Rational Equations & Inequalities (AA HL)
Solving a rational equation means clearing the fractions and checking your answers still make sense in the original expression; solving a rational inequality means finding the critical values and reading off which regions satisfy the sign you need. Both skills sit on top of the graph work in the wider Rational Functions topic - the intercepts and asymptotes you'd sketch there are exactly the critical values this page uses.
28 questions on this sub-topic.
Method, not a formula
This falls under IB syllabus reference SL2.8, which covers rational functions of the form \(f(x)=\dfrac{ax+b}{cx+d}\) and their graphs. There's no formula-booklet entry for "solving" them - it's a two-step method built on the algebra you already know.
Equations
Multiply every term by the denominator(s) to clear the fractions, then solve the polynomial that's left.
Always exclude any solution that would make an original denominator zero - it isn't a valid root of the rational equation, even though it solves the cleared-up polynomial.
Inequalities
Rearrange so one side is 0, find where the numerator and denominator are each zero, then build a sign diagram across the intervals those critical values create.
Never multiply both sides by an expression whose sign you don't know - it can flip the inequality the wrong way for part of the domain.
Want the graph shapes and asymptote formulas behind this? See Rational Functions.
Worked examples
Find the \(x\)- and \(y\)-intercepts of \(f(x)=\dfrac{2x-6}{x+1}.\)
(a)(i) State the x-intercept.
(a)(ii) State the y-intercept.
Worked solution
\(x\)-intercept: numerator \(= 0 \Rightarrow x = 3\), point \((3,0).\) M1 A1 \(y\)-intercept: \(f(0) = -6\), point \((0,-6).\) A1
State the asymptotes of \(f(x) = \dfrac{2x+1}{x-3}\).
(a)(i) State the vertical asymptote.
(a)(ii) State the horizontal asymptote.
Worked solution
Where the denominator is zero: \(x - 3 = 0 \Rightarrow x = 3.\) M1 A1
Equal degrees, so \(y = \dfrac{\text{leading coefficients}}{} = 2.\) M1 A1
Common mistakes
- Cross-multiplying an inequality. \(\dfrac{ax+b}{cx+d} > k\) does not become \(ax+b > k(cx+d)\) unless you already know \(cx+d\) is positive - if the denominator can be negative for part of the domain, this flips the inequality sign for those \(x\)-values and gives a wrong answer.
- Keeping an excluded solution. Clearing denominators can introduce a root that makes the original denominator zero. Always substitute your answers back into the original expression and reject anything undefined there.
- Drawing the sign diagram from the wrong critical values. Both the zeros of the numerator and the zeros of the denominator change the sign of a rational expression - leaving out the denominator's zero misses a sign change at the vertical asymptote.
Ready to practise properly?
28 rational equations & inequalities questions, marked instantly like the real exam.
Quick answers
How do you solve a rational equation?
Multiply both sides by the denominator(s) to clear the fractions, solve the resulting polynomial equation, then reject any solution that makes an original denominator zero.
How do you solve a rational inequality?
Move everything to one side, find the critical values where the numerator or denominator is zero, then use a sign diagram across those regions - you cannot cross-multiply an inequality by an unknown-sign denominator.