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hscscienceMaths Std · Y11
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Module 1 · L18 of 20 ~45 min ⚡ +90 XP available

Inequalities with Negatives and Fractions

Consolidate the flip rule with negative coefficients and extend to fractional coefficients. Multiplying by a positive reciprocal does NOT flip the sign, only division or multiplication by a negative does.

Today's hook, A classmate solved $-3x < 12$ and wrote $x < -4$. Is she correct? One tiny missed step leads to the wrong solution set, and potentially the wrong answer in an exam.
0/5QUESTS
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Recall, your gut answer first
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Solve $-3x < 12$. A classmate wrote $x < -4$. Before reading on, can you identify the error? What should the correct answer be?

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The flip rule, a deeper look
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The flip rule applies only when multiplying or dividing by a negative number. Multiplying by a positive reciprocal (to clear a fraction) does NOT flip the sign.

$\dfrac{x}{3} < 5$: multiply both sides by $+3$ (positive) → $x < 15$. No flip.   $\dfrac{-x}{2} \geq 3$: multiply by $-2$ (negative) OR divide $-x$ by $-1$ → flip the sign.

FLIP WHEN: ÷ or × by NEGATIVE -3x < 12 ÷(-3) → x > -4 ✓ NO FLIP: × or ÷ by POSITIVE x/3 < 5 ×3 → x < 15 ✓
$-3x < 12 \Rightarrow$ divide by $-3$ (flip): $x > -4$.   $\dfrac{x}{3} < 5 \Rightarrow$ multiply by $+3$ (no flip): $x < 15$.
Check the sign of the divisor
Before dividing, identify whether the divisor is positive or negative. Only negative divisors trigger the flip.
Fractions: find the sign
$\frac{x}{3} < 5$: multiplying by $+3$ is positive, no flip. $\frac{-x}{2} \geq 3$: multiplying both sides by $-2$ OR dividing by $-1$, flip required.
Always verify
After solving, substitute a test value. For $x > -4$, try $x = 0$: $-3(0) = 0 < 12$ ✓. Try $x = -5$: $-3(-5) = 15 \not< 12$, correctly outside the set.
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What you'll master
Know

Key facts

  • Flip only when multiplying or dividing by a negative number.
  • Multiplying by a positive reciprocal to clear a fraction does NOT flip the sign.
  • The sign of the resulting coefficient determines whether a flip is needed.
Understand

Concepts

  • Why $-x/2 \geq 3$ requires a flip but $x/3 < 5$ does not.
  • How to recognise whether a fractional coefficient is positive or negative.
  • Why verifying with a test value is essential for these problems.
Can do

Skills

  • Solve inequalities with negative coefficients, applying the flip rule correctly.
  • Solve inequalities with fractional coefficients using reciprocal multiplication.
  • Identify and correct the common error of forgetting to flip.
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Key terms
CoefficientThe number multiplying a variable. In $-3x$, the coefficient is $-3$.
ReciprocalThe multiplicative inverse of a number. The reciprocal of $\frac{1}{3}$ is $3$; the reciprocal of $-\frac{1}{4}$ is $-4$.
Flip ruleThe rule that the inequality sign must be reversed when both sides are multiplied or divided by a negative number.
Equivalent inequalityAn inequality with the same solution set as the original, produced by applying legal operations to both sides.
Fractional coefficientA coefficient that is a fraction, such as $\frac{1}{3}x$ or $-\frac{x}{2}$. Sign determines whether a flip is needed.
Test valueA specific number substituted into the original inequality to verify the solution set is correct.
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Negative coefficients, always flip
core concept

When the coefficient of $x$ is negative (e.g. $-3x$, $-5x$, $-x$), dividing both sides to isolate $x$ means dividing by a negative number. This always triggers the flip rule.

$-3x < 12$: divide by $-3$ → flip $<$ to $>$: $x > -4$. The classmate in today's hook forgot to flip, giving the wrong direction entirely.

Why it matters: Forgetting the flip gives a completely wrong solution set. Every value in the wrong set fails the original inequality. Always check one value from your answer.
Quick check: solve $-5x \leq 20$.

Negative coefficient rule: dividing both sides by a negative number always flips the inequality sign (< → >, ≤ → ≥, etc.). This is the most common error in inequality questions. Check: is the coefficient of x negative? If yes, flip when you divide.

Pause, copy the negative coefficient flip rule (−ax > b → divide by −a → flip → x < −b/a) and the checking question ("is the coefficient of x negative?"), if yes, the flip occurs when you divide by it into your book.

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Fractional coefficients
core concept

We just saw that negative whole-number coefficients always require a flip when you divide both sides, e.g. −2x > 6 → divide by −2 → flip → x < −3. That raises a question: what about fractional coefficients like x/3 > 4 or −x/2 < 5, do you multiply (not divide), and does the flip rule still apply? This card answers it → multiply both sides by the denominator to clear the fraction; the flip only applies if that denominator is negative.

For $\dfrac{x}{3} < 5$: the coefficient of $x$ is $+\frac{1}{3}$. Multiply both sides by $+3$ (positive) → no flip: $x < 15$.

For $\dfrac{-x}{2} \geq 3$: the coefficient of $x$ is $-\frac{1}{2}$. Multiply both sides by $-2$ (negative) → flip: $x \leq -6$. Or equivalently: multiply by $+2$ to get $-x \geq 6$, then multiply by $-1$ (flip): $x \leq -6$.

Decide first: Before solving, ask "is the coefficient of $x$ positive or negative?" If positive → no flip when isolating. If negative → flip required.
Which step does NOT require flipping the inequality sign?

