📊 Linear Inequalities: Multiplication & Division (Sign Flipping)

Master multiplication and division properties of inequalities, apply the golden rule of flipping the inequality sign when multiplying or dividing by a negative number, and solve one-step inequalities.

Multiplication & Division (Sign Flipping)

Learning Topics

Multiplication, Division & Sign Flipping - Linear Inequalities

1. Multiplication & Division Properties of Inequalities

To isolate a variable multiplied or divided by a coefficient, we use the Multiplication and Division Properties of Inequality:
  • Multiplying/Dividing by a Positive Number (c>0c > 0): The inequality direction remains unchanged:
    If a<b and c>0  ⟹  a⋅c<b⋅candac<bc\text{If } a < b \text{ and } c > 0 \implies a \cdot c < b \cdot c \quad \text{and} \quad \frac{a}{c} < \frac{b}{c}

    Example: 2x<10  ⟹  x<52x < 10 \implies x < 5
  • ⚠️ The Golden Rule — Multiplying/Dividing by a Negative Number (c<0c < 0): When multiplying or dividing both sides by a negative number, the inequality sign MUST FLIP direction:
    If a<b and c<0  ⟹  a⋅c>b⋅candac>bc\text{If } a < b \text{ and } c < 0 \implies a \cdot c > b \cdot c \quad \text{and} \quad \frac{a}{c} > \frac{b}{c}
  • Why does the sign flip? (Two Intuitive Explanations):
    1. Geometric / Number Line Reason: Consider 2<52 < 5. If we multiply both sides by −1-1, we get −2-2 and −5-5. On the real number line, −2-2 lies to the right of −5-5, so −2>−5-2 > -5. Multiplying or dividing by a negative number reflects positions across zero, reversing the relative order.
    2. Algebraic Explanation by Moving Both Sides: Consider −x>−3-x > -3. If we add +x+x to both sides, we get 0>−3+x0 > -3 + x. Next, adding +3+3 to both sides gives 3>x3 > x, which directly means x<3x < 3. Thus, −x>−3  ⟺  x<3-x > -3 \iff x < 3. This is exactly the same result as multiplying both sides by −1-1, which reverses the inequality. In this sense, moving both sides to the opposite side is algebraically equivalent to multiplying the entire inequality by −1-1.

2. Avoiding the Common Misconception

🚨 Crucial distinction: The sign flips ONLY when the number you multiply or divide BY is negative. Dividing a negative result by a positive coefficient does NOT flip the sign:
  • 3x≤−12  ⟹  x≤−43x \le -12 \implies x \le -4 (Divided by +3>0+3 > 0, sign stays ≤\le)
  • −3x≤12  ⟹  x≥−4-3x \le 12 \implies x \ge -4 (Divided by −3<0-3 < 0, sign flips to ≥\ge)

3. Step-by-Step Solver Method

To solve a multiplication or division inequality, follow these systematic steps:
  • Step 1 — Identify the Coefficient & Operation: Determine the number and operation affecting the variable.
  • Step 2 — Apply the Inverse Operation: Divide if the variable is multiplied, or multiply if the variable is divided.
  • Step 3 — 🚨 Sign Flip Check: If the multiplier/divisor is negative (c<0c < 0), reverse the inequality sign (<⟷>< \longleftrightarrow > and ≤⟷≥\le \longleftrightarrow \ge). If positive (c>0c > 0), keep the sign unchanged.
  • Step 4 — Simplify & Graph: Calculate the numerical boundary point, select open/solid circle, and shade the solution ray.
Worked Example: Solve −4x≥28-4x \ge 28
1. xx is multiplied by −4-4.
2. Apply inverse operation: Divide both sides by −4-4.
3. 🚨 Sign Flip Check: Since we are dividing by a negative number (−4<0-4 < 0), reverse the inequality sign (≥  ⟹  ≤\ge \implies \le):
−4x−4≤28−4\frac{-4x}{-4} \le \frac{28}{-4}

4. Simplify: x≤−7x \le -7.
5. Graph: Solid circle at −7-7, shaded to the left towards −∞-\infty.
-10-9-8-7-6-5-4-3-2-10

Mastering SealMath: Typing Inequality Symbols

When solving inequalities with math input, you can type inequality signs directly using simple keyboard shortcuts. The editor automatically converts them to mathematical symbols:
To WriteType on KeyboardMath Display
Greater Than or Equal To (≥)>=
≥\ge
Less Than or Equal To (≤)<=
≤\le
Strictly Greater (>)>
>>
Strictly Less (<)<
<<
Not Equal (≠)neq
≠\ne
⚠️
⚠️ The Golden Rule: Sign Flip
Whenever you multiply or divide both sides of an inequality by a negative number, always reverse the direction of the inequality symbol:
<<    ⟹  \implies  >>
>>    ⟹  \implies  <<
≤\le    ⟹  \implies  ≥\ge
≥\ge    ⟹  \implies  ≤\le
Learning Topics

❓ Frequently Asked Questions

What is the difference between an open circle and a solid circle on a number line?
An open circle (○) indicates a strict inequality (<< or >>) where the boundary point is not included. A solid circle (●) indicates an inclusive inequality (≤\le or ≥\ge) where the boundary point is part of the solution.
How do I know which direction to shade on the number line?
When the variable is isolated on the left side (like x>ax > a or x≥ax \ge a), shade to the right towards positive infinity (+∞+\infty). For x<ax < a or x≤ax \le a, shade to the left towards negative infinity (−∞-\infty). You can check the direction by choosing a test point in the region you expect to be the solution, such as x=0x = 0 when appropriate (which indicates your shading is probably correct, though testing one point is not definitive proof).
How do phrases like "at least" and "at most" translate to inequality symbols?
"At least" represents a minimum requirement and translates to greater than or equal to (≥\ge). "At most" represents a maximum limit and translates to less than or equal to (≤\le).
When do you flip the inequality sign when solving an inequality?
You flip the inequality sign (e.g., from << to >>, or from ≤\le to ≥\ge) whenever you multiply or divide both sides by a negative number. Adding or subtracting numbers, or multiplying/dividing by a positive number, never flips the sign.
How do you solve and graph a compound inequality like a<x≤ba < x \le b?
A compound inequality a<x≤ba < x \le b represents the interval where xx is strictly greater than aa and at most bb. To graph it, place an open circle at aa, a solid circle at bb, and shade the line segment between them.
How can I verify if my inequality solution is correct?
Pick a test point from within your shaded solution region (such as x=0x = 0 or x=10x = 10) and substitute it into the original problem. If it yields a true statement (such as 5≤125 \le 12), your solution and shading are probably correct (though a single test point is not a definitive proof).