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Algebra 2 · A.REI.A.2

Solving a rational equation

An equation has the unknown in a denominator. How do you get it out, and how do you know the answer you get is real?

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Solving a rational equation
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When you reach for it

A rational equation is an equation with a variable in a denominator, such as \displaystyle \frac{6}{x} + 1 = \frac{5}{x - 2}. They arise whenever an unknown is divided into something: the part of a road a snow plow clears in an hour is \displaystyle \frac{1}{t} when the whole road takes t hours, and the time to cover 30 km at speed v is \displaystyle \frac{30}{v}. Set such quantities equal to what is known and the equation is rational.

Fractions make an equation hard to solve, so the first move is to get rid of them. This lesson runs over two days. Part 1 clears the denominators and checks what comes out. Part 2 uses the same move on problems about work, rates and speed.

The procedure

Part 1: clearing the denominators

List the excluded values first. A value that makes any denominator zero gives no number on that side, so it can never be a solution. Factor each denominator and write down its zeros, as in excluded values of a rational expression. For \displaystyle \frac{6}{x} + 1 = \frac{5}{x - 2} the list is x \ne 0, 2.

Multiply both sides by the least common denominator. Find the LCD of every fraction in the equation, the way it is found in adding rational expressions, and multiply every term on both sides by it. Each denominator divides the LCD, so every fraction cancels and a polynomial equation is left. The move is allowed because multiplying both sides of a true equation by the same number keeps it true, one of the Inverse operations and the properties of equality.

This is exactly what may not be done to an expression. An expression has one side; multiplying it by the LCD changes its value. An equation has two sides, and multiplying both keeps the balance.

Solve what is left. It is usually linear, solved by collecting the variable on one side, or quadratic, solved by factoring and the Solving quadratics by factoring or by the quadratic formula. Its solutions are only candidates.

Check every candidate. Strike any candidate on the excluded list. Substitute the others into the original equation; this confirms them and also catches slips in the algebra. A candidate that solves the cleared equation but not the original is called an extraneous solution.

Why extraneous solutions appear. Multiplying both sides by a number L is reversible only when L \ne 0, since then you can divide by L again. Here L is an expression in x, and it is zero exactly at the excluded values. So:

  • Every solution of the original is a solution of the cleared equation, because the original solution makes L a number that is not zero. No solution is lost.
  • The cleared equation may gain a solution, but only at a value where L = 0. There the original sides have no value at all, while the cleared sides are ordinary numbers.

So an extraneous solution is always an excluded value, and comparing with the list finds it.

Proportions. An equation \displaystyle \frac{a}{b} = \frac{c}{d} is a proportion. Multiplying both sides by bd gives ad = bc, which is cross-multiplying from Equations from proportions and percents. It is the same move with bd as the common denominator, so when b or d contains x, the same check applies.

Part 2: rates

Work. If a plow clears a road alone in t hours, it clears \displaystyle \frac{1}{t} of the road each hour, a unit rate. Two plows on the same road together clear the sum of their two parts each hour. Rates add; times do not. If the plows take a and b hours alone and T hours together, then in one hour together they clear \displaystyle \frac{1}{T} of the road:

\frac{1}{a} + \frac{1}{b} = \frac{1}{T}

Distance, speed and time. From Speed, distance and time, \displaystyle \text{time} = \frac{\text{distance}}{\text{speed}}. A boat with speed b in still water, on a river whose current runs at c, moves at b - c upstream and b + c downstream. A plane in a wind works the same way. When the times are known, or compared, the unknown speed lands in a denominator.

Does the answer fit the story? A candidate can pass the excluded list and solve the equation and still be impossible: a negative time, a negative speed, a boat slower than the current it is supposed to climb. Those are rejected for the story’s reason, not the algebra’s. Also check that the size is sensible: two plows together must be quicker than the faster one alone.

Worked examples

Examples 1 to 4 go with Part 1. Examples 5 to 7 go with Part 2.

