Solving a Rational Inequality

Why you cannot multiply by the denominator

Solve (x − 1)/(x − 4) > 0. An inequality with the unknown in a denominator is called a rational inequality.

With an equation, you would clear the fraction by multiplying both sides by x − 4. With an inequality, that is not safe. Multiplying both sides of an inequality by a negative number reverses the sign, and x − 4 is negative whenever x is less than 4. You do not know in advance whether it is positive or negative, so you cannot know whether to reverse the sign.

Multiplying by x − 4 anyway gives x − 1 > 0, so x > 1, and that is wrong. Try x = 2, which is more than 1: (2 − 1)/(2 − 4) = 1/(−2) = −1/2, which is not greater than 0.

Test the sign of each factor

Find the critical values instead. The top, x − 1, is 0 when x = 1. The bottom, x − 4, is 0 when x = 4. The quotient can change sign only at these values, so they cut the number line into three parts: x < 1, 1 < x < 4 and x > 4.

In each part, work out the sign of the top and the sign of the bottom. A quotient of two numbers with the same sign is positive, and a quotient of two numbers with different signs is negative.

For x < 1, both x − 1 and x − 4 are negative, so the quotient is positive. For 1 < x < 4, x − 1 is positive and x − 4 is negative, so the quotient is negative. For x > 4, both are positive, so the quotient is positive.

The sign of the top, the sign of the bottom, and the sign of the quotient in each part of the number line.

Read off the answer

The inequality asks for a positive quotient, so the solution is x < 1 or x > 4. Check one value from each part: at x = 0 the quotient is (−1)/(−4) = 1/4, positive; at x = 2 it is −1/2, negative; at x = 5 it is 4/1 = 4, positive.

The value x = 1 is left out because there the quotient is 0, and 0 is not greater than 0.

The quotient is positive outside the critical values.

The graph of y = (x − 1)/(x − 4) is above the x-axis to the left of 1 and to the right of 4, and below it between them. The dashed line is x = 4, where the denominator is 0 and the graph has no point.

Or multiply by a square

There is a safe number to multiply by: (x − 4)². A square is never negative, and for x ≠ 4 it is positive, so multiplying by it never reverses the sign.

Multiplying (x − 1)/(x − 4) by (x − 4)² cancels one factor of x − 4 and leaves (x − 1)(x − 4). So the inequality becomes (x − 1)(x − 4) > 0, a quadratic inequality. Its curve is a bowl crossing the axis at 1 and 4, above the axis outside them, so again x < 1 or x > 4.

x = 4 is never a solution

At x = 4 the denominator is 0, so the quotient has no value there at all. Whatever the inequality asks, x = 4 cannot be part of the answer.

This matters with ≥. The solution of (x − 1)/(x − 4) ≥ 0 includes x = 1, where the quotient is 0, but not x = 4. It is x ≤ 1 or x > 4. Multiplying by (x − 4)² gives (x − 1)(x − 4) ≥ 0, whose solution x ≤ 1 or x ≥ 4 wrongly lets x = 4 in, so take it out again by hand.

Bring everything to one side first

Solve 3/(x − 2) < 1. The sign test needs 0 on one side, so subtract 1 from both sides: 3/(x − 2) − 1 < 0. Write the left side over one denominator: (3 − (x − 2))/(x − 2) < 0, which is (5 − x)/(x − 2) < 0.

The critical values are 5, where the top is 0, and 2, where the bottom is 0. For x < 2, the top is positive and the bottom negative, so the quotient is negative. For 2 < x < 5, both are positive, so it is positive. For x > 5, the top is negative and the bottom positive, so it is negative. The solution is x < 2 or x > 5.

Check: at x = 0, 3/(−2) = −1.5, which is less than 1. At x = 3, 3/1 = 3, which is not. At x = 8, 3/6 = 0.5, which is less than 1.

The usual mistakes

Multiplying both sides by the denominator as if it were positive. It is negative for some values of x, and there the inequality sign reverses.

Including the value that makes the denominator 0. The quotient has no value there, even when the inequality has ≤ or ≥.

