## Elementary Algebra

$\dfrac{7\sqrt{x}-28}{x-16}$
Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $\dfrac{7}{\sqrt{x}+4} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{7}{\sqrt{x}+4}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}-4} \\\\= \dfrac{7\sqrt{x}-28}{(\sqrt{x})^2-(4)^2} \\\\= \dfrac{7\sqrt{x}-28}{x-16} .\end{array}