Fractional equations with existence conditions

Text

Solve the following fractional equations, stating the existence conditions:

• (a) $\dfrac{2}{3+2x}-\dfrac{1}{3-2x}=0$;
• (b) $\dfrac{3}{x^2-3x+2}-\dfrac{1}{2-x}=0$.

Solution

(a) Existence conditions: $3+2x\neq0$ and $3-2x\neq0$, that is $x\neq\pm\dfrac32$. Common denominator $(3+2x)(3-2x)$: $\frac{2(3-2x)-(3+2x)}{(3+2x)(3-2x)}=\frac{6-4x-3-2x}{(3+2x)(3-2x)}=\frac{3-6x}{(3+2x)(3-2x)}=0.$ The numerator vanishes: $3-6x=0 \implies x=\frac12.$ Since $\dfrac12\neq\pm\dfrac32$, the solution is acceptable.

(b) Factor: $x^2-3x+2=(x-1)(x-2)$ and $2-x=-(x-2)$. Existence conditions: $x\neq1$ and $x\neq2$. The equation becomes: $\frac{3}{(x-1)(x-2)}-\frac{1}{-(x-2)}=\frac{3}{(x-1)(x-2)}+\frac{1}{x-2}=0.$ Common denominator $(x-1)(x-2)$: $\frac{3+(x-1)}{(x-1)(x-2)}=\frac{x+2}{(x-1)(x-2)}=0 \implies x+2=0 \implies x=-2.$ Since $-2\neq1,2$, the solution is acceptable. $\boxed{x_a=\frac12,\qquad x_b=-2}$