We solve the inequality just like we would solve an equation.

First, we apply the addition principle:

\begin{align} 2g + 12 &\le 10 - 7g \\[5pt] 2g + 12 + 7g &\le 10 - 7g + 7g \\[5pt] 9g + 12 &\le 10 \\[5pt] 9g + 12 - 12 &\le 10 - 12 \\[5pt] 9g &\le {-2} \end{align}

Then, we use the multiplication principle:

\begin{align} 9g &\le {-2} \\[5pt] \dfrac{9g}{9} &\le \dfrac{-2}{9} \\[5pt] g &\le -\dfrac{2}{9} \end{align}

We solve the inequality just like we would solve an equation.

First, we apply the addition principle:

\begin{align} -6b + 36 &> 4b - 4 \\[5pt] -6b + 36 + 6b &> 4b - 4 + 6b \\[5pt] 36 &> 10b - 4 \\[5pt] 36 + 4 &> 10b - 4 + 4 \\[5pt] 40 &> 10b \\[5pt] \end{align}

Then, we use the multiplication principle:

\begin{align} 40 &> 10b \\[5pt] \dfrac{40}{10} &> \dfrac{10b}{10} \\[5pt] 4 &> b \\[5pt] b &< 4 \end{align}

Therefore, the solution is $b < 4.$

Our solution can be expressed using a number line, as shown below.

First, we clear the rational expression by multiplying both sides by $3{:}$ \begin{align} \require{cancel} \dfrac {7x-5} {3}&\gt 3\\[5pt] 3\cdot\left (\dfrac {7x-5} {3}\right) &\gt 3\cdot 3 \\[5pt] \cancel{3}\cdot\left (\dfrac {7x-5} {\cancel{3}}\right) &\gt 9 \\[5pt] 7x-5 & \gt 9 \end{align}

Then, we apply the addition and multiplication principles:

\begin{align} 7x - 5 &\gt 9 \\[5pt] 7x - 5 + 5 &\gt 9 + 5 \\[5pt] 7x &\gt 14 \\[5pt] \dfrac{7x}{7} &\gt \dfrac{14}{7} \\[5pt] \dfrac{\cancel{7}x}{\cancel{7}} &\gt \dfrac{14}{7} \\[5pt] x &\gt 2 \end{align}

Therefore, the solution is $x > 2.$

First, we clear the rational expression by multiplying both sides by $2{:}$ \begin{align*} \require{cancel} \dfrac{7-3x}{2} &\gt 2 \\[5pt] 2\cdot \left(\dfrac {7-3x} {2}\right) &\gt 2\cdot 2 \\[5pt] \cancel{2}\cdot \left(\dfrac {7-3x} {\cancel{2}}\right) &\gt 2\cdot 2 \\[5pt] 7-3x &\gt 4 \end{align*}

Then, we apply the addition and multiplication principles.

Remember, when we divide an inequality by a negative number, we need to flip the inequality sign:

\begin{align*} 7-3x &\gt 4 \\[5pt] 7 - 3x - 7 &\gt 4 - 7 \\[5pt] -3x &\gt {-3} \\[5pt] \dfrac{-3x}{-3} &\,\,{\color{red}{\lt}}\,\, \dfrac{-3}{-3} \\[5pt] \dfrac{\cancel{-3}x}{\cancel{-3}} &\lt 1 \\[5pt] x &\lt 1 \end{align*}

Therefore, the solution is $x < 1.$

First, we clear the rational expression by multiplying both sides by $2{:}$

\begin{align} \require{cancel} \dfrac {9x+1} {2}&\geq 5\\[5pt] 2\cdot\left (\dfrac {9x+1} {2}\right) &\geq 2\cdot 5 \\[5pt] \cancel{2}\cdot\left (\dfrac {9x+1} {\cancel{2}}\right) &\geq 10 \\[5pt] 9x+1 & \geq 10 \end{align}

Then, we apply the addition and multiplication principles: \begin{align} 9x + 1 & \geq 10\\[5pt] 9x + 1 - 1& \geq 10 - 1\\[5pt] 9x &\geq 9 \\[5pt] \dfrac{9x}{9}&\geq \dfrac{9}{9} \\[5pt] \dfrac{\cancel{9}x}{\cancel{9}} &\geq 1 \\[5pt] x &\geq 1 \end{align}

Therefore, the solution is $x \geq 1.$ This solution can be expressed on a number line, as shown below.