Example 1

Two candidates, both real

Solve \displaystyle \frac{6}{x} + 1 = \frac{5}{x - 2}.

Solution

The denominators are x and x - 2, so x \ne 0, 2, and the LCD is x(x - 2). Multiply every term on both sides by it. The x cancels in the first fraction and x - 2 in the last:

\begin{aligned} 6(x - 2) + x(x - 2) &= 5x \\ 6x - 12 + x^2 - 2x &= 5x \\ x^2 - x - 12 &= 0 \\ (x - 4)(x + 3) &= 0 \end{aligned}

The candidates are x = 4 and x = -3, and neither is excluded. Check x = 4: \displaystyle \frac{6}{4} + 1 = \frac{5}{2} and \displaystyle \frac{5}{2}. Check x = -3: -2 + 1 = -1 and \displaystyle \frac{5}{-5} = -1. Both solve the equation.

Each side is a function of x. A solution is an x where the two have the same value, so it is where their graphs meet:

2 4 6 2 4 x y
(-3, -1)
\left(4, \tfrac{5}{2}\right)
\displaystyle y = \frac{6}{x} + 1
\displaystyle y = \frac{5}{x - 2}
Figure 1Each side of Example 1 drawn as its own graph. The solutions x = -3 and x = 4 are the x-coordinates of the two crossing points.
Example 2

No solution at all

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

Solution

Here x \ne 3, and the LCD is x - 3. Multiplying both sides by it:

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

The only candidate is excluded, so the equation has no solution. At x = 3 the cleared equation reads 3 = 3, but the original reads \displaystyle \frac{3}{0} = \frac{3}{0} + 2, which says nothing.

There is a plain reason no solution exists. Since \displaystyle \frac{x}{x - 3} = \frac{(x - 3) + 3}{x - 3}, the left side is \displaystyle 1 + \frac{3}{x - 3}, while the right side is \displaystyle 2 + \frac{3}{x - 3}. The right side is always exactly 1 more, so the two sides are never equal:

−6 −3 6 9 −4 −2 2 4 6 x y
x = 3
\displaystyle y = \frac{x}{x - 3}
\displaystyle y = \frac{3}{x - 3} + 2
Figure 2The two sides of Example 2. The dashed curve sits exactly 1 above the solid one at every allowed x, so the equation has no solution; the only candidate, x = 3, is where neither side exists.
Example 3

Factor first, then one candidate fails

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

Solution

Factor the last denominator: x^2 - 1 = (x - 1)(x + 1). So x \ne 1, -1, and the LCD is (x - 1)(x + 1). Multiply through. The first fraction keeps x + 1, the second keeps x - 1, and the third keeps nothing:

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

The candidates are 2 and -1. But -1 is excluded, so it is extraneous. Check x = 2: \displaystyle \frac{2}{1} - \frac{2}{3} = \frac{4}{3} and \displaystyle \frac{4}{4 - 1} = \frac{4}{3}. The solution is x = 2 only.

Example 4

A proportion with expressions

Solve \displaystyle \frac{x + 1}{4} = \frac{6}{x - 1}.

Solution

Here x \ne 1. Cross-multiplying is multiplying both sides by 4(x - 1):

\begin{aligned} (x + 1)(x - 1) &= 24 \\ x^2 - 1 &= 24 \\ x^2 &= 25 \end{aligned}

so x = 5 or x = -5. Neither is 1. Check x = 5: \displaystyle \frac{6}{4} and \displaystyle \frac{6}{4}. Check x = -5: \displaystyle \frac{-4}{4} = -1 and \displaystyle \frac{6}{-6} = -1. Both are solutions; a quadratic proportion can have two.

Example 5

Two plows on one road

Klaudia’s plow clears a mountain road alone in 9 hours, and Sigismund’s in 12 hours. How long do they take working together?

