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.
Solve $-3x < 12$. A classmate wrote $x < -4$. Before reading on — can you identify the error? What should the correct answer be?
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.
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.
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.
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.
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.
What to write in your book
- Negative coefficient: divide by the (negative) coefficient and flip the sign.
- Write the flip step explicitly, e.g. "$\div (-3)$: flip $<$ to $>$".
- Verify with a test value from the solution set.
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): $x \leq -6$.
What to write in your book
- Fractional coefficient $\frac{x}{k}$: multiply by $k$. Flip only if $k$ is negative.
- Fractional coefficient $\frac{-x}{k}$: the sign of the final multiplier determines the flip.
- Key question: "Am I multiplying/dividing by a positive or negative number?"
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 $<$).
What to write in your book
- Multi-step: isolate the variable term first, then divide/multiply by the coefficient.
- Apply the flip rule only at the step where you divide/multiply by a negative.
- Verify by substituting a value both inside and outside the solution set.
Worked examples · 3 in a row, reveal as you go
Solve $-4x + 6 > -10$.
Solve $\dfrac{x}{3} < 5$.
Solve $-\dfrac{x}{2} \geq 3$.
What to write in your book
- Negative coefficient → divide by negative → flip.
- Positive fractional coefficient → multiply by positive reciprocal → no flip.
- Negative fractional coefficient → two steps: clear fraction (no flip), then clear negative sign (flip).
- Solve $-6x > 24$. Show the flip step clearly.
- Solve $\dfrac{x}{5} \leq 4$.
- Solve $-\dfrac{x}{3} < 2$. Identify where the flip occurs.
- A classmate solved $-2x > 8$ and wrote $x > -4$. Identify and correct the 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.
Pick your answer, then rate your confidence.
Q1. Solve $-2x + 10 < 4$ and check your solution. Show each step clearly, indicating where the flip rule is applied. (4 marks)
Q2. Solve $\dfrac{x}{4} \geq -3$ and $-\dfrac{x}{4} \geq -3$ and compare the solutions. (3 marks)
Q3. A student solved $\frac{x}{-3} > 4$ and wrote $x > -12$. Show whether this is correct and explain. (2 marks)
📖 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].
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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