How do you solve a linear inequality in one variable, when do you flip the sign, and how do you graph and interpret the solution?
Solve linear inequalities in one variable, including those requiring the distributive property and those with variables on both sides, graph the solution, and interpret it in context (TEKS A.5B).
A STAAR Algebra I answer on solving one-variable linear inequalities (TEKS A.5B), the rule for flipping the sign when multiplying or dividing by a negative, graphing on a number line, and interpreting in context.
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What this topic is asking
TEKS A.5B is the inequality companion to solving equations, in the same heavily weighted Reporting Category 3. You solve a one-variable inequality almost exactly like an equation, with one extra rule, then graph the solution on a number line and interpret it in context. The redesigned test often makes the graph itself the answer through a hot-spot item.
Solve like an equation, with one rule
The steps mirror equation-solving: distribute, combine like terms, collect the variable, and divide. The one difference is the sign-flip rule.
For : add 1 to get , divide by positive 4 (no flip) to get . For : divide by and flip to get .
Graphing the solution on a number line
The solution to a one-variable inequality is a ray (or interval) on the number line.
- Circle style. Open circle for a strict inequality ( or ); closed circle for or .
- Direction. Shade toward the solutions: right for or , left for or , once the variable is alone on the left.
So is an open circle at 28 with a ray to the right; is a closed circle at with a ray to the left.
Variables on both sides
Collect the variable on one side first, then apply the rule only if the final division is by a negative.
Interpreting in context
In a real-world inequality, finish with a sentence. A solution months means "from month 6 onward"; a solution items means "at most 20 items". Where only whole numbers make sense, state the smallest or largest acceptable whole number, the interpretive credit the standard rewards.
How STAAR examines this topic
- Multiple choice. Solve an inequality; the "forgot to flip" answer is always offered.
- Hot spot. Build the number-line graph (circle style plus ray direction), both of which must be correct.
- Inline choice and number entry. State a boundary value or choose the correct symbol.
A clarifying idea is why the sign flips: multiplying by a negative reverses the order of numbers (for example but ), so to keep the statement true the symbol must turn around.
Compound inequalities and "at least / at most" language
Some inequalities have two bounds, written as a compound statement. "Between 10 and 20 inclusive" is , graphed as a segment with closed circles at both ends. You solve each part the same way, applying any operation to all three regions at once: from , add 3 throughout to get , then divide by 2 to get .
Translating the everyday phrasing is its own assessed skill. "At least" means , "at most" means , "more than" means , and "no more than" means . A budget that allows spending "no more than \\le 50$ constraint, and the solution describes every affordable amount. Reading these phrases as the correct symbol is where many context inequalities are won or lost, because the rest of the algebra is routine once the symbol is right.
Try this
Q1. Solve . [1 point]
- Cue. (flip on dividing by ).
Q2. Graph on a number line. [1 point]
- Cue. Closed circle at 2, ray to the right.
Exam-style practice questions
Practice questions written in the style of TEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
STAAR (style)1 marksMultiple choice. What is the solution to ? (A) (B) (C) (D) Show worked answer →
The correct answer is (A).
Subtract 5: . Divide both sides by , and because you divide by a negative, flip the inequality: . Forgetting to flip gives (choice B), the single most common inequality error. Adding or subtracting never flips the sign; only multiplying or dividing by a negative does.
STAAR (style)2 marksHot spot. A delivery driver must make more than 40 stops to earn a bonus. They have made 12 stops and average no fewer than the rest. Solve for , then select the graph: an open or closed circle at 28 with a ray to the right or left.Show worked answer →
The solution is : an open circle at 28 with the ray pointing right.
Subtract 12 from both sides: . The strict symbol means 28 itself is not included, so the circle is open (not filled). The solution is all values greater than 28, so the ray points right. A closed circle would require , and a leftward ray would represent ; both are distractors on the hot-spot item.
Related dot points
- Solve linear equations in one variable, including those requiring the distributive property and those with variables on both sides, and identify equations with one solution, no solution, or infinitely many (TEKS A.5A).
A STAAR Algebra I answer on solving one-variable linear equations (TEKS A.5A), the inverse-operations routine, the distributive property, variables on both sides, and recognizing one solution, no solution, or infinitely many.
- Graph the solution set of linear inequalities in two variables on the coordinate plane, using a solid or dashed boundary and shading the correct half-plane (TEKS A.3D).
A STAAR Algebra I answer on graphing linear inequalities in two variables (TEKS A.3D) - dashed versus solid boundary lines, choosing the half-plane to shade with a test point, and the hot-spot item type.
- Write linear functions that model the relationship between two quantities from a description, table, or graph, write an equation representing a functional relationship, and evaluate functions in function notation (TEKS A.2C, A.2G, A.12B).
A STAAR Algebra I answer on writing linear functions to model situations, identifying initial value and rate, function notation f(x), and evaluating functions (TEKS A.2C, A.2G, A.12B).
- Determine the domain and range of a linear function in mathematical problems, and reasonable domain and range values (continuous and discrete) for real-world situations, representing them using inequalities (TEKS A.2A).
A STAAR Algebra I answer on the domain and range of linear functions (TEKS A.2A), continuous versus discrete situations, reasonable real-world values, and representing domain and range with inequalities.
- Write systems of two linear equations given a table, a graph, a verbal description, or a real-world problem, then solve and interpret the solution in context (TEKS A.2H, A.2I).
A STAAR Algebra I answer on writing and modeling with systems of two linear equations from real-world situations (TEKS A.2H, A.2I), defining variables, building one equation per condition, and interpreting the solution.
Sources & how we know this
- STAAR Algebra I Assessed Curriculum — Texas Education Agency (2024)
- 19 TAC Chapter 111, Algebra I (TEKS), Adopted 2012 — Texas Education Agency (2012)