Algebraic Fractions

Lessons

Senior 3 · Mathematics

Algebraic Fractions

5 lessons

Lesson 5 of 5

Equations and problems with algebraic fractions

25 min

How long with both taps open?

Suppose a school tank can be filled from the mains by one tap in 6 hours, or by a bigger tap in 3 hours (illustrative figures). With both taps open, is the answer 9 hours? 4.5 hours? Neither. In one hour the first tap fills $\frac{1}{6}$ of the tank and the second fills $\frac{1}{3}$ . If together they take $t$ hours, together they fill $\frac{1}{t}$ of the tank per hour, so

\frac{1}{6} + \frac{1}{3} = \frac{1}{t}.

The unknown sits in a denominator. This lesson shows how to solve such equations, and why every answer must be checked against the excluded values.

Definition

Rational equation: an equation that contains at least one algebraic fraction with the unknown in its denominator.

Notation: its conditions of existence are the excluded values of all its fractions.

Example: $\frac{2}{x - 1} + \frac{3}{x + 1} = 1$ is a rational equation with the conditions $x \ne \pm 1$ .

Definition

Extraneous solution: a solution of the equation obtained after clearing the fractions that is an excluded value of the original equation. It is not a solution and must be rejected.

Notation: write "rejected: excluded value" next to it.

Example: clearing the fractions in $\frac{x}{x - 2} = \frac{2}{x - 2} + 3$ gives $x = 2$ , which is excluded; that equation has no solution (see Worked example 3).

Important

Rule. Multiply every term on both sides by the LCD of the fractions. The new equation has no fractions. Every solution of the original equation is a solution of the new one, but the new one may have extra solutions: the values that make the LCD zero. So solve the new equation, then reject any solution that is an excluded value.

Why the extra solutions appear: multiplying both sides by an expression is reversible only when that expression is not zero. For a value that makes the LCD zero, both sides become $0$ , so the new equation is true there even though the original has no meaning.

The method

Factorise the denominators and write the conditions of existence.
Multiply every term, on both sides, by the LCD, and cancel.
Solve the equation that is left (linear or quadratic).
Reject any solution that is an excluded value; check the others in the original equation.

Worked example 1: a linear result

Solve $\frac{5}{x} - \frac{1}{3} = \frac{1}{x}$ .

Solution Condition: $x \ne 0$ . Multiply every term by the LCD $3 x$ :

15 - x = 3 \;\Rightarrow\; x = 12.

12 is not excluded. Check: $\frac{5}{12} - \frac{4}{12} = \frac{1}{12}$ . ✓

Worked example 2: a quadratic result

Solve $\frac{2}{x - 1} + \frac{3}{x + 1} = 1$ .

Solution Conditions: $x \ne \pm 1$ . Multiply every term by $(x - 1) (x + 1)$ :

\begin{aligned} 2(x + 1) + 3(x - 1) &= (x - 1)(x + 1) \\ 5x - 1 &= x^2 - 1 \\ x^2 - 5x &= 0 \\ x(x - 5) &= 0. \end{aligned}

So $x = 0$ or $x = 5$ ; neither is excluded. Check: $x = 0$ gives $- 2 + 3 = 1$ ✓; $x = 5$ gives $\frac{2}{4} + \frac{3}{6} = 1$ ✓.

Worked example 3: extraneous solutions

Solve (a) $\frac{x + 1}{x - 3} = \frac{4}{x - 3} + \frac{x}{2}$ (b) $\frac{x}{x - 2} = \frac{2}{x - 2} + 3$ .

Solution

(a) Condition: $x \ne 3$ . Multiply by $2 (x - 3)$ :

2(x + 1) = 8 + x(x - 3) \;\Rightarrow\; x^2 - 5x + 6 = 0 \;\Rightarrow\; (x - 2)(x - 3) = 0.

$x = 3$ is rejected (excluded value), so the only solution is $x = 2$ . Check: $\frac{3}{-1} = -3$ and $\frac{4}{-1} + 1 = -3$ . ✓

(b) Condition: $x \ne 2$ . Multiply by $x - 2$ : $x = 2 + 3 (x - 2)$ , so $x = 3 x - 4$ and $x = 2$ . But 2 is excluded, so the equation has no solution.

Rate problems

Many problems about work and motion lead to rational equations, because a rate is a quantity divided by a time.

Type	Rate	Key fact
work (filling, weeding, building)	a job done in $t$ hours has rate $\frac{1}{t}$ of the job per hour	rates of people or machines working together add
motion	time $= \frac{\text{distance}}{\text{speed}}$	on a river, speed upstream $= v - c$ , downstream $= v + c$

Worked example 4: two taps

Solve the tank problem from the start of the lesson.

