We replace $x$ with $7$ in the above expression and simplify as much as possible.

\[ \begin{align} \dfrac{4}{x+1} &=\\[5pt] \dfrac{4}{7+1} &=\\[5pt] \dfrac{4}{8} &=\\[5pt] \dfrac{1}{2} \end{align} \]

We replace $x$ with $5$ in the above expression and simplify as much as possible.

\[ \begin{align} \dfrac{6x}{5-3x} &=\\[5pt] \dfrac{6(5)}{5-3(5)}&=\\[5pt] \dfrac{30}{5-15} &=\\[5pt] \dfrac{30}{(-10)}&=\\[5pt] -3 \end{align} \]

We replace $x$ with $-2$ in the expression and simplify as much as possible.

\[ \begin{align} \dfrac{x+4}{10-x}&=\\[5pt] \dfrac{(-2)+4}{10-(-2)}&=\\[5pt] \dfrac{2}{12} &=\\[5pt] \dfrac{1}{6} \end{align} \]

We replace $x$ with $2$ in the expression and simplify as much as possible.

\[ \require{cancel} \begin{align*} \dfrac{3(6-x)}{x+1}&=\\[5pt] \dfrac{3(6-2)}{2+1}&=\\[5pt] \dfrac{3(4)}{3}&=\\[5pt] \dfrac{\cancel{3}(4)}{\cancel{3}}&=\\[5pt] 4 \end{align*} \]

We replace $p$ with $-3$ in the expression and simplify as much as possible.

\[ \begin{align*} \require{cancel} \dfrac{p+5}{2(4-p)}&=\\[5pt] \dfrac{-3+5}{2(4-(-3))}&=\\[5pt] \dfrac{2}{2(7)}&=\\[5pt] \dfrac{\cancel{2}}{\cancel{2}(7)}&=\\[5pt] \dfrac{1}{7} \end{align*} \]

We replace $a$ with $3$ in the above expression and simplify as much as possible.

\[ \begin{align*} \dfrac{3(3a+1)}{2(2a-1)}&=\\[5pt] \dfrac{3(3(3)+1)}{2(2(3)-1)}&=\\[5pt] \dfrac{3(9+1)}{2(6-1)}&=\\[5pt] \dfrac{3(10)}{2(5)}&=\\[5pt] \dfrac{3\cdot 10}{10} &=\\[5pt] 3 \end{align*} \]

Replacing $w$ with $2$ in the expression we get: \begin{align*} \dfrac{6}{w}+\dfrac{w}{6} &= \\[5pt] \dfrac{6}{(2)}+\dfrac{(2)}{6} &= \\[5pt] \dfrac{3}{1}+\dfrac{1}{3} &= \\[5pt] \dfrac{(3)(3)+(1)(1)}{(1)(3)} &= \\[5pt] \dfrac{9+1}{3} &=\\[5pt] \dfrac{10}{3} \end{align*}

Replacing $r$ with $3$ in the expression we get: \begin{align*} \dfrac{9r+3}{3r+1}-2r&= \\[5pt] \dfrac{9(3)+3}{3(3)+1}-2(3)&= \\[5pt] \dfrac{27+3}{9+1}-6&= \\[5pt] \dfrac{30}{10}-6 &= \\[5pt] 3-6 &= \\[5pt] -3 \end{align*}

We replace $x$ with $-3$ in the expression and simplify as much as possible.

\[ \begin{align*} \dfrac{2}{x} - \dfrac{5}{2x} &=\\[5pt] \dfrac{2}{(-3)} - \dfrac{5}{2\cdot (-3)}&=\\[5pt] -\dfrac{2}{3} - \left( - \dfrac{5}{6} \right) &=\\[5pt] -\dfrac{2}{3} + \dfrac{5}{6} &=\\[5pt] -\dfrac{4}{6} + \dfrac{5}{6} &=\\[5pt] \dfrac{-4+5}{6} &=\\[5pt] \dfrac 1 6 \end{align*} \]

Replacing $x$ with $2$ and $y$ with $3$ in the expression, we get the following: \begin{align*} \dfrac{24}{xy}+\dfrac{x}{4y} &= \\[5pt] \dfrac{24}{(2)(3)}+\dfrac{(2)}{4(3)} &= \\[5pt] \dfrac{24}{6}+\dfrac{2}{12} &= \\[5pt] 4+\dfrac{1}{6} &= \\[5pt] \dfrac{24+1}{6} &= \\[5pt] \dfrac{25}{6} & \end{align*}

Replacing $r$ with $3$ and $t$ with $2$ in the expression, we get the following: \begin{align*} \dfrac{2r}{r+t}-\dfrac{r}{t}&= \\[5pt] \dfrac{2(3)}{3+2}-\dfrac{3}{2}&= \\[5pt] \dfrac{6}{5}-\dfrac{3}{2}&= \\[5pt] \dfrac{12}{10}-\dfrac{15}{10}&= \\[5pt] \dfrac{12-15}{10} &= \\[5pt] -\dfrac{3}{10}. \end{align*}

We replace $a$ with $3$ and $b$ with $6$ in the expression and simplify as much as possible. \begin{align*} \dfrac{a-b}{4}+\dfrac{b}{3a-5}&= \\[5pt] \dfrac{3-6}{4}+\dfrac{6}{3(3)-5}&= \\[5pt] \dfrac{-3}{4}+\dfrac{6}{4}&= \\[5pt] -\dfrac{3}{4}+\dfrac{6}{4}&= \\[5pt] \dfrac{-3+6}{4}&= \\[5pt] \dfrac{3}{4} \end{align*}