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How do you solve a linear inequality in one variable, and why does multiplying or dividing by a negative flip the inequality?

Solve linear inequalities in one variable and represent the solution on a number line and in interval form, reversing the inequality when multiplying or dividing by a negative (TN A1.A.REI.B.3).

A TNReady Algebra I answer on solving linear inequalities (TN A1.A.REI.B.3), the flip rule for negatives, graphing on a number line with open and closed circles, and compound inequalities.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. The flip rule
  3. Graphing on a number line
  4. Compound inequalities
  5. How TNReady examines this topic
  6. Why the circle type matters for credit
  7. Interval notation and writing the answer
  8. Try this

What this topic is asking

Standard A1.A.REI.B.3 also covers inequalities. Solving an inequality is almost identical to solving an equation, with one extra rule: multiplying or dividing both sides by a negative number reverses the inequality sign. You then express the solution as an inequality, on a number line (open or closed circle plus a shaded ray), or in interval notation.

The flip rule

Adding or subtracting from both sides never changes the direction. Multiplying or dividing by a negative does. The reason is concrete: 2<52 < 5 is true, but multiplying both sides by 1-1 gives 2-2 and 5-5, and 2>5-2 > -5, so the direction must reverse to stay true.

Graphing on a number line

The two graph decisions are the circle type and the direction.

  • Open circle for << or >>: the endpoint is not a solution.
  • Closed (filled) circle for \le or \ge: the endpoint is a solution.
  • Shade toward the solutions: x>ax > a shades right; x<ax < a shades left.

A reliable check is to pick a test value in the shaded region and confirm it satisfies the original inequality.

Compound inequalities

A compound inequality combines two conditions. An "and" inequality like 2<x5-2 < x \le 5 is the overlap (a segment between 2-2 open and 55 closed). An "or" inequality like x<1x < 1 or x4x \ge 4 is two separate rays. To solve a three-part inequality such as 12x+3<7-1 \le 2x + 3 < 7, apply each operation to all three parts at once.

How TNReady examines this topic

  • Multiple choice. Solve and choose the correct inequality, with unflipped-sign distractors.
  • Graphing. Place the circle (open or closed) and shade the ray on a number line.
  • Multiple select. Choose all values that satisfy the inequality.

A clarifying idea is that an inequality has infinitely many solutions (a whole region), unlike a linear equation's single value. That is why the answer is a graph or a range, and why "select all that apply" items suit inequalities so well.

Why the circle type matters for credit

On a graphing item, the open-versus-closed circle is scored, and it encodes a real distinction. For x>3x > 3, the value x=3x = 3 gives 3>33 > 3, which is false, so 33 is excluded and the circle is open. For x3x \ge 3, the value x=3x = 3 gives 333 \ge 3, which is true, so 33 is included and the circle is filled. Getting the boundary right is not a formality: in a context problem, \ge versus >> can be the difference between "at least 3 items" (3 allowed) and "more than 3 items" (3 not allowed). Read the inequality symbol and translate it faithfully to the circle.

Interval notation and writing the answer

Some TNReady items accept or display the solution in interval notation, a compact way to write a range. A bracket [[\, or ]]\, includes the endpoint (matches a closed circle); a parenthesis ((\, or ))\, excludes it (matches an open circle), and \infty always takes a parenthesis because infinity is never reached. So x>3x > 3 is (3,)(3, \infty), x3x \le -3 is (,3](-\infty, -3], and the compound 2<x5-2 < x \le 5 is (2,5](-2, 5]. Reading between the three forms, inequality, number line, and interval, is a small but testable skill, and the bracket-versus-parenthesis choice carries the same open-or-closed information as the circle.

Try this

Q1. Solve 5x<20-5x < 20. [1 point]

  • Cue. Divide by 5-5 and flip: x>4x > -4.

Q2. Solve and graph 3x+173x + 1 \le 7. [2 points]

  • Cue. 3x6x23x \le 6 \Rightarrow x \le 2; closed circle at 22, shade left.

Exam-style practice questions

Practice questions written in the style of TDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TNReady (style)2 marksMultiple choice. Solve 3x+719-3x + 7 \le 19. (A) x4x \ge -4 (B) x4x \le -4 (C) x4x \ge 4 (D) x4x \le 4
Show worked answer →

The correct answer is (A).

Subtract 77 from both sides: 3x12-3x \le 12. Now divide both sides by 3-3, and because you are dividing by a negative, the inequality sign flips: x4x \ge -4. Distractor (B) keeps the sign unflipped, the single most common error. Testing x=0x = 0 in the original: 3(0)+7=719-3(0) + 7 = 7 \le 19 is true, and 040 \ge -4 is also true, confirming the direction.

TNReady (style)2 marksGraphing. Solve 2x5>12x - 5 > 1 and describe the graph on a number line (direction and circle type).
Show worked answer →

The solution is x>3x > 3: an open circle at 33 with the ray shaded to the right.

Add 55: 2x>62x > 6. Divide by 22 (positive, no flip): x>3x > 3. Because the inequality is strict (>>, not \ge), 33 itself is not included, so the circle is open. The solution is all values greater than 33, so shade right. A closed circle would be wrong here because x=3x = 3 gives 1>11 > 1, which is false.

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