Solution

Let T be the time together, in hours. In one hour Klaudia’s plow clears \displaystyle \frac{1}{9} of the road and Sigismund’s \displaystyle \frac{1}{12}, and together they clear \displaystyle \frac{1}{T}:

\frac{1}{9} + \frac{1}{12} = \frac{1}{T}

The LCD is 36T. Multiplying through, 4T + 3T = 36, so 7T = 36 and \displaystyle T = \frac{36}{7} hours, about 5.14 hours, or about 5 hours 9 minutes.

Does it fit? Together they must beat Klaudia’s 9 hours alone, and they do. Adding the times (21 hours) or averaging them (10.5 hours) gives a time longer than either plow alone, which cannot be right.

Example 6

An unknown time for each

Ludmilla’s new plow and Marcellin’s old one together clear a parking lot in 6 hours. Alone, the old plow takes 5 hours longer than the new one. How long does each take alone?

Solution

Let the new plow take t hours alone; the old one takes t + 5. Their rates add to the rate together:

\frac{1}{t} + \frac{1}{t + 5} = \frac{1}{6}

Here t \ne 0, -5, and the LCD is 6t(t + 5):

\begin{aligned} 6(t + 5) + 6t &= t(t + 5) \\ 12t + 30 &= t^2 + 5t \\ t^2 - 7t - 30 &= 0 \\ (t - 10)(t + 3) &= 0 \end{aligned}

The candidates are 10 and -3. Neither is excluded, and -3 really does solve the equation: \displaystyle -\frac{1}{3} + \frac{1}{2} = \frac{1}{6}. It is rejected for the story’s reason, since no plow takes -3 hours. So the new plow takes 10 hours and the old one 15.

Check: \displaystyle \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{1}{6}.

Example 7

Upstream and back

Olympia steers a boat 30 km up a river and 30 km back in 8 hours of traveling. The current runs at 2 km/h. What is the boat’s speed in still water?

Solution

Let b be the speed in still water, in km/h. Upstream the boat makes b - 2 and downstream b + 2, so the two legs take \displaystyle \frac{30}{b - 2} and \displaystyle \frac{30}{b + 2} hours:

\frac{30}{b - 2} + \frac{30}{b + 2} = 8

Here b \ne 2, -2, and the LCD is (b - 2)(b + 2) = b^2 - 4:

\begin{aligned} 30(b + 2) + 30(b - 2) &= 8(b^2 - 4) \\ 60b &= 8b^2 - 32 \\ 2b^2 - 15b - 8 &= 0 \\ (2b + 1)(b - 8) &= 0 \end{aligned}

where the third line moves 60b across and divides by 4. The candidates are 8 and \displaystyle -\frac{1}{2}. A negative speed is impossible, and the boat must beat the current, so b > 2. The boat goes at 8 km/h in still water.

Check: upstream at 6 km/h takes 5 hours, downstream at 10 km/h takes 3, and 5 + 3 = 8.

Boundaries

Clearing denominators needs an equation. Both sides are multiplied by the same thing. An expression has no second side, and multiplying it by its LCD changes it.

The multiplier is zero at the excluded values. That is the only place an extra candidate can come from, so every candidate is compared with the list. When every candidate is excluded, the equation has no solution.

A solution of the equation may not answer the question. In a story, a negative time or speed, or a speed too small for the trip, is rejected even though it solves the equation.

Work rates add only when the workers help each other. The formula assumes each keeps its own steady rate on the same job. A worker that undoes the job, such as a drain emptying a tank, has its rate subtracted.

Where this goes next

Rational equations are the language of rational functions in context: average cost, mixtures and shared work, where a quantity is divided by another that changes. On a graph, a rational equation’s solutions are where two curves meet, as in the figures above; graphing rational functions explains the breaks and the leveling off those curves show. The same caution, that a step done to both sides can add a false answer, returns in radical equations, where squaring both sides can.

Exercises

Part 1: clearing the denominators

Solve, and state the excluded values.

