Fractional Equations
A fractional equation is an equation that contains at least one fraction with an unknown variable in the denominator. Solving them is extremely easy; in fact, it’s so easy that we can sum up the whole solving process in these four steps:
- Express the equation in normal form, that is, as a single fraction: $\frac{N(x)}{D(x)} = 0$;
- Clear out the denominator(s) by multiplying both sides of the equation by the denominator;
- Solve the resulting equation $N(x) = 0$;
- Check the solutions against the domain of the original equation to exclude any values that make the denominator equal to zero.
That’s it! Let’s try to solve some practical problems.
Example 1
$$ \frac{4x+2}{5x-1} = 0 $$
The equation is already in normal form; therefore, we just need to clear out the denominator by multiplying both sides by $5x - 1$:
$$ \cancel{(5x-1)} \cdot \frac{4x+2}{\cancel{5x-1}} = \cancel{0 \cdot (5x - 1)} $$
We’re left with $4x + 2 = 0$. That is: $$ 4x + 2 = 0 \implies 4x = -2 \implies x = - \frac{1}{2} $$
Finally, let’s check it against the domain of the equation: $$ 5x - 1 \neq 0 \iff x \neq \frac{1}{5} $$
Since our solution satisfies this condition, it is acceptable and the solution is $x = - \frac{1}{2}$.
Example 2
$$ \frac{x^2 - 3x}{2x} = 0 $$
The equation is already in normal form. Let’s get rid of the denominator by multiplying both sides by $2x$, obtaining: $$ x^2 - 3x = 0 $$
Now let’s factor the resulting polynomial by taking out the greatest common factor (GCF): $$ x(x-3) = 0 $$
We get two distinct solutions, namely: $x_1 = 0$ and $x_2 = 3$. Now let’s check the domain of the original equation: $$ 2x \neq 0 \iff x \neq 0 $$
Since our first solution doesn’t satisfy this condition, it is not acceptable (we’d get a division by zero!). Therefore, the solution of the original equation is $x = 3$.
Example 3
$$ \frac{1}{x+5} - \frac{1}{x+4} = - \frac{x+5}{x^2 + 9x + 20} $$
This time the equation is not in normal form since it consists of three separate fractions. Let’s start by moving all of them on the left-hand side: $$ \frac{1}{x+5} - \frac{1}{x+4} + \frac{x+5}{x^2 + 9x + 20} = 0 $$
We can now find the least common denominator. First, let’s factor the third denominator by grouping: $$ \begin{aligned} &x^2 + 9x + 20\\ &x^2 + 4x + 5x + 20\\ &x(x+4) + 5(x+4)\\ &(x+4)(x+5) \end{aligned} $$
Therefore, we can rewrite the original equation as: $$ \frac{1}{x+5} - \frac{1}{x+4} + \frac{x+5}{(x+4)(x+5)} = 0 $$
Now let’s combine these fractions: $$ \begin{aligned} \frac{x+4 - (x+5) + x +5}{(x+4)(x+5)} &= 0\\ \frac{x+4 -x - 5 +x + 5}{(x+4)(x+5)} &= 0\\ \frac{x+4}{(x+4)(x+5)} &= 0 \end{aligned} $$
The equation is now in normal form. We can therefore clear the denominator by multiplying both sides by $(x+4)(x+5)$. We get: $$ x+4 = 0 \implies x = -4 $$
Finally, let’s check the domain of the original equation: $$ (x+4)(x+5) \neq 0 $$
Which means that: $$ x \neq -4 \wedge x \neq -5 $$
The solution we’ve found, $x = -4$, does not satisfy this condition. Therefore, it cannot be accepted and the original equation has no solution.