Method for simplifying equations
In mathematics, the method of clearing denominators, also called clearing fractions, is a technique for simplifying an equation equating two expressions that each are a sum of rational expressions – which includes simple fractions.
Example
Consider the equation
![{\displaystyle {\frac {x}{6}}+{\frac {y}{15z}}=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de7f32c128a62a96a1f368f8051aa70d6fbd6a61)
The smallest common multiple of the two denominators 6 and 15z is 30z, so one multiplies both sides by 30z:
![{\displaystyle 5xz+2y=30z.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75241a93d8342d118b583e082f4697a2094ec634)
The result is an equation with no fractions.
The simplified equation is not entirely equivalent to the original. For when we substitute y = 0 and z = 0 in the last equation, both sides simplify to 0, so we get 0 = 0, a mathematical truth. But the same substitution applied to the original equation results in x/6 + 0/0 = 1, which is mathematically meaningless.
Description
Without loss of generality, we may assume that the right-hand side of the equation is 0, since an equation E1 = E2 may equivalently be rewritten in the form E1 − E2 = 0.
So let the equation have the form
![{\displaystyle \sum _{i=1}^{n}{\frac {P_{i}}{Q_{i}}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04f22dfd1e4c53e965deeb55526c64a9a06e772c)
The first step is to determine a common denominator D of these fractions – preferably the least common denominator, which is the least common multiple of the Qi.
This means that each Qi is a factor of D, so D = RiQi for some expression Ri that is not a fraction. Then
![{\displaystyle {\frac {P_{i}}{Q_{i}}}={\frac {R_{i}P_{i}}{R_{i}Q_{i}}}={\frac {R_{i}P_{i}}{D}}\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6f2b8d098919655791b76165785aa28dd5a1d59)
provided that RiQi does not assume the value 0 – in which case also D equals 0.
So we have now
![{\displaystyle \sum _{i=1}^{n}{\frac {P_{i}}{Q_{i}}}=\sum _{i=1}^{n}{\frac {R_{i}P_{i}}{D}}={\frac {1}{D}}\sum _{i=1}^{n}R_{i}P_{i}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7789a319a5d102c2180a7fc7925d8098187e1e8)
Provided that D does not assume the value 0, the latter equation is equivalent with
![{\displaystyle \sum _{i=1}^{n}R_{i}P_{i}=0\,,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8a4a5be59e3263dc2e742624b12cfe21d828f57)
in which the denominators have vanished.
As shown by the provisos, care has to be taken not to introduce zeros of D – viewed as a function of the unknowns of the equation – as spurious solutions.
Example 2
Consider the equation
![{\displaystyle {\frac {1}{x(x+1)}}+{\frac {1}{x(x+2)}}-{\frac {1}{(x+1)(x+2)}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aab636250c2c75d0db8cfbcd46c9155098a54edd)
The least common denominator is x(x + 1)(x + 2).
Following the method as described above results in
![{\displaystyle (x+2)+(x+1)-x=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a2211b458e3f1bacedad1ca29a84363f56da505)
Simplifying this further gives us the solution x = −3.
It is easily checked that none of the zeros of x(x + 1)(x + 2) – namely x = 0, x = −1, and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.
References
- Richard N. Aufmann; Joanne Lockwood (2012). Algebra: Beginning and Intermediate (3 ed.). Cengage Learning. p. 88. ISBN 978-1-133-70939-8.