2. Rational equations · Equations and systems

What is a rational equation?

Definition

A rational equation is an equation in which the unknown appears in the denominator of one or more algebraic fractions.

For example:

\dfrac{x-1}{x} - \dfrac{x+1}{x-5} = \dfrac{x+2}{x^{2}-5x}.

The idea for solving them is the same as with numerical fractions: clear the denominators so that we are left with a polynomial equation that we already know how to solve.

Strategy (5 steps)

Factor all the denominators.
Compute the LCM of the denominators.
Build up each fraction so that they all have the LCM as their denominator.
Clear the denominators (you are left with an equality of polynomials).
Solve the resulting polynomial equation.

The final check is essential

When we multiply by a denominator that contains the unknown, we may introduce false solutions (called extraneous) that make some original denominator zero and are therefore not valid.

That is why, once the solutions have been found, we must always check that none of them makes any denominator zero. If one does, it is discarded.

Exercises

1 Equation with three fractions

Solve $\;\dfrac{x-1}{x} - \dfrac{x+1}{x-5} = \dfrac{x+2}{x^{2}-5x}$.

Step 1 — Factor the denominators

The first two already are. The third factors as:

x^{2}-5x = x\,(x-5).

We rewrite the equation:

\dfrac{x-1}{x} - \dfrac{x+1}{x-5} = \dfrac{x+2}{x(x-5)}.

Step 2 — LCM of the denominators

The denominators are $x$, $(x-5)$ and $x(x-5)$. The LCM is $\;x(x-5)$.

Step 3 — Build up each fraction

\dfrac{(x-1)(x-5)}{x(x-5)} - \dfrac{(x+1)\cdot x}{x(x-5)} = \dfrac{x+2}{x(x-5)}.

Step 4 — Clear the denominators

Since all three fractions have the same denominator, we equate the numerators:

(x-1)(x-5) - (x+1)\cdot x = x+2.

Step 5 — Solve the equation

We expand the products:

x^{2}-5x-x+5 \;-\; x^{2}-x \;=\; x+2.

The $x^{2}$ terms cancel out:

$$-7x + 5 = x + 2.$$

5 - 2 = x + 7x \;\Longrightarrow\; 3 = 8x \;\Longrightarrow\; \boxed{\,x = \tfrac{3}{8}.\,}

Check

We verify that $x = \tfrac{3}{8}$ makes no denominator zero:

$x = \tfrac{3}{8} \neq 0$ ✓
$x - 5 = \tfrac{3}{8} - 5 = -\tfrac{37}{8} \neq 0$ ✓
$x(x-5) \neq 0$ ✓ (since neither factor is)

The solution is valid.

2 Equation with a term without a denominator

Solve $\;\dfrac{3}{x+2} - \dfrac{x}{x-2} = 2$.

Step 1 — Factor the denominators

All the denominators are already irreducible factors: $\;(x+2)$, $\;(x-2)$ and the $1$ on the right-hand side (which we treat as $\dfrac{2}{1}$).

Step 2 — LCM of the denominators

$\operatorname{lcm}\bigl(x+2,\; x-2,\; 1\bigr) = (x+2)(x-2)$.

Step 3 — Build up each fraction

\dfrac{3(x-2)}{(x+2)(x-2)} - \dfrac{x(x+2)}{(x+2)(x-2)} = \dfrac{2(x+2)(x-2)}{(x+2)(x-2)}.

Step 4 — Clear the denominators

$$3(x-2) - x(x+2) = 2(x+2)(x-2).$$

Step 5 — Solve the equation

We expand the three products:

3x-6 \;-\; x^{2}-2x \;=\; 2\,(x^{2}-4) = 2x^{2}-8.

We group everything on one side:

3x-6-x^{2}-2x-2x^{2}+8 = 0 \;\Longrightarrow\; -3x^{2}+x+2 = 0.

We multiply by $-1$ to get a positive leading coefficient and apply the formula:

3x^{2}-x-2 = 0 \;\Longrightarrow\; x = \dfrac{1 \pm \sqrt{1-4\cdot 3\cdot(-2)}}{6} = \dfrac{1 \pm \sqrt{25}}{6} = \dfrac{1 \pm 5}{6}.

So:

x_{1} = \dfrac{6}{6} = 1, \qquad x_{2} = \dfrac{-4}{6} = -\tfrac{2}{3}.

Check

Neither of the two values makes $\,x+2\,$ or $\,x-2$ zero:

$x=1$: $\;1+2 = 3$, $\;1-2 = -1$. ✓
$x=-\tfrac{2}{3}$: $\;-\tfrac{2}{3}+2 = \tfrac{4}{3}$, $\;-\tfrac{2}{3}-2 = -\tfrac{8}{3}$. ✓

\boxed{\;x = 1 \quad\text{or}\quad x = -\tfrac{2}{3}.\;}