  1. \displaystyle \frac{3}{x} + \frac{1}{4} = \frac{5}{x}
  2. \displaystyle x + \frac{8}{x} = 6
  3. \displaystyle \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2}
  4. \displaystyle \frac{4}{x - 1} - 1 = \frac{6}{x + 1}
  5. \displaystyle \frac{2}{x - 7} + \frac{1}{x + 7} = \frac{14}{x^2 - 49}

Part 1: extraneous solutions

Solve, and say which candidates are extraneous.

  1. \displaystyle \frac{x}{x - 5} = \frac{5}{x - 5} + 4
  2. \displaystyle \frac{x}{x - 6} - \frac{6}{x + 6} = \frac{72}{x^2 - 36}
  3. \displaystyle \frac{x}{x + 2} + \frac{3}{x} = \frac{6}{x(x + 2)}

Part 1: proportions

  1. \displaystyle \frac{7}{x + 1} = \frac{2}{x - 4}
  2. \displaystyle \frac{x - 2}{3} = \frac{5}{x + 2}
  3. \displaystyle \frac{x}{x + 4} = \frac{x - 3}{x}

Part 2: work

  1. Marcellin’s plow clears a parking lot in 4 hours, and Olympia’s in 6 hours. How long do they take together?
  2. Two plows together clear a road in 8 hours. One alone takes 12 hours. How long does the other take alone?
  3. At the plow depot, a pipe fills a meltwater tank in 3 hours and a drain empties it in 5 hours. Starting empty with both open, how long does the tank take to fill?
  4. One plow takes 8 hours longer than another to clear a road alone. Together they take 3 hours. How long does each take alone?
  5. Solve \displaystyle \frac{1}{a} + \frac{1}{b} = \frac{1}{T} for T, then find T when a = 20 and b = 30.

Part 2: speed

  1. A light plane flies 600 km with a 25 km/h wind in the same time it flies 400 km against it. What is its speed in still air?
  2. Sigismund paddles at 5 km/h in still water. He goes 12 km up a river and 12 km back in 5 hours. How fast is the current?
  3. Klaudia drives a plow 36 km to a depot, and drives back 6 km/h faster. The two drives take 5 hours. How fast did she drive out?

Reasoning

Exercises 20 and 21 go with Part 1, and 22 and 23 with Part 2.

