Solving Rational Equations

A rational equation has a variable in a denominator, like $\dfrac{3}{x} + \dfrac{4}{x + 1} = 2$. These equations come up whenever rates are involved: how long two people take to finish a job together, how fast a canoe moves against a current, or when a chemical reaches a certain concentration. The method is simple (clear the fractions), but there’s one trap you always have to check for.

Key ideas

Roots and x-intercepts

The real roots of $f(x) = 0$ are the $x$-intercepts of the graph of $y = f(x)$, just as for polynomials. For a single fraction, that means:

A fraction equals $0$ exactly when its numerator is $0$ (and its denominator isn’t).

For example, $\dfrac{x + 1}{x - 4} = 0$ has the root $x = -1$, and the graph of $y = \dfrac{x + 1}{x - 4}$ crosses the $x$-axis at $-1$. But $\dfrac{1}{x - 4} = 0$ has no roots: the numerator is never $0$, and the graph of $y = \dfrac{1}{x - 4}$ never meets the $x$-axis.

Restrictions first

Before you do anything, write down the values that make any denominator $0$. Those values can never be solutions.

Multiply by the lowest common denominator

Multiply every term on both sides by the lowest common denominator (LCD). The fractions cancel, leaving a polynomial equation you already know how to solve. When the equation is one fraction equal to another, $\dfrac{a}{b} = \dfrac{c}{d}$, this is the same as cross-multiplying: $ad = bc$.

Extraneous roots

Multiplying by an expression that contains $x$ can create extraneous roots: answers to the polynomial equation that don’t work in the original, because they make a denominator $0$. Always compare your answers with the restrictions, and check them in the original equation.

Checking with a graph

To check $\text{left side} = \text{right side}$ with graphing technology (for example, Desmos), either graph both sides and look for intersection points, or move everything to one side and find the $x$-intercepts.

Setting up word problems

Situation	Key relationship
Work rate	A job that takes $t$ hours is done at a rate of $\frac{1}{t}$ of the job per hour. Rates add: $\frac{1}{a} + \frac{1}{b} = \frac{1}{\text{time together}}$.
Speed, distance, time	$\text{time} = \dfrac{\text{distance}}{\text{speed}}$
Concentration	$\text{concentration} = \dfrac{\text{amount of substance}}{\text{total volume}}$

On the SAT

On the SAT, graph each side of the equation as y = ... in Desmos and click the intersections; their x-coordinates are the solutions. This is also a good guard against extraneous roots: a value that makes a denominator zero isn’t on the graph, so it won’t show up as a real intersection. Multiplying by the LCD by hand is still the way to find exact answers, so use the graph to check and to count solutions. See using Desmos on the SAT.

Worked examples

Example 1: One fraction equals another

Solve $\dfrac{5}{x - 2} = \dfrac{3}{x + 4}$.

Solution. Restrictions: $x \ne 2$ and $x \ne -4$.

Cross-multiply (that is, multiply both sides by $(x - 2)(x + 4)$):

$$\begin{aligned} 5(x + 4) &= 3(x - 2) \\ 5x + 20 &= 3x - 6 \\ 2x &= -26 \\ x &= -13 \end{aligned}$$

$-13$ isn’t a restriction. Check: $\dfrac{5}{-15} = -\dfrac{1}{3}$ and $\dfrac{3}{-9} = -\dfrac{1}{3}$ ✓

Example 2: An equation that becomes a quadratic

Solve $\dfrac{3}{x} + \dfrac{4}{x + 1} = 2$.

Solution. Restrictions: $x \ne 0$ and $x \ne -1$. The LCD is $x(x + 1)$. Multiply every term by it:

$$\begin{aligned} 3(x + 1) + 4x &= 2x(x + 1) \\ 7x + 3 &= 2x^2 + 2x \\ 0 &= 2x^2 - 5x - 3 \\ 0 &= (2x + 1)(x - 3) \end{aligned}$$

So $x = -\dfrac{1}{2}$ or $x = 3$. Neither is a restriction.

Check $x = 3$: $\dfrac{3}{3} + \dfrac{4}{4} = 1 + 1 = 2$ ✓

Check $x = -\dfrac{1}{2}$: $\dfrac{3}{-1/2} + \dfrac{4}{1/2} = -6 + 8 = 2$ ✓

The graph of $y = \dfrac{3}{x} + \dfrac{4}{x + 1} - 2$ (everything moved to one side) crosses the $x$-axis at exactly these two values:

The roots $-\frac{1}{2}$ and $3$ are the $x$-intercepts of $y = \dfrac{3}{x} + \dfrac{4}{x + 1} - 2$.

Example 3: An extraneous root

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

Solution. Factor the last denominator: $x^2 - 2x = x(x - 2)$. Restrictions: $x \ne 0$ and $x \ne 2$. The LCD is $x(x - 2)$:

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

So $x = 0$ or $x = -2$. But $x = 0$ is a restriction: it makes $\dfrac{1}{x}$ undefined. It’s extraneous, so reject it.