First, we clear the rational expression by multiplying both sides by $3{:}$

\begin{align*} \require{cancel} \dfrac {1-6x} {3}&\geq 2x\\[5pt] 3\cdot\left (\dfrac {1-6x} {3}\right) &\geq 3\cdot(2x) \\[5pt] \cancel{3}\cdot\left (\dfrac {1-6x} {\cancel{3}}\right) &\geq 6x \\[5pt] 1-6x & \geq 6x \end{align*} Next, we use the addition principle:

\begin{align*} 1-6x & \geq 6x \\[5pt] 1-6x+6x & \geq 6x + 6x\\[5pt] 1 & \geq 12x\\[5pt] \end{align*}

Finally, we use the multiplication principle:

\begin{align*} 1 & \geq 12x\\[5pt] \dfrac{1}{12} & \geq \dfrac {12x}{12}\\[5pt] \dfrac{1}{12} & \geq \dfrac {\cancel{12}x}{\cancel{12}}\\[5pt] \dfrac{1}{12} & \geq x\\[5pt] x&\leq \dfrac{1}{12} \end{align*}

Therefore, the solution is $x\leq \dfrac{1}{12}.$

First, we clear the rational expression by multiplying both sides by $2{:}$

\begin{align*} \require{cancel} 5y + 2 &\le \dfrac {3y - 1} {2} \\[5pt] 2\cdot(5y + 2) &\le 2 \cdot \left(\dfrac{3y - 1}{2}\right) \\[5pt] 2\cdot(5y + 2) &\le \cancel{2} \cdot \left(\dfrac{3y - 1}{\cancel{2}}\right) \\[5pt] 2\cdot(5y + 2) &\le 3y-1 \\[5pt] 10y+4 &\le 3y-1 \end{align*}

Next, we use the addition principle:

\[ \eqalign{ 10y+4 &\le 3y-1 \\[5pt] 10y+4-3y &\le 3y-1-3y \\[5pt] 7y+4 &\le {-1} \\[5pt] 7y+4 - 4 &\le {-1} - 4 \\[5pt] 7y &\le {-5} } \]

Finally, we use the multiplication principle:

\begin{align*} 7y &\le {-5} \\[5pt] \dfrac{7y}{7} &\le \dfrac{-5}7 \\[5pt] \dfrac{\cancel{7}y}{\cancel{7}} &\le \dfrac{-5}7 \\[5pt] y &\le -\dfrac57 \end{align*}

Therefore, the solution is $y \le -\dfrac57.$

First, we clear the rational expression by multiplying both sides by $2{:}$

\begin{align*} \require{cancel} \dfrac{1-3a}{2} &< 4a - 2 \\[5pt] 2\cdot \left(\dfrac{1-3a}{2}\right) &< 2\cdot\left(4a - 2\right) \\[5pt] \cancel{2}\cdot \left(\dfrac{1-3a}{\cancel{2}}\right) &< 2\cdot\left(4a - 2\right) \\[5pt] 1-3a&< 2\cdot\left(4a - 2\right) \\[5pt] 1-3a&< 8a - 4 \end{align*}

Next, we use the addition principle:

\begin{align*} 1-3a&< 8a - 4 \\[5pt] 3a - 1 + 8a &> -8a + 4 + 8a \\[5pt] 11a - 1 &> 4 \\[5pt] 11a - 1 + 1 &> 4 + 1 \\[5pt] 11a &> 5 \end{align*}

Finally, we use the multiplication principle:

\begin{align*} 11a &> 5 \\[5pt] \dfrac{11a}{11} &> \dfrac{5}{11} \\[5pt] \dfrac{\cancel{11}a}{\cancel{11}} &> \dfrac{5}{11} \\[5pt] a &> \dfrac{5}{11} \end{align*}

Therefore, the solution is $a > \dfrac{5}{11}.$

First, we clear the rational expressions by cross-multiplying:

\begin{align*} \dfrac{4x+7}{3} &\geq \dfrac{5x}{2}\\[5pt] 2\cdot (4x+7) &\geq 3 \cdot 5x \\[5pt] 8x + 14 &\geq 15x \end{align*}

Next, we use the addition principle:

\begin{align*} 8x + 14 &\geq 15x \\[5pt] 8x + 14 - 8x &\geq 15x - 8x \\[5pt] 14 &\geq 7x \\[5pt] \end{align*}

Finally, we use the multiplication principle:

\begin{align*} \require{cancel} 14 &\geq 7x \\[5pt] \dfrac{14}{7} &\geq \dfrac{7x}{7} \\[5pt] 2 &\geq \dfrac{\cancel{7}x}{\cancel{7}} \\[5pt] 2 &\geq x \\[5pt] x&\leq 2 \end{align*}

Therefore, the solution is $x\leq 2.$

First, we clear the rational expressions by cross-multiplying:

\begin{align*} \dfrac{x+10}{4} &> \dfrac{6-2x}{5} \\[5pt] 5\cdot (x+10) &> 4 \cdot (6-2x) \\[5pt] 5x + 50 &> 24 - 8x \end{align*}

Next, we use the addition principle:

\begin{align*} 5x + 50 &> 24 - 8x \\[5pt] 5x + 50 + 8x &> 24 - 8x + 8x \\[5pt] 13x + 50 &> 24 \\[5pt] 13x + 50 - 50 &> 24 - 50 \\[5pt] 13x &> {-26} \end{align*}

Finally, we use the multiplication principle:

\begin{align*} \require{cancel} 13x &> {-26} \\[5pt] \dfrac{13x}{13} &> \dfrac{-26}{13} \\[5pt] \dfrac{\cancel{13}x}{\cancel{13}} &> -2 \\[5pt] x &> {-2} \end{align*}

Therefore, the solution is $x > -2.$