Engineers design bridges that must withstand variable loads, doctors dose medications so concentrations stay within safe ranges, and economists model revenue functions that must stay positive. All of these require solving inequalities involving fractions, rational inequalities. In this lesson you'll master the sign-table method that makes these problems systematic and foolproof.
Why is it dangerous to multiply both sides of an inequality by an expression containing $x$? Think before you read on what could go wrong?
There are only two core ideas in this lesson. First: always bring everything to one side so the other side is zero. Second: use a sign table test one point per interval, never multiply by unknowns.
Every rational inequality lives on one road: bring all terms to one side, combine into a single fraction, then find where that fraction is positive or negative using a sign table. The denominator's zeros are always excluded no exceptions.
To solve a rational inequality such as $\dfrac{x - 2}{x + 3} \le 0$, follow these six steps every time:
Six steps: (1) move all terms to one side, (2) single fraction, (3) critical values, (4) sign table, (5) determine signs, (6) select intervals and exclude denominator zeros; Never multiply both sides by an expression containing x, you don't know its.
Pause, copy the six-step rational inequality method into your book: (1) move all terms to one side; (2) combine into a single fraction; (3) find critical values; (4) draw a sign table; (5) determine sign of each interval; (6) select correct intervals, excluding denominator zeros.
Quick check: To solve $\dfrac{x+1}{x-3} > 0$, what are the critical values?
We just saw the six-step method: combine into a single fraction, find critical values, build a sign table, and exclude denominator zeros. That raises a question: why can we test just one point per interval, what guarantees the sign doesn't change between critical values? This card answers it → a rational expression changes sign only at a zero of numerator or denominator; between consecutive critical values, the sign is constant throughout.
A rational expression can only change sign at a critical value. Between any two consecutive critical values, the expression maintains the same sign throughout. This is the foundation of the whole method.
Consider $\dfrac{x-2}{x+3}$. The critical values divide the number line into three intervals:
We test just one point in each interval, say $x = -4$, $x = 0$, $x = 3$, to confirm the sign. Then we select the intervals where the sign matches the inequality.
Sign of f(x){g(x)}: positive when f and g have the same sign; negative when they have opposite signs; Between critical values, the sign is constant, test one representative point per interval
Pause, copy the sign-table rule into your book: $\frac{f(x)}{g(x)}$ is positive when $f$ and $g$ have the same sign; between any two consecutive critical values the expression's sign is constant, test one point per interval.
Did you get this? True or false: if $x = 5$ makes the denominator of a rational expression zero, then $x = 5$ must be excluded from the solution even when solving $\dfrac{f(x)}{x-5} \le 0$.
Worked examples · 3 in a row, reveal as you go
Solve: $\dfrac{x - 2}{x + 3} \le 0$
| Interval | $x+3$ | $x-2$ | Expression |
|---|---|---|---|
| $x < -3$ | $-$ | $-$ | $+$ |
| $-3 < x < 2$ | $+$ | $-$ | $-$ |
| $x > 2$ | $+$ | $+$ | $+$ |
Solve: $\dfrac{2x - 1}{x + 3} > 1$
Solve: $\dfrac{x^2 - 4}{x} > 0$
| Interval | $x-2$ | $x+2$ | $x$ | Result |
|---|---|---|---|---|
| $x < -2$ | $-$ | $-$ | $-$ | $-$ |
| $-2 < x < 0$ | $-$ | $+$ | $-$ | $+$ |
| $0 < x < 2$ | $-$ | $+$ | $+$ | $-$ |
| $x > 2$ | $+$ | $+$ | $+$ | $+$ |
Fill the gap: To solve $\dfrac{3x}{x+2} > 2$, the first step is to rewrite it as $\dfrac{3x}{x+2} - 2 > 0$, then combine over a common denominator to get $\dfrac{x-4}{x+2}$ $0$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for the inequality $\dfrac{x+1}{x-2} \le 0$, the value $x = 2$ must be included in the solution because the inequality uses $\le$.