Solution $\frac{1}{6} + \frac{1}{3} = \frac{1}{t}$ , with $t \gt 0$ . Multiply by $6 t$ : $t + 2 t = 6$ , so $t = 2$ . With both taps open the tank fills in 2 hours, less than either tap alone, as it should.

Worked example 5: weeding a field

Working together, Uwase and Mugisha weed a field in 4 days. Working alone, Mugisha would take 6 days longer than Uwase. How long would each take alone? (Illustrative figures.)

Solution Let Uwase take $x$ days alone, so Mugisha takes $x + 6$ days, with $x \gt 0$ . Their daily rates add up to the joint rate:

\frac{1}{x} + \frac{1}{x + 6} = \frac{1}{4}.

Multiply by $4 x (x + 6)$ :

4(x + 6) + 4x = x(x + 6) \;\Rightarrow\; x^2 - 2x - 24 = 0 \;\Rightarrow\; (x - 6)(x + 4) = 0.

Reject $x = - 4$ . Uwase takes 6 days and Mugisha 12 days. Check: $\frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$ . ✓

Worked example 6: a boat against the current

A motorboat goes 24 km upstream on a river and 24 km back downstream in a total of 5 hours. Its speed in still water is $10\ \mathrm{km\,h^{-1}}$ . Find the speed of the current. (Illustrative figures.)

Solution Let the current be $c\ \mathrm{km\,h^{-1}}$ , with $0 \lt c \lt 10$ . Upstream speed $10 - c$ , downstream $10 + c$ :

\frac{24}{10 - c} + \frac{24}{10 + c} = 5.

Multiply by $(10 - c) (10 + c) = 100 - c^{2}$ :

24(10 + c) + 24(10 - c) = 5(100 - c^2) \;\Rightarrow\; 480 = 500 - 5c^2 \;\Rightarrow\; c^2 = 4.

$c = 2$ (reject $- 2$ ). The current flows at $2\ \mathrm{km\,h^{-1}}$ . Check: $\frac{24}{8} + \frac{24}{12} = 3 + 2 = 5$ hours. ✓

Warning

Multiply every term, then check every answer. In $\frac{5}{x} - \frac{1}{3} = \frac{1}{x}$ , the term $\frac{1}{3}$ must be multiplied by $3 x$ too; forgetting a term (especially a whole number such as the 1 in Worked example 2) is the most common mistake. And an answer that is an excluded value is not a solution, however carefully it was found.

Tip

In rate problems, write the rates, not the times. Times of people working together do not add; rates do. Write each rate as $\frac{1}{\text{time}}$ , add them, and set the sum equal to the joint rate.

Khan Academy: Extraneous solutions to rational equations (3 min). Watch how the fractions are cleared, the resulting equation is solved, and one of the answers is rejected because it makes a denominator zero.

✦

Practice

State the conditions first, and check your answers. The last three are harder.

$\frac{6}{x} = \frac{3}{x - 2}$
$\frac{4}{x} + \frac{1}{2} = \frac{3}{x} + 1$
$x + \frac{6}{x} = 5$
$\frac{3}{x - 1} - \frac{2}{x + 1} = 1$
$\frac{x}{x - 4} = \frac{4}{x - 4} + 2$
$\frac{x^2}{x + 1} = \frac{1}{x + 1}$
One tap fills a tank in 4 hours and another in 12 hours. How long do they take together?
Working together, two pumps empty a pond in 6 hours. Alone, one pump takes 5 hours longer than the other. How long does each take alone?

Show answersHide answers

$x \ne 0, 2$ . $6 (x - 2) = 3 x$ , so $x = 4$ .
$x \ne 0$ . Multiply by $2 x$ : $8 + x = 6 + 2 x$ , so $x = 2$ .
$x \ne 0$ . $x^{2} - 5 x + 6 = 0$ : $x = 2$ or $x = 3$ .
$x \ne \pm 1$ . $3 (x + 1) - 2 (x - 1) = x^{2} - 1$ gives $x^{2} - x - 6 = 0$ : $x = 3$ or $x = - 2$ .
$x \ne 4$ . $x = 4 + 2 (x - 4)$ gives $x = 4$ , which is excluded: no solution.
$x \ne -1$ . $x^{2} = 1$ gives $x = \pm 1$ ; reject $- 1$ , so $x = 1$ .
$\frac{1}{4} + \frac{1}{12} = \frac{1}{t}$ , $\frac{4}{12} = \frac{1}{t}$ : 3 hours.
$\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}$ gives $x^{2} - 7 x - 30 = 0$ , $(x - 10) (x + 3) = 0$ : 10 hours and 15 hours.

1How long with both taps open?

2The method

3Rate problems

How long with both taps open?

The method

Rate problems