Grade 12 · MHF4U · Advanced Functions

MHF4U: Rational Equations and Inequalities

A rational equation is solved by clearing its denominators; a rational inequality is not, because multiplying by something that can be negative flips the sign in some intervals and not others. The two look alike and the methods are different, which is exactly what a test checks.

How It Works

Equations. Write down the restrictions first: every value that makes a denominator zero. Then multiply both sides by the lowest common denominator, solve the polynomial equation that is left, and throw out any answer that is a restriction. An answer on the restricted list is not a solution, even if the algebra produced it.

Inequalities. Get zero on one side and combine everything into a single fraction. Factor top and bottom. The critical values are the zeros of the numerator and the zeros of the denominator, and together they cut the number line into intervals. Test one point per interval, or read the sign from the factors, and keep the intervals that match the inequality.

The endpoints need care. A zero of the numerator is included when the inequality has $\le$ or $\ge$ , because the fraction is $0$ there. A zero of the denominator is never included, whatever the sign, because the function does not exist there.

Why not multiply through? In $\frac{x + 3}{x - 1} \le 2$ , multiplying by $x - 1$ keeps the inequality when $x > 1$ and reverses it when $x < 1$ . You would have to split into cases. Moving the $2$ across and combining into one fraction avoids that entirely.

A quick check at the end: pick a number from your solution and one from outside it, and substitute both into the original. The first should work and the second should not.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Solve $\frac{x + 2}{x - 3} = 4$ .
Show solution
1. Restriction: $x \ne 3$ .
2. Multiply both sides by $x - 3$ : $x + 2 = 4(x - 3) = 4x - 12$ .
3. So $14 = 3x$ and $x = \frac{14}{3}$ , which is not the restricted value.
Answer: $x = \frac{14}{3}$
Easy
Solve $\frac{x - 2}{x + 1} > 0$ .
Show solution
1. One fraction already, with zero on the right. Critical values: the numerator is zero at $x = 2$ , the denominator at $x = -1$ .
2. Three intervals. For $x < -1$ , say $x = -2$ : $\frac{-4}{-1} > 0$ . For $-1 < x < 2$ , say $x = 0$ : $\frac{-2}{1} < 0$ . For $x > 2$ , say $x = 3$ : $\frac{1}{4} > 0$ .
3. The inequality is strict, so neither endpoint is included; $x = -1$ could not be anyway.
Answer: $x < -1$ or $x > 2$
Medium
Solve $\frac{x + 3}{x - 1} \le 2$ .
Show solution
1. Do not multiply by $x - 1$ . Move the $2$ over: $\frac{x + 3}{x - 1} - 2 \le 0$ .
2. Combine: $\frac{x + 3 - 2(x - 1)}{x - 1} = \frac{5 - x}{x - 1} \le 0$ .
3. Critical values $x = 5$ (numerator) and $x = 1$ (denominator). Test $x = 0$ : $\frac{5}{-1} < 0$ . Test $x = 2$ : $\frac{3}{1} > 0$ . Test $x = 6$ : $\frac{-1}{5} < 0$ .
4. Keep the negative intervals. $x = 5$ is included because the fraction is $0$ there; $x = 1$ is excluded.
Answer: $x < 1$ or $x \ge 5$
Medium
Solve $\frac{x}{x - 3} = 2 + \frac{3}{x - 3}$ .
Show solution
1. Restriction: $x \ne 3$ .
2. Multiply every term by $x - 3$ : $x = 2(x - 3) + 3 = 2x - 3$ .
3. So $x = 3$ . But $3$ is the restriction: it makes both original denominators zero.
4. The only candidate is extraneous, so the equation has no solution. A rewrite agrees: the left side is $1 + \frac{3}{x - 3}$ and the right side is $2 + \frac{3}{x - 3}$ , which can never be equal.
Answer: No solution.
Test level
Solve $\frac{2}{x - 1} \le \frac{1}{x + 2}$ .
Show solution
1. Restrictions: $x \ne 1$ and $x \ne -2$ . Move everything to one side: $\frac{2}{x - 1} - \frac{1}{x + 2} \le 0$ .
2. Combine over $(x - 1)(x + 2)$ : the numerator is $2(x + 2) - (x - 1) = x + 5$ . So $\frac{x + 5}{(x - 1)(x + 2)} \le 0$ .
3. Critical values $-5$ , $-2$ and $1$ give four intervals. Signs of $(x + 5)$ , $(x - 1)$ , $(x + 2)$ : for $x < -5$ they are $-, -, -$ , so the fraction is negative. For $-5 < x < -2$ : $+, -, -$ , positive. For $-2 < x < 1$ : $+, -, +$ , negative. For $x > 1$ : all positive.
4. Keep the negative intervals. Include $x = -5$ , where the fraction is $0$ ; exclude $-2$ and $1$ .
Answer: $x \le -5$ or $-2 < x < 1$

Learn It on Martian Lab

Martian Lab teaches this in the Rational Inequalities lesson of its MHF4U Rational Functions unit. Watch it, then take the Rational Functions practice tests to see if it stuck.

Watch the Rational Inequalities lesson free

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Free Rational Functions practice test, with solutions

More Rational Functions topics: Vertical and Horizontal Asymptotes · Holes in Rational Functions · Graphing Rational Functions

More MHF4U topics: Long and Synthetic Division · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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