First, we apply the addition principle, adding $5$ to both sides:

\begin{align*} 3x - 5 &\le 4 \\[5pt] 3x-5+5 &\le 4+5 \\[5pt] 3x+0 &\le 9 \\[5pt] 3x &\le 9 \end{align*}

Second, we apply the multiplication principle, dividing both sides by $3{:}$

\begin{align*} \require{cancel} 3x &\le 9 \\[5pt] \dfrac{3x}{3} &\le \dfrac{9}{3} \\[5pt] \dfrac{\cancel{3}x}{\cancel{3}} &\le 3 \\[5pt] x &\le 3 \end{align*}

Therefore, the solution is $x \le 3.$

First, we apply the addition principle, subtracting $1$ from both sides:

\begin{align*} 3c+1 &\le 0 \\[5pt] 3c+1-1 &\le 0-1 \\[5pt] 3c+0 &\le -1 \\[5pt] 3c &\le -1 \end{align*}

Second, we apply the multiplication principle, dividing both sides by $3{:}$

\begin{align*} \require{cancel} 3c &\le -1 \\[5pt] \dfrac{3c}{3} &\le -\dfrac{1}{3} \\[5pt] \dfrac{\cancel{3}c}{\cancel{3}} &\le -\dfrac{1}{3} \\[5pt] c &\le -\dfrac{1}{3} \end{align*}

Therefore, the solution is $c \le -\dfrac{1}{3}.$

First, we apply the addition principle, adding $30$ to both sides:

\begin{align*} 4t - 30 &> -20 \\[5pt] 4t - 30+30 &>-20+30\\[5pt] 4t+0 &>10 \\[5pt] 4t &>10 \end{align*}

Second, we apply the multiplication principle, dividing both sides by $4{:}$

\begin{align*} \require{cancel} 4t &>10 \\[5pt] \dfrac{4t}{4} &> \dfrac{10}{4} \\[5pt] \dfrac{\cancel{4}t}{\cancel{4}} &> \dfrac52 \\[5pt] t &> \dfrac52 \end{align*}

Therefore, the solution is $t > \dfrac{5}{2}.$

First, we apply the addition principle, subtracting $3$ from both sides: \begin{align} -2x + 3 &\le 7 \\[5pt] -2x + 3 - 3 &\le 7 - 3 \\[5pt] -2x + 0 &\le 4 \\[5pt] -2x &\le 4 \end{align}

Then, we apply the multiplication principle, dividing both sides by $-2.$

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

\begin{align} \require{cancel} -2x \,\,&{\color{blue}\le\,\,} 4 \\[5pt] \dfrac{-2x}{-2} \,&{\color{red}\ge\,\,} \dfrac{4}{-2} \\[5pt] \dfrac{\cancel{-2}x}{\cancel{-2}} \,&{\color{red}\ge\,\,} {-2} \\[5pt] x \,\,&{\color{red}\ge\,} {-2} \end{align}

Therefore, the solution is $x\ge -2.$

First, we apply the addition principle, adding $5$ to both sides: \begin{align} -6x - 5 &\ge -8 \\[5pt] -6x - 5+5 &\ge -8+5 \\[5pt] -6x + 0 &\ge -3 \\[5pt] -6x &\ge -3 \end{align}

Then, we apply the multiplication principle, dividing both sides by $-6.$

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

\begin{align} \require{cancel} -6x \,\,&{\color{blue}\ge\,\,} -3 \\[5pt] \dfrac{-6x}{-6} \,&{\color{red}\le\,\,} \dfrac{-3}{-6} \\[5pt] \dfrac{\cancel{-6}x}{\cancel{-6}} \,&{\color{red}\le\,\,} \dfrac12 \\[5pt] x \,\,&{\color{red}\le\,} \dfrac12 \end{align}

Therefore, the solution is $x\le \dfrac12.$

First, we apply the addition principle, subtracting $6$ from both sides: \begin{align} -5x + 6 &> -9 \\[5pt] -5x + 6-6 &> -9-6 \\[5pt] -5x + 0 &> -15 \\[5pt] -5x &> -15 \end{align}

Then, we apply the multiplication principle, dividing both sides by $-5.$

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

\begin{align} \require{cancel} -5x \,\,&{\color{blue}>\,\,} -15 \\[5pt] \dfrac{-5x}{-5} \,&{\color{red}< \,\,} \dfrac{-15}{-5} \\[5pt] \dfrac{\cancel{-5}x}{\cancel{-5}} \,&{\color{red}< \,\,} {3} \\[5pt] x \,\,&{\color{red}< \,} {3} \end{align}

Therefore, the solution is $x< 3.$

First, we apply the addition principle, subtracting $7$ from both sides:

\begin{align} -5 &< 6w+7 \\[3pt] -5 - 7 &< 6w+7 - 7 \\[3pt] -12 &< 6w + 0 \\[3pt] -12 &< 6w \end{align}

Second, we apply the multiplication principle, dividing both sides by $6{:}$

\begin{align} \require{cancel} -12 &< 6w \\[5pt] -\dfrac{12}{6} &< \dfrac{6w}{6} \\[5pt] -2 &< \dfrac{\cancel{6}w}{\cancel{6}} \\[5pt] -2 &< w \end{align}

Finally, let's swap the left and right-hand sides so that the variable is on the left-hand side. Remember, we also need to flip the inequality.

\[ w > -2 \]

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

First, we apply the addition principle, subtracting $7$ from both sides:

\begin{align} 6 &< -2y + 7 \\[5pt] 6 - 7 &< -2y + 7 - 7 \\[5pt] -1 &< -2y + 0 \\[5pt] -1 &< -2y \end{align}

Then, we apply the multiplication principle, dividing both sides by $-2.$

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

\begin{align} \require{cancel} -1 \,\,&{\color{blue}< \,\,} {-2y} \\[5pt] \dfrac{-1}{-2} \,&{\color{red}>\,\,} \dfrac {-2y}{-2} \\[5pt] \dfrac{-1}{-2} \,&{\color{red}>\,\,} \dfrac {\cancel{-2}y}{\cancel{-2}} \\[5pt] \dfrac {1} {2} \,&{\color{red}>\,\,} y \end{align}

Finally, let's swap the left and right-hand sides so that the variable is on the left-hand side. Remember, we also need to flip the inequality.

\[ y < \dfrac12 \]

Therefore, the solution is $y < \dfrac12.$

First, we apply the addition principle, subtracting $12$ from both sides:

\begin{align} -8 &> 4y + 12 \\[3pt] -8 - 12 &> 4y + 12 - 12 \\[3pt] -20 &> 4y + 0 \\[3pt] -20 &> 4y \end{align}

Second, we apply the multiplication principle, dividing both sides by $4{:}$

\begin{align} \require{cancel} -20 &> 4y \\[5pt] -\dfrac{20}{4} &> \dfrac{4y}{4} \\[5pt] -5 &> \dfrac{\cancel{4}y}{\cancel{4}} \\[5pt] -5 &> y \end{align}

Finally, let's swap the left and right-hand sides so that the variable is on the left-hand side. Remember, we also need to flip the inequality.

\[ y < -5 \]

Therefore, the solution is $y < -5.$