  1. Explain why multiplying both sides by the LCD never loses a solution, and why any extra candidate it produces must be an excluded value.
  2. Explain why cross-multiplying \displaystyle \frac{a}{b} = \frac{c}{d} is a case of clearing denominators, and what must still be checked when b or d contains x.
  3. Ludmilla’s plow takes 8 hours alone and Marcellin’s takes 24. Ludmilla says that together they take 16 hours, the average. Find the true time, and explain why it must be less than 8 hours and more than 4.
  4. In Example 2 the candidate 3 was rejected, and in Example 6 the candidate -3. Explain the different reason for each.
Answers
  1. x = 8, with x \ne 0. The LCD 4x gives 12 + x = 20.
  2. x = 2 or x = 4, with x \ne 0. Multiply by x: x^2 - 6x + 8 = (x - 2)(x - 4) = 0.
  3. x = 3 or x = -2, with x \ne 0, -3. The LCD 2x(x + 3) gives 2(x + 3) + 2x = x(x + 3), so x^2 - x - 6 = 0.
  4. \displaystyle x = -1 \pm 2\sqrt{3}, with x \ne 1, -1. The LCD (x - 1)(x + 1) gives 4(x + 1) - (x^2 - 1) = 6(x - 1), so x^2 + 2x - 11 = 0, and the quadratic formula gives \displaystyle \frac{-2 \pm \sqrt{48}}{2}.
  5. \displaystyle x = \frac{7}{3}, with x \ne 7, -7. The LCD x^2 - 49 gives 2(x + 7) + (x - 7) = 14, so 3x + 7 = 14.
  6. No solution. Multiplying by x - 5 gives x = 5 + 4(x - 5), so x = 5, which is excluded and extraneous.
  7. No solution. The LCD x^2 - 36 gives x(x + 6) - 6(x - 6) = 72, so x^2 = 36; both 6 and -6 are excluded, so both are extraneous.
  8. x = -3; the candidate 0 is extraneous. The LCD x(x + 2) gives x^2 + 3(x + 2) = 6, so x^2 + 3x = 0. Check -3: \displaystyle \frac{-3}{-1} + \frac{3}{-3} = 2 and \displaystyle \frac{6}{3} = 2.
  9. x = 6, with x \ne -1, 4. 7(x - 4) = 2(x + 1), so 5x = 30.
  10. \displaystyle x = \pm\sqrt{19}, with x \ne -2. (x - 2)(x + 2) = 15, so x^2 = 19.
  11. x = 12, with x \ne 0, -4. x^2 = (x + 4)(x - 3) = x^2 + x - 12, so x = 12. Check: \displaystyle \frac{12}{16} = \frac{9}{12} = \frac{3}{4}.
  12. \displaystyle \frac{12}{5} hours, which is 2 hours 24 minutes. \displaystyle \frac{1}{4} + \frac{1}{6} = \frac{5}{12}.
  13. 24 hours. \displaystyle \frac{1}{12} + \frac{1}{t} = \frac{1}{8} gives \displaystyle \frac{1}{t} = \frac{1}{24}.
  14. \displaystyle \frac{15}{2} hours, which is 7.5 hours. The drain’s rate is subtracted: \displaystyle \frac{1}{3} - \frac{1}{5} = \frac{2}{15} of a tank each hour.
  15. 4 hours and 12 hours. \displaystyle \frac{1}{t} + \frac{1}{t + 8} = \frac{1}{3} gives t^2 + 2t - 24 = (t - 4)(t + 6) = 0; -6 is rejected, as no time is negative.
  16. \displaystyle T = \frac{ab}{a + b}, and T = 12. Multiply by abT: bT + aT = ab, so T(a + b) = ab. Then \displaystyle \frac{600}{50} = 12.
  17. 125 km/h. \displaystyle \frac{600}{p + 25} = \frac{400}{p - 25} gives 600p - 15000 = 400p + 10000, so 200p = 25000. Check: both flights take 4 hours.
  18. 1 km/h. \displaystyle \frac{12}{5 - c} + \frac{12}{5 + c} = 5 gives 120 = 5(25 - c^2), so c^2 = 1; -1 is rejected. Check: 3 + 2 = 5 hours.
  19. 12 km/h. \displaystyle \frac{36}{s} + \frac{36}{s + 6} = 5 gives 5s^2 - 42s - 216 = 0, which is (s - 12)(5s + 18) = 0; the negative candidate is rejected. Check: 3 + 2 = 5 hours.
  20. A solution of the original makes every denominator non-zero, so the LCD L is a non-zero number there, and multiplying a true equation by it keeps it true. Where L \ne 0 the step can be undone by dividing by L, so a solution of the cleared equation with L \ne 0 also solves the original. The only values that can slip in are those where L = 0, which are the excluded values.
  21. Multiplying both sides by bd cancels b on the left and d on the right, leaving ad = bc. When b or d contains x, bd is zero at some values of x, so the excluded values are listed and every candidate is compared with them.
  22. 6 hours, since \displaystyle \frac{1}{8} + \frac{1}{24} = \frac{4}{24} = \frac{1}{6}. Marcellin’s plow adds clearing to Ludmilla’s, so together they beat her 8 hours. But his plow is slower than hers, so together they do less than two of hers would, and two of hers would take 4 hours.
  23. In Example 2, 3 makes a denominator zero, so the original equation has no value there: an algebraic reason, and 3 is extraneous. In Example 6, -3 does solve the equation, but a time cannot be negative: a reason from the story.
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