To isolate $x,$ we need to perform the opposite of adding $4,$ which is subtracting $4.$ Remember to perform the same operation on both sides.

\begin{align*} x + 4 &< 5 \\[3pt] x + 4 - 4 &< 5 - 4 \\[3pt] x + 0 &< 1 \\[3pt] x &< 1 \end{align*}

Therefore, the solution is $x < 1.$

The variable $x$ is being multiplied by $4.$ To isolate $x$, we perform the opposite operation, which is dividing by $4.$ Remember to perform the same operation on both sides.

\begin{align*} 4x &\le 12 \\[5pt] \dfrac{4x}{4} &\le \dfrac{12}{4} \\[5pt] \dfrac{4x}{4} &\le 3 \\[5pt] \end{align*}

Now, we can cancel a common factor of $4$ from the numerator and denominator.

\begin{align*} \require{cancel} \dfrac{4x}{4} &\le 3 \\[5pt] \dfrac{\cancel{4}x}{\cancel{4}} &\le 3 \\[5pt] x &\leq 3 \end{align*}

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

The variable $x$ is being multiplied by $-3.$ To isolate $x,$ we perform the opposite operation, which is dividing by $-3.$ Remember to perform the same operation on both sides.

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

\begin{align} -3x \,\,&{\color{blue}>}\,\, 9 \\[5pt] \dfrac{-3x}{-3} \,\,&{\color{red}< }\,\, \dfrac{9}{-3} \\[5pt] \dfrac{-3x}{-3} \,\,&{\color{red}< }\,\, {-3} \end{align}

Now, we can cancel a common factor of $-3$ from both the numerator and denominator.

\begin{align} \require{cancel} \dfrac{-3x}{-3} &< {-3} \\[5pt] \dfrac{\cancel{-3}x}{\cancel{-3}} &< {-3} \\[5pt] x &< {-3} \end{align}

Therefore, the solution is $x< -3.$

The variable $z$ is being multiplied by $-4.$ To isolate $z,$ we perform the opposite operation, which is dividing by $-4.$ Remember to perform the same operation on both sides.

Also, when multiplying or dividing an inequality by a negative number, we need to flip the inequality sign: \begin{align} -4z \,\,&{\color{blue}>}\,\, -1 \\[5pt] \dfrac{-4z}{-4} \,\,&{\color{red}< }\,\, \dfrac{-1}{-4} \\[5pt] \dfrac{-4z}{-4} \,\,&{\color{red}< }\,\, {\dfrac{1}{4}} \end{align}

Now, we can cancel a common factor of $-4$ from both the numerator and denominator. \begin{align} \require{cancel} \dfrac{-4z}{-4} &< {\dfrac{1}{4}} \\[5pt] \dfrac{\cancel{-4}z}{\cancel{-4}} &< {\dfrac{1}{4}} \\[5pt] z &< {\dfrac{1}{4}} \end{align}

Therefore, the solution is $z < \dfrac{1}{4}.$

The variable $x$ is being divided by $2.$ To isolate $x$ and remove the negative sign, we multiply by $-2.$

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

\begin{align} -\dfrac x2 \,\,&{\color{blue}{ \leq }}\,\, 4 \\[5pt] (-2)\cdot \left(-\dfrac x2\right)\,\, & {\color{red}{ \geq}}\,\, (-2)\cdot 4 \\[5pt] (-2)\cdot \left(-\dfrac x2\right)\,\, & {\color{red}{ \geq}}\,\, -8 \\[5pt] \end{align}

Since two negative numbers multiply to give a positive number, we first cancel the negatives:

\begin{align} (-2)\cdot \left(-\dfrac x2\right)&\geq -8 \\[5pt] (\cancel{-}2)\cdot \left(\cancel{-}\dfrac x2\right)&\geq -8 \\[5pt] 2\cdot \dfrac x2&\geq -8 \\[5pt] \end{align}

Now, we can cancel a common factor of $2$ from both the numerator and denominator.

\begin{align} \require{cancel} 2\cdot \dfrac x2 &\geq -8 \\[5pt] \dfrac{2x}{2} &\geq -8 \\[5pt] \dfrac{\cancel{2}x}{\cancel{2}}&\geq -8 \\[5pt] x & \geq-8 \end{align}

Therefore, the solution is $x \geq -8.$

The variable $z$ is being divided by $5.$ To isolate $z$ and remove the negative sign, we multiply by $-5.$

