Fractional equation

Statement

Solve the fractional equation: $\frac{2x^3-x^2-18x+9}{2x^3+5x^2-3x} = 0$

Solution

Factorisation of numerator and denominator: $\begin{aligned} \text{Num.:}\quad & x^2(2x-1)-9(2x-1) = (x^2-9)(2x-1) = (x+3)(x-3)(2x-1) \\ \text{Den.:}\quad & x(2x^2+5x-3) = x(2x-1)(x+3) \end{aligned}$ Conditions of existence: the denominator must not vanish, hence $x\neq 0$, $x\neq \dfrac{1}{2}$, $x\neq -3$.

Simplifying the common factors: $\frac{(x+3)(x-3)(2x-1)}{x(2x-1)(x+3)} = \frac{x-3}{x} = 0 \implies \boxed{x=3}$ The value $x=3$ satisfies all the conditions of existence, so it is acceptable.

Links

Topics: First-degree equations
Concepts: Conditions of existence · Fractional equation
Skills: Solving equations · Factorising
Exercise type: Solving an equation