## College Algebra (10th Edition)

$\displaystyle \frac{22x+5}{10\sqrt{(x-5)(4x+3)}}$
$\displaystyle \sqrt{4x+3}\cdot\frac{1}{2\sqrt{x-5}}+\sqrt{x-5}\cdot\frac{1}{5\sqrt{4x+3}},\quad x>5$ $=\displaystyle \frac{\sqrt{4x+3}}{2\sqrt{x-5}}+\frac{\sqrt{x-5}}{5\sqrt{4x+3}}$ ... LCD =$2\sqrt{x-5}\cdot 5\sqrt{4x+3}$= $10\sqrt{x-5}\sqrt{4x+3}$ $=\displaystyle \frac{\sqrt{4x+3}\cdot 5\cdot\sqrt{4x+3} +\sqrt{x-5}\cdot 2\sqrt{x-5}}{10\sqrt{x-5}\sqrt{4x+3}}$ ... apply $\sqrt{a}\cdot\sqrt{a}=\sqrt{a^{2}}=|a|$ $=\displaystyle \frac{5|4x+3|+|x-5|}{10\sqrt{x-5}\sqrt{4x+3}}$ ... since $x>5$, both absolute brackets can be replaced with parentheses $=\displaystyle \frac{5(4x+3)+2(x-5)}{10\sqrt{x-5}\sqrt{4x+3}}$ ... numerator: distribute and simplify ... in the denominator, $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ $=\displaystyle \frac{20x+15+2x-10}{10\sqrt{(x-5)(4x+3)}}$ $=\displaystyle \frac{22x+5}{10\sqrt{(x-5)(4x+3)}}$