Remember, when multiplying or dividing an inequality by a negative number, we need to flip the inequality sign: \begin{align} -\dfrac{z}{5} \,\, & {\color{blue}{ \gt }}\,\, -5 \\[5pt] (-5) \cdot \left(-\dfrac{z}{5}\right)\,\, & {\color{red}{ \lt}}\,\, (-5) \cdot (-5) \\[5pt] (-5) \cdot \left(-\dfrac{z}{5}\right)\,\, & {\color{red}{ \lt}}\,\, 25 \end{align}

Since two negative numbers multiply to give a positive number, we first cancel the negatives: \begin{align} (-5) \cdot \left(-\dfrac{z}{5}\right) & \lt 25 \\[5pt] (\cancel{-}5) \cdot \left(\cancel{-}\dfrac{z}{5}\right) & \lt 25 \\[5pt] 5 \cdot \dfrac{z}{5} & \lt 25 \end{align}

Now, we can cancel a common factor of $5$ from both the numerator and denominator. \begin{align} \require{cancel} 5 \cdot \dfrac{z}{5} & \lt 25\\[5pt] \dfrac{5z}{5} & \lt 25\\[5pt] \dfrac{\cancel{5}z}{\cancel{5}} & \lt 25\\[5pt] z & \lt 25 \end{align}

Therefore, the solution is $z \lt 25.$

The variable $y$ is being multiplied by $3.$ To isolate $y,$ we perform the opposite operation, which is dividing by $3.$ Remember to perform the same operation on both sides. \begin{align} 18 &> 3y \\[5pt] \dfrac{18}{3} &> \dfrac{3y}{3} \\[5pt] 6 &> \dfrac{3y}{3} \end{align}

Now we can cancel the common factor $3$ from the numerator and denominator. \begin{align} \require{cancel} 6 &> \dfrac{3y}{3} \\[5pt] 6 &> \dfrac{\cancel{3}y}{\cancel{3}} \\[5pt] 6 &> y \end{align}

Finally, we rewrite the inequality with the variable on the left-hand side: \[ y < 6 \]

Therefore, the solution is $y < 6.$

The variable $x$ is being multiplied by $-4.$ To isolate $x,$ we can perform the opposite operation, which is dividing by $-4.$ Remember to perform the same operation on both sides.

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

\begin{align} 12 \,\,&{\color{blue}\leq }\,\, -4x \\[5pt] \dfrac{12}{-4} \,\,&{\color{red}\geq }\,\, \dfrac{-4x}{-4} \\[5pt] -3 \,\,&{\color{red}\geq }\,\, \dfrac{-4x}{-4} \end{align}

Now, we can cancel the common factor $-4$ from both the numerator and denominator. \begin{align} \require{cancel} -3 &\geq \dfrac{-4x}{-4} \\[5pt] -3 &\geq\dfrac{\cancel{-4}x}{\cancel{-4}} \\[5pt] -3 &\geq x \end{align}

Finally, we rewrite the inequality with the variable on the left-hand side: \[ x\leq -3 \]

Therefore, the solution is $x \leq -3.$

The variable $x$ is being divided by $4.$ To isolate $x$ and remove the negative sign, we multiply by $-4$

Remember, when multiplying or dividing an inequality by a negative number, we need to flip the inequality sign: \begin{align} 9 \,\,&{\color{blue}{ \geq }}\,\, -\dfrac{x}{4} \\[5pt] (-4)\cdot 9 \,\, & {\color{red}{ \leq}}\,\, (-4)\cdot \left(-\dfrac x4\right) \\[5pt] -36 \,\, & {\color{red}{ \leq}}\,\, (-4)\cdot \left(-\dfrac x4\right) \\[5pt] \end{align}

Since two negative numbers multiply to give a positive number, we first cancel the negatives: \begin{align} -36 \,\, &\leq (-4)\cdot \left(-\dfrac x4\right) \\[5pt] -36 &\leq (\cancel{-}4)\cdot \left(\cancel{-}\dfrac x4\right) \\[5pt] -36 &\leq 4\cdot \dfrac x4 \end{align}

Now, we can cancel the common factor $4$ from both the numerator and denominator. \begin{align} \require{cancel} -36 &\leq 4\cdot \dfrac x4 \\[5pt] -36&\leq \dfrac{4x}{4} \\[5pt] -36 &\leq \dfrac{\cancel{4}x}{\cancel{4}} \\[5pt] -36 &\leq x \end{align}

Finally, we rewrite the inequality with the variable on the left-hand side: \[ x \geq -36 \]

Therefore, the solution is $x \geq -36.$