First, we clear the rational expression by multiplying both sides by $5\mathbin{:}$ \begin{align*} \require{cancel} \dfrac {2x + 6} 5 &= 4 \\[5pt] 5\left(\dfrac {2x + 6} 5\right )&= 5 \cdot 4 \\[5pt] \cancel{5}\left(\dfrac{2x + 6}{\cancel{5}}\right)&= 20 \\[5pt] 2x + 6 &= 20 \end{align*}

Next, we solve the remaining equation using the addition and multiplication principles: \begin{eqnarray} \eqalign{ 2x + 6 &= 20 \\[5pt] 2x + 6 - 6 &= 20 - 6 \\[5pt] 2x &= 14 \\[5pt] \dfrac{2x}{2} &= \dfrac{14}{2} \\[5pt] x &= 7 } \end{eqnarray}

First, we clear the rational expression by multiplying both sides by $5\mathbin{:}$ \[ \require{cancel} \eqalign{ -4 &=\dfrac{2y+10}{5} \\[5pt] 5 \cdot (-4) &= 5 \cdot \left(\dfrac{2y+10}{5}\right) \\[5pt] -20 &= \cancel{5} \cdot \left(\dfrac{2y+10}{\cancel{5}}\right) \\[5pt] -20 &= 2y+10 }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \eqalign{ -20 &= 2y+10 \\[5pt] -20-10 &= 2y+10-10 \\[5pt] -30 &= 2y \\[5pt] \dfrac{-30}{2} &= \dfrac{2y}{2} \\[5pt] -15 &= y }\]

Thus, $y=-15.$

First, we clear the rational expression by multiplying the equation through by $3$: \[ \require{cancel} \eqalign{ \dfrac{5x-1}{3} &=3x\\[5pt] 3 \left( \dfrac{5x-1}{3} \right) &=3\cdot 3x \\[5pt] \cancel{3} \cdot \left( \dfrac{5x-1}{\cancel{3}} \right) &=9x \\[5pt] 5x-1 &=9x }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \require{cancel} \eqalign{ 5x-1 &=9x \\[5pt] 5x-1 - 5x &=9x - 5x \\[5pt] -1 &= 4x \\[5pt] \dfrac{-1}{4} &= \dfrac{4x}{4} \\[5pt] -\dfrac{1}{4} &= x }\]

Thus, $x = -\dfrac{1}{4}.$

First, we clear the rational expression by multiplying the equation through by $3$: \[ \require{cancel} \eqalign{ \dfrac{8y+5}{3} &=6y\\[5pt] 3 \left(\dfrac{8y+5}{3}\right) &=3\cdot 6y \\[5pt] \cancel{3} \cdot \left(\dfrac{8y+5}{\cancel{3}}\right) &=18y \\[5pt] 8y+5 &=18y }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \require{cancel} \eqalign{ 8y+5 &= 18y \\[5pt] 8y+5-8y &= 18y-8y \\[5pt] 5 &= 10y \\[5pt] \dfrac{5}{10} &= \dfrac{10y}{10} \\[5pt] \dfrac{1}{2} &= y }\]

Thus, $y=\dfrac{1}{2}.$

First, we clear the rational expression by multiplying both sides by $2.$ We must make sure that all of the terms on the right-hand side are multiplied by $2:$ \[ \require{cancel} \eqalign{ \dfrac{21-g}{2} &=21+3g \\[5pt] 2 \cdot \left(\dfrac{21-g}{2}\right) &=2 \cdot (21+3g) \\[5pt] \cancel{2} \cdot \left(\dfrac{21-g}{\cancel{2}}\right) &=2 \cdot 21 + 2 \cdot 3g \\[5pt] 21-g&=42+6g }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \eqalign{ 21-g &=42+6g \\[5pt] 21- g -6 g &=42+6g -6g \\[5pt] 21 - 7g &= 42 \\[5pt] 21 - 7g - 21 &= 42 - 21 \\[5pt] -7g &= 21 \\[5pt] \dfrac{-7g}{-7} &= \dfrac{21}{-7} \\[5pt] g &= -3 }\]

First, we clear the rational expression by multiplying both sides by $4.$ We must make sure that all of the terms on the right-hand side are multiplied by $4.$ \[ \require{cancel} \eqalign{ \dfrac{10+a}{4} &=4 - 2a \\[5pt] 4 \cdot \left( \dfrac{10+a}{4} \right) &= 4 \cdot \left( 4 - 2a \right) \\[5pt] \cancel{4} \cdot \left(\dfrac{10+a}{\cancel{4}}\right) &=4 \cdot 4 - 4 \cdot 2a \\[5pt] 10+a &=16-8a }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \eqalign{ 10+a &=16-8a \\[5pt] 10+a+8a &=16-8a+8a \\[5pt] 10 + 9a &= 16 \\[5pt] 10 + 9a - 10 &= 16 - 10 \\[5pt] 9a &= 6 \\[5pt] \dfrac{9a}{9} &= \dfrac{6}{9} \\[5pt] a &= \dfrac{2}{3} }\]

First, we clear the rational expression by multiplying the equation through by $5$: \[ \require{cancel} \eqalign{ \dfrac{1+z}{5} &= 2(z-2) \\[5pt] 5 \cdot \left( \dfrac{1+z}{5} \right) &= 5 \cdot 2(z-2) \\[5pt] \cancel{5} \cdot \left( \dfrac{1+z}{ \cancel{5} } \right) &= 10(z-2) \\[5pt] 1+z &= 10z-20 }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \eqalign{ 1+z &= 10z-20 \\[5pt] 1+z-1 &= 10z-20-1 \\[5pt] z &= 10z - 21 \\[5pt] z - 10z &= 10z-21-10z \\[5pt] -9z &= -21 \\[5pt] \dfrac{-9z}{-9} &= \dfrac{-21}{-9} \\[5pt] z &= \dfrac{21}{9} \\[5pt] z &= \dfrac{7}{3} }\]

First, we clear the rational expression by multiplying the equation through by $4$: \[ \require{cancel} \eqalign{ -2(y+10) &= \dfrac{-9+y}{4} \\[5pt] 4 \cdot (-2)(y+10) &= 4\left(\dfrac{-9+y}{4}\right) \\[5pt] -8(y+10) &= \cancel{4}\left(\dfrac{-9+y}{\cancel{4}}\right) \\[5pt] -8y-80 &= -9+y }\]

Then, we solve the remaining equation using the addition and multiplication principles: \[ \eqalign{ -8y-80 &=-9+y\\[5pt] -8y-80 - y &= -9 +y-y\\[5pt] -9y - 80 &= -9 \\[5pt] -9y - 80 +80 &= -9 + 80 \\[5pt] -9y &= 71 \\[5pt] \dfrac{-9y}{-9} &= \dfrac{71}{-9} \\[5pt] y &= -\dfrac{71}{9} }\]