The variable $b$ is being multiplied by $10.$ To isolate $b$, we can perform the opposite operation, which is dividing by $10.$ So, we divide both sides of the equation by $10.$

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

Now, we can cancel a common factor of $10$ from the numerator and denominator.

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

Therefore, the solution is $b < 4.$ The solution can be expressed on a number line, as shown below.

To isolate $a,$ we need to perform the opposite of adding $5,$ which is subtracting $5.$ Remember to perform the same operation on the right-hand side.

\begin{align} a + 5 &\geq 11 \\[3pt] a + 5 - 5 &\geq 11 - 5 \\[3pt] a + 0 &\geq 6 \\[3pt] a &\geq 6 \end{align}

Therefore, the solution is $a \geq 6.$ The solution can be expressed on a number line, as shown below.

The variable $q$ is being multiplied by $-4.$ To isolate $q,$ we can perform the opposite operation, which is dividing by $-4.$ So, we divide both sides of the equation by $-4.$

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

\begin{align} -4q \,\,&{\color{blue}>}\,\, {-8} \\[5pt] \dfrac{-4q}{-4} \,&{\color{red}< }\,\, \dfrac{-8}{-4} \\[5pt] \dfrac{-4q}{-4} \,&{\color{red}< }\,\, 2 \end{align}

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

\begin{align} \require{cancel} \dfrac{-4q}{-4} &< 2 \\[5pt] \dfrac{\cancel{-4}q}{\cancel{-4}} &< 2 \\[5pt] q &< 2 \end{align}

Therefore, the solution is $q< 2.$ The solution can be expressed on a number line, as shown below.

The variable $x$ is being divided by $3.$ To isolate $x$ and remove the negative sign, we multiply by $-3.$

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

\begin{align} -\dfrac x3 \,\,&{\color{blue}{ > }}\,\, -1 \\[5pt] (-3)\cdot \left(-\dfrac x3\right)\,\, & {\color{red}{ < }}\,\, (-3)\cdot (-1) \\[5pt] (-3)\cdot \left(-\dfrac x3\right)\,\, & {\color{red}{ < }}\,\, 3 \\[5pt] \end{align}

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

\begin{align} (-3)\cdot \left(-\dfrac x3\right)&< 3 \\[5pt] (\cancel{-}3)\cdot \left(\cancel{-}\dfrac x3\right)&< 3 \\[5pt] 3\cdot \dfrac x3&< 3 \\[5pt] \end{align}

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

\begin{align} \require{cancel} 3\cdot \dfrac x3 &< 3 \\[5pt] \dfrac{3x}{3} &< 3 \\[5pt] \dfrac{\cancel{3}x}{\cancel{3}}&< 3 \\[5pt] x & < 3 \end{align}

Therefore, the solution is $x < 3.$ The solution can be expressed on a number line, as shown below.

First, we apply the addition principle, subtracting $2$ from both sides:

\begin{align} 4x + 2 &\le 10 \\[5pt] 4x + 2 - 2 &\le 10 - 2 \\[5pt] 4x + 0 &\le 8 \\[5pt] 4x &\le 8 \end{align}

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

\begin{align} \require{cancel} 4x &\le 8 \\[5pt] \dfrac{4x}{4} &\le \dfrac84 \\[5pt] \dfrac{\cancel{4}x}{\cancel{4}} &\le \dfrac84 \\[5pt] x &\le 2 \end{align}

Therefore, the solution is $x \le 2.$ The solution can be expressed on a number line, as shown below.

First, we apply the addition principle, subtracting $2$ from both sides:

\begin{align} -3x + 2 &\ge 5 \\[5pt] -3x + 2 - 2 &\ge 5 - 2 \\[5pt] -3x + 0 &\ge 3 \\[5pt] -3x &\ge 3 \end{align}

Then, we apply the multiplication principle, dividing by $-3.$

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

\begin{align} \require{cancel} -3x \,\,&{\color{blue}\ge\,\,} 3 \\[5pt] \dfrac{-3x}{-3} \,&{\color{red}\le\,\,} \dfrac{3}{-3} \\[5pt] \dfrac{\cancel{-3}x}{\cancel{-3}} \,&{\color{red}\le\,\,} -1 \\[5pt] x \,\,&{\color{red}\le\,\,} -1 \end{align}

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

\[\]

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

\begin{align} 7 &\le 3x + 1 \\[5pt] 7 - 1 &\le 3x + 1 - 1 \\[5pt] 6 &\le 3x + 0 \\[5pt] 6 &\le 3x \end{align}

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

\begin{align} \require{cancel} 6 &\le 3x \\[5pt] \dfrac{6}{3} &\le \dfrac{3x}3 \\[5pt] \dfrac 63 &\le \dfrac{\cancel{3}x}{\cancel{3}} \\[5pt] 2 &\le x \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. \[ x \ge 2 \]

Therefore, the solution is $x \ge 2.$ The solution can be expressed on a number line, as shown below.