Answer
The answer is $=\frac{2x^2-9x-1}{2(x-4)(x+3)}$.
Work Step by Step
The given expression is
$\frac{x}{x+3}-\frac{x+1}{2x^2-2x-24}$
Factor denominator of the second term.
$=2x^2-2x-24$
$=2(x^2-x-12)$
$=2(x^2-4x+3x-12)$
$=2[x(x-4)+3(x-4)]$
$=2(x-4)(x+3)$ Substitute into the expression.
$=\frac{x}{x+3}-\frac{x+1}{2(x-4)(x+3)}$
Make both term's denominator equal to least common factor $2(x-4)(x+3)$.
$=\frac{x}{x+3}\cdot \frac{2(x-4)}{2(x-4)}-\frac{x+1}{2(x-4)(x+3)}$
$=\frac{2x(x-4)}{2(x-4)(x+3)}-\frac{x+1}{2(x-4)(x+3)}$
Now subtract both numerator.
$=\frac{2x(x-4)-(x+1)}{2(x-4)(x+3)}$
$=\frac{2x^2-8x-x-1}{2(x-4)(x+3)}$
$=\frac{2x^2-9x-1}{2(x-4)(x+3)}$.