Check $x = -2$: left side $\dfrac{1}{-4} - \dfrac{1}{-2} = -\dfrac{1}{4} + \dfrac{1}{2} = \dfrac{1}{4}$; right side $\dfrac{2}{4 + 4} = \dfrac{1}{4}$ ✓

The only solution is $x = -2$.

Example 4: Shovelling the driveway

Working together, Maya and Sam can shovel their driveway in $20$ minutes. Working alone, Maya takes $30$ minutes less than Sam. How long does each take alone?

Solution. Let Maya’s time alone be $t$ minutes, so Sam’s is $t + 30$. In one minute, Maya does $\frac{1}{t}$ of the job, Sam does $\frac{1}{t + 30}$, and together they do $\frac{1}{20}$:

$$\frac{1}{t} + \frac{1}{t + 30} = \frac{1}{20}$$

Multiply by the LCD, $20t(t + 30)$:

$$\begin{aligned} 20(t + 30) + 20t &= t(t + 30) \\ 40t + 600 &= t^2 + 30t \\ 0 &= t^2 - 10t - 600 \\ 0 &= (t - 30)(t + 20) \end{aligned}$$

$t = 30$ or $t = -20$. A time can’t be negative, so $t = 30$.

Maya takes $30$ minutes alone and Sam takes $60$ minutes. Check: $\dfrac{1}{30} + \dfrac{1}{60} = \dfrac{2}{60} + \dfrac{1}{60} = \dfrac{3}{60} = \dfrac{1}{20}$ ✓

Common mistakes

Skipping the restrictions. Write them down first. Otherwise it’s easy to “solve” an equation and keep an answer that makes a denominator $0$, like $x = 0$ in Example 3.

Multiplying only some of the terms by the LCD. Every term on both sides gets multiplied, including whole numbers. In Example 2, the $2$ on the right becomes $2x(x + 1)$.

Cross-multiplying when there are more than two terms. $\dfrac{a}{b} + c = \dfrac{d}{e}$ is not ready for cross-multiplying. Multiply every term by the LCD instead.

Setting the denominator equal to zero to “solve”. The zeros of the denominator are where the function is undefined (asymptotes or holes), not where it equals $0$. Roots come from the numerator.

Keeping answers that don’t make sense in context. A negative time or a speed slower than the current makes no sense, even if it satisfies the equation. Say why you reject it.

Practice

1. (Warm-up) Find the real roots, if any, of each equation, and say what this means for the graph of the left side.

• (a) $\dfrac{x - 5}{x + 2} = 0$
• (b) $\dfrac{4}{x + 2} = 0$

Solution

(a) The numerator is $0$ at $x = 5$ (the denominator is $7 \ne 0$ there), so $x = 5$. The graph of $y = \dfrac{x - 5}{x + 2}$ crosses the $x$-axis at $5$.

(b) The numerator $4$ is never $0$, so there are no real roots. The graph of $y = \dfrac{4}{x + 2}$ never meets the $x$-axis, which is its horizontal asymptote.

2. (Warm-up) Solve $\dfrac{6}{x} = \dfrac{2}{x - 4}$.

Solution

Restrictions: $x \ne 0, 4$. Cross-multiply:

$$6(x - 4) = 2x \quad\Rightarrow\quad 6x - 24 = 2x \quad\Rightarrow\quad x = 6$$

Check: $\dfrac{6}{6} = 1$ and $\dfrac{2}{2} = 1$ ✓

3. (Core) Solve $x + \dfrac{8}{x} = 6$.

Solution

Restriction: $x \ne 0$. Multiply every term by $x$:

$$x^2 + 8 = 6x \quad\Rightarrow\quad x^2 - 6x + 8 = 0 \quad\Rightarrow\quad (x - 2)(x - 4) = 0$$

$x = 2$ or $x = 4$. Check: $2 + 4 = 6$ ✓ and $4 + 2 = 6$ ✓

4. (Core) Solve $\dfrac{1}{x} + \dfrac{1}{x + 2} = \dfrac{5}{12}$.

Solution

Restrictions: $x \ne 0, -2$. Multiply by the LCD, $12x(x + 2)$:

$$\begin{aligned} 12(x + 2) + 12x &= 5x(x + 2) \\ 24x + 24 &= 5x^2 + 10x \\ 0 &= 5x^2 - 14x - 24 \\ 0 &= (5x + 6)(x - 4) \end{aligned}$$

$x = 4$ or $x = -\dfrac{6}{5}$.

Check $x = 4$: $\dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}$ ✓

Check $x = -\dfrac{6}{5}$: $-\dfrac{5}{6} + \dfrac{1}{4/5} = -\dfrac{5}{6} + \dfrac{5}{4} = -\dfrac{10}{12} + \dfrac{15}{12} = \dfrac{5}{12}$ ✓

5. (Core) Solve $\dfrac{x}{x - 3} - 2 = \dfrac{3}{x - 3}$.

Solution

Restriction: $x \ne 3$. Multiply every term by $x - 3$:

$$x - 2(x - 3) = 3 \quad\Rightarrow\quad -x + 6 = 3 \quad\Rightarrow\quad x = 3$$

But $x = 3$ is a restriction, so it’s extraneous. The equation has no solution.