Activities · practice with the ideas
Solve $\dfrac{x + 1}{x - 2} \le 0$ and express your answer in interval notation.
Solve $\dfrac{3x}{x + 2} > 2$. Show all steps including how you bring terms to one side.
Solve $\dfrac{x^2 - 9}{x} \ge 0$. There are three critical values, identify them all.
A student claims the solution to $\dfrac{x-3}{x+1} > 0$ is $x > 3$. What error did they make, and what is the correct solution?
Explain in your own words why the denominator zero must always be excluded from the solution, even when the inequality sign includes equality ($\le$ or $\ge$).
Earlier you were asked: why is it dangerous to multiply both sides of an inequality by an expression containing $x$?
The answer is that the expression in $x$ can be either positive or negative depending on the value of $x$. When you multiply an inequality by a negative number, the inequality sign flips. Since you don't know whether the expression is positive or negative, you don't know whether to flip the sign, and any assumption you make will give the wrong answer for at least some values of $x$.
The correct approach: always bring all terms to one side and use a sign table to determine the sign in each interval.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Solve $\dfrac{x + 1}{x - 2} \le 0$, clearly stating which critical values are included and which are excluded. (2 marks)
Q2. Solve $\dfrac{3x}{x + 2} > 2$. Show all working including how you bring terms to one side and combine into a single fraction. (3 marks)
Q3. Solve $\dfrac{x^2 - 4}{x} > 0$. Your solution should include a sign table with all three critical values, and express the final answer in interval notation. (3 marks)
Activity drills: 1. CVs: $x = -1$, $x = 2$. Sign: $+$ on $x < -1$, $-$ on $-1 < x < 2$, $+$ on $x > 2$. Solution: $[-1, 2)$, include $x=-1$ (numerator zero, $\le$), exclude $x=2$ (denominator zero). · 2. Subtract 2: $\frac{3x - 2(x+2)}{x+2} = \frac{x-4}{x+2} > 0$. CVs: $x=4$, $x=-2$. Positive when $x < -2$ or $x > 4$. · 3. $\frac{(x-3)(x+3)}{x}$. CVs: $-3, 0, 3$. Positive on $(-3,0)\cup(3,\infty)$; zero at $\pm 3$ included. Solution: $[-3, 0) \cup [3, \infty)$. · 4. Error: multiplied both sides by $(x+1)$ without knowing its sign. Correct: $x < -1$ or $x > 3$. · 5. At a denominator zero, the expression is undefined, division by zero is not defined, so it cannot satisfy any inequality.
Q1 (2 marks): CVs: $x = -1$ (numerator zero) and $x = 2$ (denominator zero) [0.5]. Sign table: $+$ for $x < -1$, $-$ for $-1 < x < 2$, $+$ for $x > 2$ [0.5]. Select negative or zero: $-1 \le x < 2$, i.e. $[-1, 2)$ [1]. Include $x = -1$ (numerator zero, $\le$); exclude $x = 2$ (denominator zero always excluded) [correct notation: +0.5 if both inclusions stated correctly].
Q2 (3 marks): $\frac{3x}{x+2} - 2 > 0$ [1]. $\frac{3x - 2(x+2)}{x+2} = \frac{x-4}{x+2} > 0$ [1]. CVs: $x = 4$ (excluded, $>$) and $x = -2$ (excluded, denominator). Positive when $x < -2$ or $x > 4$. Solution: $(-\infty, -2) \cup (4, \infty)$ [1].
Q3 (3 marks): $\frac{(x-2)(x+2)}{x} > 0$ [0.5]. CVs: $x = -2, 0, 2$ [0.5]. Four intervals; sign pattern: $-,+,-,+$ (for $x<-2$, $-2<x<0$, $0<x<2$, $x>2$) [1]. Select positive: $(-2, 0) \cup (2, \infty)$, all endpoints excluded (strict inequality or denominator zero) [1].
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