Fractional Equations

A fractional equation is an equation in which one or more unknown variables appear in the denominator of at least one term.

For instance, the following is a fractional equation because the unknown variable x appears in the denominator:

$$ \frac{x}{x+1} = 3 $$

Fractional equations can be classified as follows:

• Numerical (or whole) when all coefficients of the unknown variable are numbers.

  Example. This equation is numerical because the unknown x has numerical coefficients - in this case, integers. $$ \frac{4x}{x-1} = \frac{1}{x+1} $$

• Literal when at least one coefficient of the unknown variable contains one or more letters.

  Example. This equation is literal because the unknown x has a literal coefficient (k). $$ \frac{4x}{x-1} = \frac{1}{kx+1} $$

How do we solve a fractional equation?

To solve a fractional equation, we must first identify the condition of existence (C.E.) - the set of values that make all denominators non-zero - and then determine which solutions satisfy that condition.

In practice, the solution process follows these steps:

1. Determine the condition of existence (C.E.) of the algebraic fractions.
2. Rewrite both sides of the equation with a common denominator.
3. Multiply both sides of the equation by this common denominator to obtain an ordinary (whole) equation.
4. Find the solution set S of the resulting equation.
5. Remove from S any values that violate the condition of existence (these are non-acceptable solutions).
6. The remaining values that satisfy the initial condition are the acceptable solutions of the fractional equation.

Note. It’s crucial to remember that only solutions satisfying the existence conditions of the fractions are valid.

A practical example

Consider the equation:

$$ \frac{3}{x} = -2 $$

First, determine the condition of existence (C.E.) - the values of x for which the denominators are non-zero.

Here, the fraction is defined only if x ≠ 0.

$$ C.E. \ : \ x \ne 0 $$

Now, rewrite both sides of the equation with a common denominator:

$$ \frac{3}{x} + 2 = 0 $$

$$ \frac{3 + 2x}{x} = 0 $$

Multiply both sides by the common denominator and simplify:

$$ \frac{3 + 2x}{x} \cdot x = 0 \cdot x $$

$$ \require{cancel} \frac{3 + 2x}{\cancel{x}} \cdot \cancel{x} = 0 $$

$$ 3 + 2x = 0 $$

Now isolate the variable x using the principles of equation equivalence.

Subtract 3 from both sides:

$$ 3 - 3 + 2x = 0 - 3 $$

$$ 2x = -3 $$

Divide both sides by 2:

$$ \frac{2x}{2} = \frac{-3}{2} $$

$$ x = \frac{-3}{2} $$

The solution x = -3/2 satisfies the condition of existence (x ≠ 0).

Therefore, x = -3/2 is an acceptable solution of the fractional equation.

Example 2

Consider the equation:

$$ \frac{x}{x-1} = \frac{-1}{1-x} $$

Determine the condition of existence (C.E.) of the algebraic fractions.

In this case, the fractions are defined only if x ≠ 1.

$$ C.E. \ : \ x \ne 1 $$

Next, rewrite both sides of the equation with a common denominator.

Using the invariant property of fractions, multiply both the numerator and the denominator of the second fraction by -1:

$$ \frac{x}{x-1} = \frac{-1}{1-x} \cdot \frac{-1}{-1} $$

$$ \frac{x}{x-1} = \frac{-1 \cdot (-1)}{(1-x) \cdot (-1)} $$

$$ \frac{x}{x-1} = \frac{1}{x-1} $$

Now both sides have the same denominator.

Thus, we can multiply both sides by the common denominator and simplify:

$$ \frac{x}{x-1} \cdot (x-1) = \frac{1}{x-1} \cdot (x-1) $$

$$ \require{cancel} \frac{x}{\cancel{x-1}} \cdot \cancel{x-1} = \frac{1}{\cancel{x-1}} \cdot (\cancel{x-1}) $$

This yields a simple whole equation with an apparent solution:

$$ x = 1 $$

However, this solution is not valid because x = 1 violates the initial condition of existence (x ≠ 1).

Therefore, the fractional equation has no valid solution - it is impossible.

And so on.