Solving equations involving algebraic fractions

An equation with algebraic fractions can be cleared of fractions by multiplying both sides by the LCD of all its denominators. Before you do this, state the values of the unknown that make a denominator zero; a solution equal to one of them must be rejected.

Why this is allowed for an equation but not for an expression: multiplying both sides of an equation by the same non-zero quantity keeps it balanced. An expression such as \(\dfrac{3}{x-1}+\dfrac{2}{x+1}\) has no second side, so to simplify it you keep the denominator.

Recap from Secondary 1, number denominators: for \(\dfrac{x+1}{4}-\dfrac{x-2}{6}=1\), multiply by \(12\): \(3(x+1)-2(x-2)=12\), so \(x+7=12\) and \(x=5\).

Method

1. Factorise the denominators and write down the excluded values.
2. Multiply every term on both sides by the LCD, including terms without a denominator.
3. Expand, collect and solve the resulting linear equation.
4. Reject any excluded value, and check the answer in the original equation.

Example

Solve \(\dfrac{3}{x-1}+\dfrac{2}{x+1}=\dfrac{5}{x^2-1}\).

\(x^2-1=(x-1)(x+1)\), so \(x\ne1\) and \(x\ne-1\). Multiply both sides by \((x-1)(x+1)\):

\(\begin{aligned}3(x+1)+2(x-1)&=5 \\ 3x+3+2x-2&=5 \\ 5x+1&=5 \\ x&=\dfrac45\end{aligned}\)

\(\dfrac45\) is not an excluded value. Check: the left side is \(\dfrac{3}{-\frac15}+\dfrac{2}{\frac95}=-15+\dfrac{10}{9}=-\dfrac{125}{9}\), and the right side is \(\dfrac{5}{-\frac{9}{25}}=-\dfrac{125}{9}\).

One fraction equal to one fraction: multiplying both sides by both denominators clears them in one step, often called cross-multiplying. For \(\dfrac{3}{x+2}=\dfrac{5}{2x}\) (\(x\ne-2\), \(x\ne0\)): \(3(2x)=5(x+2)\), so \(6x=5x+10\) and \(x=10\).

Common error: forgetting to multiply a whole-number term by the LCD; in \(\dfrac{2}{x}+1=\dfrac{5}{x}\), multiplying by \(x\) gives \(2+x=5\), not \(2+1=5\).