Question

The solution set of the equation, $\frac{x}{x-4}-\frac{1}{x+3}=\frac{28}{x^2-x-12}$ is

• A. $-4$
• B. $4,6$
• C. $-6$
• D. $4$

Answer 1

The correct option is C $-6$

Given,

$\frac{x}{x-4}-\frac{1}{x+3}=\frac{28}{x^2-x-12}$

$\Rightarrow \frac{x(x+3)-1(x-4)}{(x-4)(x+3)}=\frac{28}{x^2-x-12}$

$\Rightarrow \frac{x(x+3)-1(x-4)}{x^2+3x-4x-12}=\frac{28}{x^2-x-12}$

$\Rightarrow \frac{x^2+3x-x+4}{x^2-x-12}=\frac{28}{x^2-x-12}$

$\Rightarrow x^2+2x+4=28$

$\Rightarrow x^2+2x-24=0$

$\Rightarrow (x+6)(x-4)=0$

$\Rightarrow x=-6,4$

But for $x=4$ denominator of one of the term in the given expression is $0$.

Hence the only solution is $x=-6$