MHF4U Rational Equations and Inequalities: Notes and Practice

Key things to remember

1State the restrictions — every zero of every denominator — before solving anything.

2To solve an equation, multiply every term by the LCD, then solve the resulting polynomial equation.

3A root that equals a restriction is extraneous: reject it and say so. All roots rejected means "no solution".

4Never multiply or cross-multiply an inequality by an expression containing the variable — its sign is unknown.

5For an inequality, rewrite it as a single rational expression compared to

0

, then build a sign chart.

6Critical values come from the numerator (square brackets allowed) and the denominator (always round brackets).

Everything in this topic comes down to one uncomfortable fact: the moment you multiply both sides of an equation by something containing $x$ , you have multiplied by a quantity that might be zero. Doing that can create answers that solve the new equation but not the original one. The fix is simple and it is worth marks on every test — write the restrictions before you start, and check every root against them at the end.

Solving rational equations

Factor every denominator and state the restrictions.
Find the lowest common denominator (LCD) and multiply every term on both sides by it.
Simplify — every fraction should disappear — leaving a linear or quadratic equation.
Solve that polynomial equation by any method you know.
Compare each root to the restrictions. Reject any root that equals a restriction and say explicitly that it is extraneous.
If every root is rejected, the correct answer is "no solution" — that is a real answer, not a mistake.

The most expensive habit in this unit

Students solve the algebra perfectly, get a root that happens to be a restricted value, and hand it in. Full algebra marks, zero communication marks, wrong final answer. Write the restrictions on the first line so the check at the end is automatic.

Fig 1Restrictions come from the denominators and must be written down before any solving begins.

Why extraneous roots appear

Multiplying both sides by $(x - 2)$ is a legitimate move only when $x \neq = 2$ . If $x = 2$ , you multiplied both sides by zero, which turns any false statement into the true statement $0 = 0$ . That is why the transformed equation can have a solution the original does not. The equation below is the classic example: the algebra is flawless and the answer is still rejected.

\frac{x}{x - 3} = \frac{3}{x - 3} \times (x - 3) x = 3, but x \neq = 3, so there is no solution

VideoKhan Academy solves a rational equation by multiplying through by the denominator and then shows why one of the answers has to be rejected because it makes a denominator zero.Watch on YouTube

Rational inequalities

An inequality looks similar but must be handled completely differently. With an equation you may multiply both sides by anything non-zero. With an inequality, multiplying by a negative number reverses the inequality sign — and an expression like $x - 3$ is positive for some values of $x$ and negative for others. You do not know which, so you are not allowed to multiply by it at all.

Never multiply an inequality by a variable expression

Cross-multiplying an inequality is the same error wearing a disguise. If you multiply by something whose sign you do not know, you do not know whether to flip the sign, and half of your answer will be wrong.

Instead, move everything to one side, combine over a single common denominator, and compare the result to zero. Now the question is only "where is this expression positive / negative?", which a sign chart answers. The critical values are the zeros of the numerator and the zeros of the denominator, because the expression can only change sign by passing through zero or by being undefined.

Fig 2Critical values from both the numerator and the denominator split the number line into test intervals.

Interval	x − 1	x + 2	Quotient	Satisfies $\geq 0$ ?
$(- \infty, - 2)$	−	−	+	yes
$(- 2, 1)$	−	+	−	no
$(1, \infty)$	+	+	+	yes

Sign chart for (x − 1)/(x + 2) ≥ 0. Solution: x < −2 or x ≥ 1.

Read the brackets off carefully. A numerator zero satisfies $\geq$ or $\leq$ , so it gets a square bracket. A denominator zero is undefined, so it always gets a round bracket, no matter which inequality symbol appears. For the chart above the solution is $(- \infty, - 2) \cup [1, \infty)$ : round at $- 2$ because the function is undefined there, square at $1$ because $f (1) = 0$ and $0 \geq 0$ is true.

One rule for each kind of problem

Equation: multiply through by the LCD, solve, then reject extraneous roots. Inequality: collect on one side over a common denominator, then sign chart. Never mix the two methods.

VideoMove everything to one side, combine into a single fraction, factor, and use the critical values from both the numerator and denominator on a number line, the same method as this section.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Solve

\frac{x + 3}{x - 1} = 2

Example 2

Solve

\frac{1}{x - 2} + \frac{1}{x + 2} = \frac{4}{x ^{2} - 4}

Example 3

Solve

\frac{2}{x} + \frac{3}{x + 1} = 1

, leaving exact values.

Example 4

Solve the inequality

\frac{3}{x - 2} < 1

Where marks get lost

Solving correctly and then reporting a root that was restricted. Always compare every root to the restriction list before writing the final answer.

Multiplying an inequality by

(x - 3)

as though it were positive. When

x < 3

that expression is negative and the inequality sign must flip, so the method silently loses half the solution set.

Cross-multiplying

\frac{a}{b} < \frac{c}{d}

. This is the same illegal move as multiplying by a variable expression.

Leaving the denominator's zeros off the sign chart. The expression can change sign there (it does at any odd-multiplicity zero) even though it is never equal to zero there.

Using a square bracket at a restricted value, for example writing

[2, \infty)

when

x = 2

makes the denominator zero. Restricted endpoints are always round.

Key things to remember

Worked examples

Where marks get lost

Flashcards 8

Practice