Solving Rational Equations

A rational equation contains one or more rational expressions, and it is solved by multiplying both sides by the least common denominator to clear the fractions, solving the resulting equation and rejecting any solution that makes a denominator zero. In brief: to solve 2/x + 3/(x + 1) = 1, multiply by x(x + 1): 2(x + 1) + 3x = x(x + 1), so x² − 4x − 2 = 0 and x = 2 ± √6. Both values are allowed because neither is 0 or −1.

Rational equations describe work, mixture and motion problems and appear in physics formulas for resistance and lenses. The method is straightforward, but it can produce extraneous solutions, because multiplying by an expression that is zero at some x introduces a false answer. This guide gives a step-by-step method, examples with linear and quadratic results, word problems and a table of the usual steps and checks.

How do you solve a rational equation step by step?

1. Factor all the denominators and note the values that make any denominator zero. These are excluded.
2. Find the least common denominator (LCD) of all the fractions.
3. Multiply every term on both sides by the LCD and cancel.
4. Solve the resulting polynomial equation.
5. Discard solutions that are excluded values. Check the rest in the original equation.

Worked examples

Example 1: a linear result

3/x + 1/2 = 5/x − 1/4. Excluded: x = 0. The LCD is 4x. Multiply: 12 + 2x = 20 − x, so 3x = 8 and x = 8/3. Check: 3/(8/3) + 1/2 = 9/8 + 4/8 = 13/8 and 5/(8/3) − 1/4 = 15/8 − 2/8 = 13/8 ✓.

Example 2: a quadratic result

2/x + 3/(x + 1) = 1. Excluded: x ≠ 0, −1. The LCD is x(x + 1). Multiply: 2(x + 1) + 3x = x(x + 1), so 5x + 2 = x² + x, giving x² − 4x − 2 = 0. The discriminant is 16 + 8 = 24, so x = (4 ± √24)/2 = 2 ± √6 ≈ 4.449 or −0.449. Both are allowed.

Example 3: an extraneous solution

x/(x − 2) = 2/(x − 2) + 3. Excluded: x = 2. Multiply by x − 2: x = 2 + 3(x − 2), so x = 3x − 4 and x = 2. But x = 2 is excluded, so the equation has no solution. The solution x = 2 is extraneous, since both sides are undefined there.

Example 4: a difference of squares

1/(x − 3) + 1/(x + 3) = 6/(x² − 9). The LCD is (x − 3)(x + 3) with x ≠ ±3. Multiply: (x + 3) + (x − 3) = 6, so 2x = 6 and x = 3. This is excluded. No solution.

Example 5: a solution with checking

x/(x + 1) − 1/(x − 1) = 2/(x² − 1). Excluded x ≠ ±1. LCD: (x + 1)(x − 1). Multiply: x(x − 1) − (x + 1) = 2, so x² − x − x − 1 = 2, giving x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0. The roots x = 3 and x = −1. But x = −1 is excluded. The only solution is x = 3. Check: 3/4 − 1/2 = 1/4 and 2/8 = 1/4 ✓.

How do you solve proportions?

When the equation has the form a/b = c/d, cross-multiply: ad = bc. For (x + 2)/(x − 1) = 3/2: 2(x + 2) = 3(x − 1), so 2x + 4 = 3x − 3 and x = 7. Check: 9/6 = 3/2 ✓. Cross-multiplication is the same as multiplying by both denominators, and it needs the same exclusions.

How do you solve word problems?

Example 6: work

One pipe fills a tank in 6 hours and another in 3 hours. Together, the fraction per hour is 1/6 + 1/3 = 1/2, so they need 2 hours. If a third pipe empties the tank in 12 hours, then 1/6 + 1/3 − 1/12 = 5/12 per hour, taking 12/5 = 2.4 hours.

Example 7: finding the unknown time

Anna paints a fence twice as fast as Ben. Together they finish in 4 hours. How long would Ben need alone? Let Ben need t hours. Then Anna needs t/2. 1/t + 2/t = 1/4, so 3/t = 1/4 and t = 12. Ben needs 12 hours and Anna 6 hours.

Example 8: motion

A boat goes 24 km upstream and returns in a total time of 5 hours. The speed of the current is 2 km/h. Let the boat’s speed in still water be v. Then 24/(v − 2) + 24/(v + 2) = 5. The LCD is (v − 2)(v + 2): 24(v + 2) + 24(v − 2) = 5(v² − 4), so 48v = 5v² − 20, which gives 5v² − 48v − 20 = 0. The discriminant is 2304 + 400 = 2704 = 52², so v = (48 ± 52)/10, giving v = 10 or v = −0.4. The negative speed is rejected. So v = 10 km/h. Check: 24/8 + 24/12 = 3 + 2 = 5 ✓.

Example 9: lenses

The lens equation 1/f = 1/d_o + 1/d_i: for f = 10 cm and d_o = 30 cm, 1/d_i = 1/10 − 1/30 = 2/30, so d_i = 15 cm.

Method summary table

Where do these methods come from?

Problems about combined work and motion, the usual sources of rational equations, appear in ancient Chinese and Indian mathematics, for example, “pipes filling a cistern”, as found in the writings of Heron of Alexandria and Indian authors. The algebraic method of clearing denominators became standard with symbolic algebra. The idea of extraneous solutions was clarified with a better understanding of equivalent transformations.

Where are they used today?

Parallel circuits, mixtures, average rates and optics lead to rational equations. A resistor of 6 ohms in parallel with an unknown R₂ giving a total of 2 ohms satisfies 1/6 + 1/R₂ = 1/2, so R₂ = 3 ohms. In economics, break-even analysis with average cost leads to equations of this type.

Which mistakes are common?

Not checking for extraneous solutions

Always substitute your candidates, or at least compare them with the excluded values.

Multiplying only some terms by the LCD

Every term on both sides must be multiplied, including terms without fractions.

Forgetting parentheses

When multiplying by a binomial, use brackets: 5 · (x + 1), not 5 · x + 1.

Cancelling terms across the equals sign

Do not cancel x in 3/x = 5/x by dividing, which would suggest 3 = 5. Instead, multiply by x to get 3 = 5, which is false, so there is no solution.