**What are fractional equations?**

The **fractional equations** are those that contain fractions in one or more of their terms. Such fractions can be numeric or algebraic, where the unknown can be found in the numerator and/or in the denominator of any term.

Here are some examples of fractional equations with only one unknown:

The first example is a linear equation with fractional coefficients; in the second example, the unknown is in the denominator of each of the terms, and in the last, the unknown is in both the numerator and the denominator.

To solve them, it is necessary to carry out some algebraic transformations and thus obtain an equivalent equation, in which the unknown does not appear in the denominator. Once this procedure is done, the solution is found using the appropriate techniques.

The solution consists of the set of values of «x» that satisfy the equality. It can be a single value, or several, but, in any case, it is very important to keep in mind that not all solutions of the equivalent equation are acceptable for the original equation.

Indeed, if it is an equation whose unknown is in the denominator, it is necessary to avoid the values of “x” that cancel it out, even if they belong to the set of solutions of the equivalent equation. This is because division by 0 is undefined.

If the equivalent equation has a unique solution, and it turns out that it cancels the denominator of one of the terms of the original equation, then it has no solution.

**How to Solve a Fractional Equation**

The operations carried out to solve non-fractional equations are valid, as long as the equality is maintained. In this way, in a fractional equation you can add or subtract the same amount to both sides of the equality, multiply all the terms by the same amount, or divide each term by the same amount (other than 0).

But since it is necessary to transform the fractional equation into an equivalent one without denominators, the following general indications are also followed:

Find the least common multiple of the denominators (lcm). Multiply each term by the lcm, in order to eliminate the denominators. Solve the equivalent equation obtained. Verify that the solutions found satisfy the original equality.

**Equivalent Equation Types**

The equivalent equations obtained following the indicated procedure can be:

Linear or first degree Quadratic Higher order

**worked examples**

**Example 1**

Solve the following equation:

Note that the equation is of the first degree in «x», since «x» is raised to 1. The coefficients of the equation are fractions and one way to eliminate them, to work with integers, is by multiplying all the terms by the least common multiple of the denominators (lcm).

lcm (2,3,6) = 6

So:

3x – 2x=1

x=1

The reader can check the validity of this solution, substituting x = 1 in the original equation and verifying that an equality is obtained.

**Example 2**

Determine the values of «x» that satisfy:

Unlike the previous example, in this case the unknown is in the denominator. Note that the denominators vanish for the values x = 2 and x = −1, a detail that should be taken into account, since if the equivalent equation admits these solutions, they must be discarded, since they are not admissible in the original equation.

Now we have to transform the equation into another without denominators, the first step is to add the terms to the left of the equality:

Since the denominators are equal, for the equality to hold, the numerators must also be equal:

4(x+1) – 3(x–2) = 8

It is enough to solve this equation, which turns out to be of first degree:

4x + 4 – 3x + 6 = 8

x = 8 – 6 – 4 = – 2

x = –2

Since this value is different from the prohibited values, it is admitted as a solution of the original equation.

**Example 3**

Find the solution of:

In this equation, the value x = 4 cancels out the denominators, therefore it is excluded from the solution set of the transformed equation, if it appears at all.

The transformed equation is easy to find, just multiply all the terms by the factor (x–4):

remaining:

2x – 4 = 4

2x = 8

x = 4

**Example 4**

Solve the equation:

In this case, the denominators have quadratic terms, so it’s a good idea to factor them first:

x2 + 8x + 7 = (x + 7)(x + 1) x2 − 49 = (x + 7)(x − 7) x2 − 6x − 7 = (x − 7)(x + 1)

The equation looks like this:

The values of x that cancel out any of the denominators are: x = −7, x = 7, x = −1. Therefore, even if these values are part of the solution set of the modified equation, they cannot be a solution of the original equation.

Now comes the process of transforming the equation. The first step is to find the least common multiple of the denominators:

lcm = (x + 7)(x − 7)(x + 1)

Multiplying both sides of the equality by the lcm is:

Resulting:

(x−7)(x− 2) = (x + 1)(2x – 5) – (x+7)(x−2)

By means of the distributive property the products are developed:

x2 – 9x +14 = 2×2 – 3x – 5 – (x2 + 5x – 14)

Reducing like terms on the right hand side:

x2 – 9x + 14 = x2 – 8x + 9

The quadratic terms cancel, because they meet with the same sign on different sides of the equality:

–9x + 14 = –8x + 9

–x = –5 ⇒ x = 5

This result is supported as a solution, since it is not one of the prohibited values.

**Fractional equations application exercise**

The denominator of a fraction is four more than the numerator. If 5 units are subtracted from the numerator and the denominator as well, the resulting fraction is 3/5. Determine the original fraction.

**Solution**

Let x be the value of the numerator.

Since the denominator of the fraction is four more than the numerator, the original fraction is:

Now we have to subtract 5 units, both from the numerator and the denominator:

Since the fraction resulting from carrying out the previous procedure is equal to 3/5, they are equal:

This is a fractional equation with the unknown in the numerator and denominator, which vanish at x = 1. Therefore, this value must be excluded, if it were among the solutions of the transformed equation.

Next, multiply both sides by the least common multiple, which is 5(x−1):

Resulting in the following equivalent equation:

5(x−5) =3(x−1)

Applying distributive property:

5x –25 = 3x – 3 ⇒ 2x = 22

x = 11

The original fraction is found by substituting x = 11 into the expression:

Which results in the fraction 11/15. This is the answer to the problem posed.

**References**

Fractional Equations. Retrieved from: mathepower.com Mathematics Portal. Fractional Equations. Problem resolution. Recovered from: silvioduarte.com. Stewart, J. (2007). Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning. Sullivan, M. (1997). precalculus. 4th. Edition. Pearson Education. Zill, D. (2008). Precalculus with Calculus advances. 4th. Edition. McGraw Hill.