Running the sign test before bringing everything to one side. The test tells you where a quotient is positive or negative, so the other side must be 0.

Worked example: A Tour Boat Against a River Current: the Speeds for a Trip Upstream, and a Round Trip, Under a Time Limit

Question A tour boat runs 12 km up a river from a town to a lake, and then back. The river flows at 4 km/h along the whole stretch, and the boat's speed in still water is v km/h. (a) For which speeds v does the trip upstream take less than 1 hour? (b) For which speeds v does the round trip, up to the lake and back, take less than 1 hour 15 minutes, not counting the stop at the lake?

1.Let the boat's speed in still water be v km/h. Against the current it moves upstream at (v − 4) km/h, so the trip takes 12v − 4 hours. The boat makes headway only if v − 4 is positive, so v > 4. The trip takes less than 1 hour when 12v − 4 < 1.

Against the current the boat moves at (v − 4) km/h, so the trip upstream takes 12v − 4 hours, and it must be less than 1.
2.The sign of v − 4 decides whether multiplying by it keeps or reverses the inequality, so do not multiply. Subtract 1 from both sides and write over one denominator: 12 − (v − 4)v − 4 < 0, which is 16 − vv − 4 < 0. The critical values are v = 16, where the top is zero, and v = 4, where the bottom is zero.

The sign of v − 4 is not known, so do not multiply by it. Subtract 1 and write over one denominator: 16 − vv − 4 < 0.
3.Test the sign of each factor. For v < 4: the top is positive and the bottom negative, so the quotient is negative. For 4 < v < 16: both are positive, so the quotient is positive. For v > 16: the top is negative and the bottom positive, so the quotient is negative. The inequality holds when v < 4 or v > 16.

Test the sign of each factor on each side of the critical values 4 and 16. The quotient is negative when v < 4 or v > 16.
4.Below 4 km/h the boat is carried downstream and never reaches the lake; the formula gives a negative time there, which is less than 1 but means nothing. Those speeds are rejected. (a) The trip upstream takes less than 1 hour when v > 16 km/h. Check: at v = 20 it takes 1216 = 0.75 hour.

(a) Below 4 km/h the boat is carried downstream and the formula gives a negative time, so those speeds are rejected: v > 16.
5.(b) Downstream the boat moves at (v + 4) km/h, and 1 hour 15 minutes is 54 hours, so 12v − 4 + 12v + 4 < 54. Adding the fractions gives 24v(v − 4)(v + 4) < 54. Subtract 54 and write over one denominator: 96v − 5(v² − 16)4(v − 4)(v + 4) < 0, which is −5v² + 96v + 804(v − 4)(v + 4) < 0. Multiply both sides by −4, a negative number, and reverse the sign: (5v + 4)(v − 20)(v − 4)(v + 4) > 0.

Downstream the boat moves at (v + 4) km/h. Add the two times, subtract 54, and multiply by −4, which reverses the sign.
6.For v > 4 the factors 5v + 4, v − 4 and v + 4 are all positive, so the quotient has the sign of v − 20, and it is positive when v > 20. (b) The round trip takes less than 1 hour 15 minutes when v > 20 km/h. Check: at v = 20 it takes 1216 + 1224 = 0.75 + 0.5 = 1.25 hours, exactly 1 hour 15 minutes, and at v = 24 it takes 0.6 + 1228 ≈ 1.03 hours.

(b) For v > 4 the quotient has the sign of v − 20, so the round trip takes less than 1 hour 15 minutes when v > 20.

Answer: (a) v > 16 km/h; (b) v > 20 km/h

Common mistakes

Multiplying both sides of 12v − 4 < 1 by v − 4 as if it were positive. For speeds below 4 km/h it is negative and the inequality reverses; the sign test covers both cases without guessing.
Giving the answer to (a) as v < 4 or v > 16. Below 4 km/h the formula gives a negative time, which passes the inequality but means the boat never arrives, so those speeds are rejected.

More quadratic equations problems, worked step by step →