Fractional coefficient x/k: multiply both sides by k. If k > 0, no flip. If k < 0 (e.g. −x/2), multiplying by k means multiplying by a negative → flip. Always identify the sign of the coefficient of x before deciding whether to flip.

Pause, copy the fractional coefficient method: multiply both sides by the denominator to clear the fraction, then solve; the flip rule applies only if the multiplier is negative (e.g. −x/2 < 5: multiply by −2 → flip → x > −10) into your book.

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Mixed problems, putting it together
core concept

We just saw fractional coefficients, multiply by the denominator, and flip only if the denominator is negative. That raises a question: a mixed problem has both fractions and a negative coefficient on x, what order do you tackle the steps to avoid a sign error? This card answers it → isolate the variable term first (add/subtract constants from both sides), then deal with the coefficient (multiply to clear fractions, check sign → flip or no flip).

Multi-step problems may involve both a negative coefficient and a fraction. Isolate the variable term first (add/subtract), then deal with the coefficient. Apply the flip rule exactly once, only when the final division/multiplication is by a negative.

Example: $-4x + 6 > -10 \Rightarrow -4x > -16 \Rightarrow x < 4$ (divide by $-4$, flip $>$ to $<$).

Fill the blank: dividing both sides of an inequality by $-4$ _______ the inequality sign.

Multi-step inequality with both negative coefficient and fractions: isolate the variable term first (add/subtract to remove constants), then deal with the coefficient (multiply/divide, check sign for flip). Work one step at a time; show each operation explicitly.

Pause, copy the multi-step inequality strategy: (1) isolate the variable term by adding/subtracting constants first; (2) then multiply or divide by the coefficient, checking its sign; (3) if the coefficient is negative, flip, this order prevents sign errors into your book.

PROBLEM 1 · NEGATIVE COEFFICIENT

Solve $-4x + 6 > -10$.

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Subtract 6 from both sides: $-4x > -16$
Isolate the variable term. Subtracting a positive, no flip yet.
PROBLEM 2 · POSITIVE FRACTIONAL COEFFICIENT

Solve $\dfrac{x}{3} < 5$.

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Coefficient of $x$ is $+\frac{1}{3}$ (positive). Multiply both sides by $+3$.
Multiplying by a positive number, no flip required.
PROBLEM 3 · NEGATIVE FRACTIONAL COEFFICIENT

Solve $-\dfrac{x}{2} \geq 3$.

1
Multiply both sides by $2$: $-x \geq 6$
Multiplying by positive 2, no flip yet. Now we have coefficient $-1$.
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Quick-fire practice
  1. Solve $-6x > 24$. Show the flip step clearly.
  2. Solve $\dfrac{x}{5} \leq 4$.
  3. Solve $-\dfrac{x}{3} < 2$. Identify where the flip occurs.
  4. A classmate solved $-2x > 8$ and wrote $x > -4$. Identify and correct the error.
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Revisit the classmate's error

The classmate solved $-3x < 12$ and wrote $x < -4$, forgetting to flip. The correct answer is $x > -4$ (divide by $-3$, flip $<$ to $>$).

Test: does $x = 0$ satisfy $-3x < 12$? $-3(0) = 0 < 12$ ✓. Is $0 > -4$? Yes ✓. Does it satisfy the classmate's answer $x < -4$? No. The flip error leads to a completely opposite solution set.

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01
Multiple choice
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Pick your answer, then rate your confidence.

02
Short answer
ApplyBand 44 marks

Q1. Solve $-2x + 10 < 4$ and check your solution. Show each step clearly, indicating where the flip rule is applied. (4 marks)

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ApplyBand 33 marks

Q2. Solve $\dfrac{x}{4} \geq -3$ and $-\dfrac{x}{4} \geq -3$ and compare the solutions. (3 marks)

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AnalyseBand 42 marks

Q3. A student solved $\frac{x}{-3} > 4$ and wrote $x > -12$. Show whether this is correct and explain. (2 marks)

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📖 Comprehensive answers (click to reveal)

Practice: 1. $-6x > 24 \Rightarrow x < -4$ (flip). 2. $x \leq 20$ (no flip). 3. $-x/3 < 2 \Rightarrow x > -6$ (flip when multiplying by $-3$). 4. Error: should be $x < -4$ not $x > -4$.

Q1 (4 marks): $-2x + 10 < 4 \Rightarrow -2x < -6$ [1]. Divide by $-2$, flip $<$ to $>$: $x > 3$ [1]. Check: $x = 4$: $-2(4)+10 = 2 < 4$ ✓ [1]. $x = 2$: $-2(2)+10 = 6 \not< 4$ ✓ (outside) [1].

Q2 (3 marks): $x/4 \geq -3 \Rightarrow x \geq -12$ (multiply by $+4$, no flip) [1]. $-x/4 \geq -3 \Rightarrow -x \geq -12 \Rightarrow x \leq 12$ (flip when $\times -1$) [1]. The solutions are completely different because one coefficient is positive and the other is negative [1].

Q3 (2 marks): $x/(-3) > 4$ means dividing by $-3$; multiply by $-3$ (negative) and flip: $x < -12$, NOT $x > -12$ [1]. The student forgot to flip, check: $x = 0$: $0/(-3) = 0 \not> 4$, so 0 should not be a solution. $x < -12$ is correct [1].

01
Boss battle · Flip Rule Gauntlet
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Solve negative and fractional coefficient inequalities at speed, always identifying whether the flip applies. Beat the boss to bank a tier.

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02
Science Jump · platform challenge

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