First, we decompose all of the terms into their various factors. The $12(a-4)$ in the numerator becomes $2\cdot 2 \cdot 3 \cdot (a-4),$ and the $18$ in the denominator becomes $2 \cdot 3 \cdot 3.$

\[ \require{cancel} \begin{align*} \dfrac {12(a-4)} {18} = \dfrac {2 \cdot 2 \cdot 3 \cdot (a-4)}{2 \cdot 3 \cdot 3} \end{align*} \]

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $2$'s and one pair of $3$'s.

\[ \require{cancel} \begin{align*} \dfrac {2 \cdot 2 \cdot 3 \cdot (a-4)}{2 \cdot 3 \cdot 3} &=\\[5pt] \dfrac {\cancel{2} \cdot 2 \cdot \cancel{3} \cdot (a-4)}{\cancel{2} \cdot \cancel{3} \cdot 3} &=\\[5pt] \dfrac{2 \cdot (a-4)}{3} &=\\[5pt] \dfrac{2(a-4)}{3} \end{align*} \]

Thus, the expression simplifies to $\dfrac{2(a-4)}{3}.$

First, we decompose all of the terms into their various factors. The $10x$ in the numerator becomes $2\cdot 5 \cdot x,$ and the $12(x+1)$ in the denominator becomes $2 \cdot 2 \cdot 3 \cdot (x+1).$

\[ \require{cancel} \begin{align*} \dfrac {10x} {12(x+1)} = \dfrac {2 \cdot 5 \cdot x} {2 \cdot 2 \cdot 3 \cdot (x+1)} \end{align*} \]

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $2$'s.

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

Thus, the expression simplifies to $\dfrac{5x}{6(x+1)}.$

First, we decompose all of the terms into their various factors.

\[ \require{cancel} \begin{align*} \dfrac {x(x + 1)} {7x} = \dfrac {x \cdot (x + 1)} {7 \cdot x} \end{align*} \]

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $x$'s.

\[ \require{cancel} \begin{align*} \dfrac {x \cdot (x + 1)} {7 \cdot x} &=\\[5pt] \dfrac {\cancel{x} \cdot (x + 1)} {7 \cdot \cancel{x} } &=\\[5pt] \dfrac{x+1}{7} \end{align*} \]

Thus, the expression simplifies to $\dfrac{x+1}{7}.$

First, we decompose all of the terms into their various factors.

\[ \require{cancel} \begin{align*} \dfrac{xy}{xz} = \dfrac{x \cdot y}{x \cdot z} \end{align*} \]

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $x$'s.

\[ \require{cancel} \begin{align*} \dfrac{x \cdot y}{x \cdot z} &=\\[5pt] \dfrac{\cancel{x} \cdot y}{ \cancel{x} \cdot z} &=\\[5pt] \dfrac{y}{z} \end{align*} \]

Thus, the expression simplifies to $\dfrac{y}{z}.$

First, we decompose all of the terms into their various factors. The $14y(y+5)$ in the numerator becomes $2\cdot 7 \cdot y \cdot (y+5),$ and the $21y$ in the denominator becomes $3 \cdot 7\cdot y.$

\[ \require{cancel} \begin{align*} \dfrac {14y(y+5)} {21} = \dfrac {2 \cdot 7 \cdot y \cdot (y+5) } {3 \cdot 7\cdot y} \end{align*} \]

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $7$'s and one pair of $y$'s.

\[ \require{cancel} \begin{align*} \dfrac {2 \cdot 7 \cdot y \cdot (y+5) } {3 \cdot 7\cdot y} &=\\[5pt] \dfrac {2 \cdot \cancel{7} \cdot \cancel{y} \cdot (y+5) } {3 \cdot \cancel{7}\cdot \cancel{y} } &=\\[5pt] \dfrac{2\cdot (y+5) }{3} &=\\[5pt] \dfrac{2(y+5)}{3} \end{align*} \]

Thus, the expression simplifies to $\dfrac{2(y+5)}{3}.$

First, we decompose all of the terms into their various factors. The $6ab$ in the numerator becomes $2\cdot 3 \cdot a \cdot b,$ and the $15a$ in the denominator becomes $3 \cdot 5 \cdot a.$

\[ \require{cancel} \begin{align*} \dfrac {6ab } {15 a} = \dfrac {2 \cdot 3 \cdot a \cdot b } {3 \cdot 5 \cdot a} \end{align*} \]

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $3$'s and one pair of $a$'s.

\[ \require{cancel} \begin{align*} \dfrac {2 \cdot 3 \cdot a \cdot b } {3 \cdot 5 \cdot a} &=\\[5pt] \dfrac {2 \cdot \cancel{3} \cdot \cancel{a} \cdot b } { \cancel{3} \cdot 5 \cdot \cancel{a} } &=\\[5pt] \dfrac{2\cdot b}{5} &=\\[5pt] \dfrac{2b}{5} \end{align*} \]

Thus, the expression simplifies to $\dfrac{2b}{5}.$

First, we decompose all of the terms into their various factors. The $18b(3-c)$ in the numerator becomes $2 \cdot 3 \cdot 3 \cdot (3-c),$ and the $12(3-c)$ in the denominator becomes $2 \cdot 2 \cdot 3 \cdot (3-c).$

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac {18b(3-c)}{12(3-c)} = \dfrac {2 \cdot 3 \cdot 3 \cdot b\cdot (3-c)}{2 \cdot 2 \cdot 3 \cdot (3-c)} } \end{eqnarray}

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $2$'s, one pair of $3$'s, and one pair of $(3-c)$'s.

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac {2 \cdot 3 \cdot 3 \cdot b \cdot (3-c)}{2 \cdot 2 \cdot 3 \cdot (3-c)} &=\\[5pt] \dfrac {\cancel{2} \cdot \cancel{3} \cdot 3 \cdot b\cdot \cancel{(3-c)}}{ \cancel{2} \cdot 2 \cdot \cancel{3} \cdot \cancel{(3-c)}} &=\\[5pt] \dfrac {3 \cdot b}{2} &= \\[5pt] \dfrac {3b}{2} } \end{eqnarray}

Thus, the expression simplifies to $\dfrac {3b}{2}.$

First, we decompose all of the terms into their various factors. The $10(y+3)$ in the numerator becomes $2\cdot 5 \cdot (y+3),$ and the $5(y+3)$ in the denominator becomes $5 \cdot (y+3).$

\begin{align*} \require{cancel} \dfrac {10 (y + 3)} {5 (y + 3)} =\dfrac {2 \cdot 5 \cdot ( y+3)} {5 \cdot (y+3)} \end{align*}

Then, we cancel out all the pairs of common factors in the numerator and denominator. In this case, there is one pair of $5$'s and one pair of $(y+3)$'s.

\begin{align*} \require{cancel} \dfrac {10 (y + 3)} {5 (y + 3)} &=\dfrac {2 \cdot 5 \cdot ( y+3)} {5 \cdot (y+3)} \\[5pt] &=\dfrac {2 \cdot \cancel{5} \cdot \cancel{( y+3)} } { \cancel{5} \cdot \cancel{(y+3)} } \\[5pt] &=\dfrac {2} {1} \\[5pt] &= 2 \end{align*}

Thus, the expression simplifies to $2.$