−5 < x < 1.
Real-World Context
Florida EOC word problems translate phrases into inequality symbols: 'at least' → ≥, 'no more than' → ≤, 'fewer than' → <, 'more than' → >. A budget constraint like 'spend no more than $30 on x notebooks at $4 each and y pens at $3 each' becomes 4x + 3y ≤ 30—a non-strict two-variable inequality with a solid boundary line.
Master the boundary-line decision and the sign-flip rule, and you have covered the two most-tested skills on this benchmark.
A student solves −3x + 6 ≤ 18 and writes the solution as x ≤ −4. Which error did the student make?
The student forgot to flip the inequality when dividing by −3; the correct solution is x ≥ −4.
The student should have added 6 to both sides; the correct solution is x ≤ 4.
The student failed to distribute; the correct solution is x ≤ 14.
The student used the wrong starting inequality; the correct solution is x ≤ −4.
Which graph correctly represents the inequality y > 2x − 5?
A solid boundary line with the region above the line shaded.
A solid boundary line with the region below the line shaded.
A dashed boundary line with the region above the line shaded.
A dashed boundary line with the region below the line shaded.
What is the solution set of the compound inequality −5 ≤ 3x + 1 < 10?
−2 < x ≤ 3
−2 ≤ x < 3
x ≤ −2 or x > 3
−5 ≤ x < 10
Sections you finish are checked off in the contents.