6. (Core) A canoeist on a river in Algonquin Park paddles $12$ km upstream and then $12$ km back downstream. The current flows at $1$ km/h, and the whole trip takes $5$ hours of paddling. How fast does the canoeist paddle in still water?

Solution

Let $v$ be the paddling speed in still water, in km/h. Upstream the speed is $v - 1$; downstream it’s $v + 1$. Using time $= \frac{\text{distance}}{\text{speed}}$:

$$\frac{12}{v - 1} + \frac{12}{v + 1} = 5$$

Multiply by $(v - 1)(v + 1)$:

$$\begin{aligned} 12(v + 1) + 12(v - 1) &= 5(v^2 - 1) \\ 24v &= 5v^2 - 5 \\ 0 &= 5v^2 - 24v - 5 \\ 0 &= (5v + 1)(v - 5) \end{aligned}$$

$v = 5$ or $v = -\frac{1}{5}$. A negative speed makes no sense, so $v = 5$ km/h.

Check: upstream $\frac{12}{4} = 3$ h, downstream $\frac{12}{6} = 2$ h, total $5$ h ✓

7. (Core) A large tank contains $500$ L of pure water. Salt water containing $30$ g of salt per litre is pumped in at $25$ L/min. After $t$ minutes, the concentration of salt in the tank, in grams per litre, is

$$C(t) = \frac{30t}{20 + t}$$

• (a) When does the concentration reach $12$ g/L?
• (b) What value does the concentration approach over a long time? Why does that make sense?

Solution

(a) Restriction $t \ne -20$, which doesn’t matter since $t \ge 0$. Multiply both sides by $20 + t$:

$$30t = 12(20 + t) \quad\Rightarrow\quad 30t = 240 + 12t \quad\Rightarrow\quad 18t = 240 \quad\Rightarrow\quad t = \frac{40}{3}$$

That’s about $13.3$ minutes (to one decimal place). Check: $C\left(\frac{40}{3}\right) = \dfrac{400}{20 + 40/3} = \dfrac{400}{100/3} = 12$ ✓

(b) The horizontal asymptote is $C = \frac{30}{1} = 30$ g/L. Over time the tank fills with more and more of the incoming salt water, so its concentration gets closer and closer to the $30$ g/L of the incoming water, but never quite reaches it.

8. (Challenge) The yearbook club has a large printer and a small printer. The large printer takes $4$ hours less than the small one to print all the yearbooks. Together they take $4.8$ hours. How long would each printer take alone?

Solution

Let the small printer take $t$ hours, so the large one takes $t - 4$:

$$\frac{1}{t} + \frac{1}{t - 4} = \frac{1}{4.8}$$

Multiply by $4.8t(t - 4)$:

$$\begin{aligned} 4.8(t - 4) + 4.8t &= t(t - 4) \\ 9.6t - 19.2 &= t^2 - 4t \\ 0 &= t^2 - 13.6t + 19.2 \end{aligned}$$

Multiply by $5$ to clear decimals: $5t^2 - 68t + 96 = 0$, which factors as $(5t - 8)(t - 12) = 0$. So $t = 12$ or $t = 1.6$.

If $t = 1.6$, the large printer would take $1.6 - 4 = -2.4$ hours, which is impossible. So reject it, even though it’s positive.

The small printer takes $12$ hours and the large printer takes $8$ hours. Check: $\dfrac{1}{12} + \dfrac{1}{8} = \dfrac{2}{24} + \dfrac{3}{24} = \dfrac{5}{24}$, and $\dfrac{1}{4.8} = \dfrac{10}{48} = \dfrac{5}{24}$ ✓

9. (Challenge) Priya drives $60$ km to her cottage and $60$ km back. On the way back she drives $20$ km/h faster than on the way there. Her average speed for the whole round trip is $48$ km/h. Find her speed in each direction. Why isn’t the average speed just the average of the two speeds?

Solution

The round trip is $120$ km at an average of $48$ km/h, so it takes $\frac{120}{48} = 2.5$ h. Let $v$ be her speed on the way there:

$$\frac{60}{v} + \frac{60}{v + 20} = 2.5$$

Multiply by $v(v + 20)$:

$$\begin{aligned} 60(v + 20) + 60v &= 2.5v(v + 20) \\ 120v + 1200 &= 2.5v^2 + 50v \\ 0 &= 2.5v^2 - 70v - 1200 \\ 0 &= v^2 - 28v - 480 \\ 0 &= (v - 40)(v + 12) \end{aligned}$$

$v = 40$ (reject $v = -12$). She drives $40$ km/h there and $60$ km/h back.

Check: $\frac{60}{40} + \frac{60}{60} = 1.5 + 1 = 2.5$ h ✓

The average of $40$ and $60$ is $50$, not $48$, because she spends more time at the slower speed ($1.5$ h versus $1$ h). Average speed is total distance divided by total time, so the slower part counts for more.