Fractional Equations

The unknown in the denominator

Solve 6/x = x − 1. The unknown is in the denominator of a fraction, and an equation like this is called a fractional equation.

First note that x cannot be 0, because 6/0 has no value. Then clear the fraction: multiply both sides by x. On the left, 6/x × x = 6. On the right, every term is multiplied by x: x(x − 1) = x² − x. So 6 = x² − x.

The equation now has an x² term, so it is a quadratic equation. Gather everything on one side: x² − x − 6 = 0.

Solve the quadratic, then check

Two numbers that multiply to −6 and add to −1 are −3 and 2, so (x − 3)(x + 2) = 0, and x = 3 or x = −2.

Check both in the original equation. For x = 3: 6/3 = 2 and 3 − 1 = 2. For x = −2: 6/(−2) = −3 and −2 − 1 = −3. Neither root is 0, the one value that was ruled out, so both are solutions.

The curve is y = 6/x. It comes in two separate pieces, with no point where x = 0. The straight line y = x − 1 crosses it twice, where x = 3 and where x = −2: the two solutions of 6/x = x − 1.

Two denominators

Solve 2/x + 3/(x + 1) = 2. Here x cannot be 0 or −1. To clear both fractions at once, multiply every term by the product of the denominators, x(x + 1).

The first term becomes 2/x × x(x + 1) = 2(x + 1). The second becomes 3/(x + 1) × x(x + 1) = 3x. The right side becomes 2x(x + 1). So 2(x + 1) + 3x = 2x(x + 1), which is 5x + 2 = 2x² + 2x.

Gather on one side: 2x² − 3x − 2 = 0. The pair that multiplies to 2 × (−2) = −4 and adds to −3 is −4 and 1, so split the middle term: 2x² − 4x + x − 2 = 2x(x − 2) + 1(x − 2) = (2x + 1)(x − 2). So x = 2 or x = −1/2.

Check x = 2: 2/2 + 3/3 = 1 + 1 = 2. Check x = −1/2: 2 ÷ (−1/2) = −4 and 3 ÷ (1/2) = 6, and −4 + 6 = 2. Both are solutions.

A root that must be rejected

Solve x²/(x − 2) = 4/(x − 2). The denominator is 0 when x = 2, so x = 2 is not allowed. Multiply both sides by x − 2: x² = 4, so x = 2 or x = −2.

The root x = 2 came out of the working, but it makes both denominators 0, so neither side of the original equation has a value there. Reject it. The only solution is x = −2. Check: (−2)²/(−2 − 2) = 4/(−4) = −1, and 4/(−4) = −1.

This is why you note the values that are not allowed before you start, and compare every root against them at the end.

The usual mistakes

Multiplying only the fraction by x. Both sides, and every term on each side, are multiplied, so the right side becomes x(x − 1) = x² − x and does not stay as x − 1.

Multiplying 6/x by x to get 6x. The x you multiply by cancels the x in the denominator, so 6/x × x = 6.

Keeping a root that makes a denominator 0. Such a value was never allowed in the original equation.

Worked example: A Journey Shortened by an Hour When the Speed Rises by 10 km/h

Question A lorry makes a journey of 300 km. If its average speed were 10 km/h faster than usual, the journey would take 1 hour less. (a) Find the usual average speed of the lorry. (b) How long would the journey take at the faster speed?

1.Let the usual speed be v km/h. Time is distance divided by speed, so the usual time is 300v hours and the faster time is 300v + 10 hours. The faster journey is 1 hour shorter: 300v − 300v + 10 = 1.

The usual time is 300v hours and the faster time is 300v + 10 hours. Their difference is 1 hour.
2.Multiply both sides by v(v + 10) to clear the fractions: 300(v + 10) − 300v = v(v + 10).

Multiply both sides by v(v + 10) to clear the fractions.
3.Expand: 300v + 3000 − 300v = v² + 10v, so 3000 = v² + 10v. Subtract 3000 from both sides: v² + 10v − 3000 = 0.

Expand both sides, then subtract 3000 from both sides: v² + 10v − 3000 = 0.
4.Factorize: two numbers with a product of −3000 and a sum of 10 are 60 and −50, so (v + 60)(v − 50) = 0, and v = −60 or v = 50. A speed cannot be negative, so v = −60 is rejected. (a) The usual speed is 50 km/h.

(a) (v + 60)(v − 50) = 0. A speed cannot be negative, so the usual speed is 50 km/h.
5.(b) The faster speed is 50 + 10 = 60 km/h, so the journey would take 30060 = 5 hours. Check: the usual time is 30050 = 6 hours, and 6 − 5 = 1 hour.

(b) At 60 km/h the journey takes 30060 = 5 hours, which is 1 hour less than 6 hours.

Answer: (a) 50 km/h; (b) 5 hours

Common mistakes

Writing 300v + 10 − 300v = 1. The faster journey takes less time, so the usual time is the larger fraction, and the smaller one is subtracted from it.
Multiplying only the two fractions by v(v + 10) and leaving the right-hand side as 1. Every term on both sides is multiplied, so the right-hand side becomes v(v + 10).

More quadratic equations problems, worked step by step →